AP Inter 1st Year

Referring to the AP Inter 1st Year Maths Study Material Chapter 10 Conic Sections Exercise 10d Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Conic Sections Solutions Exercise 10d

I.

Question 1.
Find the eccentricity of hyperbola \(\frac{\mathbf{y}^2}{\mathbf{b}^2}-\frac{\mathbf{x}^2}{\mathbf{a}^2}\) = 1.
Solution:
Given hyperbola is conjugative hyperbola.
The eccentricity e is e = \(\sqrt{1+\frac{a^2}{b^2}}\)
⇒ e = \(\sqrt{\frac{b^2+a^2}{b^2}}\)

Question 2.
Find the eccentricity of rectangular hyperbola.
Solution:
Since the length of transverse axis is equal to length of the conjugative axis i.e.,
2a = 2b ⇒ a = b.
Eccentricity e = \(\sqrt{\frac{a^2+b^2}{a^2}}=\sqrt{\frac{b^2+b^2}{b^2}}=\sqrt{\frac{2 b^2}{b^2}}\) = √2

Question 3.
Find the length of conjugate axis of the hyperbola 3x² – 4y² = 12.
Solution:
Given hyperbola is 3x² – 4y² = 12.
\(\frac{3 x^2}{12}-\frac{4 y^2}{12}\) = 1
⇒ \(\frac{x^2}{4}-\frac{y^2}{3}\) = 1 which is comparing to \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1
a² = 4, b² = 3
Since the length of the conjugative axis = 2b = 2√3

Question 4.
Find the length of transverse axis of hyperbola x² – 4y² = 4.
Solution:
Given hyperbola is \(\frac{x^2}{4}-\frac{y^2}{1}\) = 1
a² = 4, b² = 1
Since the length of the transverse axis is 2a = 2(2) = 4

II. In each of the Exercises 1 to 6, fInd the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

Question 1.
\(\frac{x^2}{16}-\frac{y^2}{9}\) = 1
Solution:
The given equation is \(\frac{x^2}{16}-\frac{y^2}{9}\) = 1 ⇒ \(\frac{x^2}{4^2}-\frac{y^2}{3^2}\) = 1
On comparing this equation with the standard equation of hyperbola
i.e., \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1, we have a = 4 and b = 3
We know that a² + b² = c²
⇒ c² = 4² + 3²
⇒ c² = 25
∴ The coordinates of foci are (± 5, 0)
The coordinates of the vertices are (± 4, 0), e = \(\frac{c}{a}=\frac{5}{4}\)
Length of the latus rectum \(\frac{2 b^2}{a}=\frac{2 \times 9}{4}=\frac{9}{2}\)

Question 2.
\(\frac{y^2}{9}-\frac{x^2}{27}\)
Solution:
The given equation is \(\frac{y^2}{9}-\frac{x^2}{27}\) = 1
⇒ \(\frac{y^2}{3^2}-\frac{x^2}{(\sqrt{27})^2}\) = 1
On comparing with \(\frac{y^2}{a^2}-\frac{x^2}{b^2}\) = 1, we have a = 3 and b = \(\sqrt{27}\)
We know that a² + b² = c²
⇒ c² = 3² + (√27)²
⇒ c² = 9 + 27
⇒ c² 36
⇒ c = 6
The coordinates of the foci are (0, ± 6)
The coordinates of the vertices are (0, ± 3)
Eccentncity e = \(\frac{c}{a}=\frac{6}{3}\) = 2.
Length of the latus rectum = \(\frac{2 b^2}{a}=\frac{2 \times 27}{3}\) = 18.

Question 3.
9y² – 4x² = 36
Solution:
The given equation is 9y² – 4x² = 36
It can be written as \(\frac{y^2}{4}-\frac{x^2}{9}\) = 1
⇒ \(\frac{y^2}{2^2}-\frac{x^2}{3^2}\) = 1 …………(1)
On comparing equation (1) with equation of hyperbola \(\frac{y^2}{a^2}-\frac{x^2}{b^2}\) = 1
We have a = 2 and b = 3
We know that a² + b² = c²
⇒ c² = 4 + 9 = 13
⇒ c = \(\sqrt{13}\)
∴ The coordinates of the foci are (0, ±\(\sqrt{13}\))
The coordinates of the vertices are (0, ± 2)
Eccentricity e = \(\frac{c}{a}=\frac{\sqrt{13}}{2}\), Length of the latus rectum

Question 4.
16x² – 9y² = 576
Solution:
The given equation is 16x² – 9y² = 576
It can be written as \(\frac{x^2}{36}-\frac{y^2}{64}\) = 1
⇒ \(\frac{x^2}{6^2}-\frac{y^2}{8^2}\) = 1
On comparing equation (1) with equation of hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1
We have a = 6 and b = 8
We know that a² + b² = c²
c² = 36 + 64
⇒ c² = 100
⇒ c = 10
∴ The coordinates of the foci are (± 10, 0)
The coordinates of the vertices are (±6, 0), Length of the latus rectum = \(\frac{2 b^2}{a}=\frac{2 \times 64}{6}=\frac{64}{3}\)

Question 5.
5y² – 9x² = 36
Solution:
The given equation is 5y² – 9x² = 36
\(\frac{y^2}{\left(\frac{36}{5}\right)}-\frac{x^2}{4}\) = 1
⇒ \(\frac{y^2}{\left(\frac{6}{\sqrt{5}}\right)^2}-\frac{x^2}{2^2}\) = 1 ……… (1)
On comparing equation (1) with equation of hyperbola \(\frac{y^2}{a^2}-\frac{x^2}{b^2}\) = 1
We have a = \(\frac{6}{\sqrt{5}}\), b = 2

We know that a² + b² = c²
⇒ c² = \(\frac{36}{5}\) + 4 = \(\frac{56}{5}\)
⇒ c = \(\frac{\sqrt{56}}{\sqrt{5}}\)
⇒ c = \(\frac{2 \sqrt{14}}{\sqrt{5}}\)
∴ The coordinates of the foci are (0, ± \(\frac{2 \sqrt{14}}{\sqrt{5}}\))
The coordinates of the vertices are (0, ±\(\frac{6}{\sqrt{5}}\)
Eccentricity e = \(\frac{c}{a}=\frac{\left(\frac{2 \sqrt{14}}{\sqrt{5}}\right)}{\left(\frac{6}{\sqrt{5}}\right)}=\frac{\sqrt{14}}{3}\)
Length of the latus rectum = \(\frac{2 b^2}{a}=\frac{2 \times 4}{\left(\frac{6}{\sqrt{5}}\right)}=\frac{4 \sqrt{5}}{3}\)

Question 6.
49y² – 16x² = 784
Solution:
The given equation is 49y² – 16x² = 784
It can be written as \(\frac{y^2}{16}-\frac{x^2}{49}\) = 1
⇒ \(\frac{y^2}{4^2}-\frac{x^2}{7^2}\) = 1 ………… (1)
On comparing equation (1) with the equation of hyperbola \(\frac{1}{2}\) = 1
We have a = 4 and b = 7 we known that
a² + b² = c²
⇒ c² = 16 + 49
⇒ c² = 65
⇒ c = \(\sqrt{65}\)
∴ The coordinates of the foci are (0, ±\(\frac{1}{2}\))
The coordinates of the vertices are (0, ± 4)
Eccentncity e = \(\frac{c}{a}=\frac{\sqrt{65}}{4}\), Length of the latus rectum= \(\frac{2 b^2}{a}=\frac{2 \times 49}{4}=\frac{49}{2}\).

In each of the Exercises 7 to 16, find the equations of the hyperbola satisfying the given conditions.

Question 7.
Vertices (±2, 0), foci (±3, 0).
Solution:
Vertices (± 2, 0), foci (± 3, 0)
Here, the vertices are on the X-axis.
The equation of the hyperbola is of the form \(\frac{1}{2}\) = 1
Here, the vertices are (± 2, 0), ∴ a = 2
since the foci are (±3, 0), ∴ c = 3
We know that a² + b² = c²
⇒ 2² + b² = 3²
⇒ b² = 9 – 4
⇒ b² = 5
∴ Equation of the hyperbola is \(\frac{x^2}{4}-\frac{y^2}{5}\) = 1

Question 8.
Vertices (0, ±5), foci (0, ±8)
Solution:
Vertices (0, ±5), foci (0, ±8)
Here the vertices are on the Y-axis.
∴ The equation of the hyperbola is of the form \(\frac{y^2}{a^2}-\frac{x^2}{b^2}\) = 1.
Here, the vertices are (0, ±5), ∴ a = 5
Since, the foci are (0, ±8), ∴ C = 8
We know that c² = a² + b²
⇒ 8² = 5² + b²
⇒ b² = 64 – 25
⇒ b² = 39
The equation of the hyperbola is \(\frac{y^2}{25}-\frac{x^2}{39}\) = 1

Question 9.
Vertices (0, ± 3), foci (0, ± 5)
Solution:
Vertices (O, ± 3), foci (0, ± 5). Here, the vertices are on the Y-axis.
∴ The equation of the hyperbola is of the form \(\frac{y^2}{a^2}-\frac{x^2}{b^2}\) = 1
Here the vertices are (0, ±3), ∴ a = 3
Since the foci are (O, ± 5), ∴ c = 5
We know that, a² + b² = c²
⇒ 3² + b² = 5²
⇒ b² = 25 – 9
⇒ b² = 16
Equation of the hyperbola is \(\frac{y^2}{9}-\frac{x^2}{16}\) = 1

Question 10.
Foci (±5, 0), transverse axis is of length 8.
Solution:
Foci (± 5, 0), the transverse axis is of length 8
Here, the foci are on the X-axis
∴ The equation of the hyperbola is of the form \(\frac{1}{2}\) = 1
Since the foci are (± 5, 0) ⇒ c = 5
Since the length of the transverse axis is 8 2a = 8 ⇒ a = 4
We know that a² + b² = c²
⇒ 4² + b² = 5²
⇒ b² = 25 – 16
⇒ b² = 9
∴ The equation of the hyperbola is \(\frac{y^2}{9}-\frac{x^2}{16}\) = 1.

Question 11.
Foci (0, ± 13), the conjugate axis is of length 24.
Solution:
Foci (0, ± 13), the conjugate axis is of length 24. Here the foci are on Y-axis.
∴ The equation of the hyperbola is of the form \(\frac{1}{2}\) = 1
Here, the foci are (0, ±13)
Since the length of the conjugate axis is 24 ⇒ 2b = 24 ⇒ b = 12
We know that a² + b² = c²
⇒ a² + 12² = 13²
⇒ a² = 169 – 144
⇒ a² = 25
The equation of the hyperbola is \(\frac{x^2}{25}-\frac{y^2}{20}\) = 1.

Question 12.
Foci (± 3√5, O). the lattis rectum is of length 8.
Solution:
Foci (± 3√5, 0), the latus rectum is of length 8. Here the foci are on the X-axis.
∴ The equation of the hyperbola is of the form \(\frac{1}{2}\) = 1
Since the foci are (±3√5, 0) ⇒ c = 3
Length of the latus recturn = 8
⇒ \(\frac{2 b^2}{a}\) = 8
⇒ b² = 4a
We know that a² + b² = c²
= a² + 4a= 45
= a² + 4a – 45 = 0
⇒ a² + 9a – 5a – 45 = O
⇒ (a + 9)(a – 5) = 0
⇒ a = -9, 5
Since, a is non-negative ∴ a = 5 ⇒ b² – 4a = 4 × 5 = 20
∴ The equation of the hyperbola is \(\frac{x^2}{4}-\frac{y^2}{12}\) = 1.

Question 13.
Foci (±4, 0), the latus rectum is of length 12.
Solution:
Foci (± 4, 0), the latus rectum is of length 12. Here, the foci are on the X-axis.
∴ The equation of the hyperbola is of the form 1
Since the foci are (± 4, 0) ⇒ c = 4
Length of the latus rectum 12 ⇒ \(\frac{2 b^2}{a}\) = 12 ⇒ b² = 6a
We know that a² + b² = c²
⇒ a² + 6a – 16= 0
⇒ a² + 8a – 2a – 16= 0
⇒ (a + 8)(a – 2)0
a = – 8, 2
Since a is non-negative; a = 2
⇒ b² = 6a = 6 × 2 = 12
The equation of the hyperbola is \(\frac{x^2}{4}-\frac{y^2}{12}\) = 1

Question 14.
Vertices (±7, 0), e = \(\frac{4}{3}\)
Solution:
Given that vertices (± 7, 0), e = \(\frac{4}{3}\)
Here the vertices are on the X-axis.
∴ The equation of the hyperbola is of the form \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1
Since the vertices are (± 7, 0) = a = 7
It is given that e = \(\frac{4}{3} \Rightarrow \frac{c}{a}=\frac{4}{3} \Rightarrow \frac{c}{7}=\frac{4}{3}\)
c = \(\frac{28}{3}\) [∵ e = \(\frac{c}{a}\)]
We know that a² + b² = c²
7² + b² = \(\left(\frac{28}{3}\right)^2\)
⇒ b² = \(\frac{784}{9}\) – 49
⇒ b² = \(\frac{784-441}{9}\)
⇒ b² = \(\frac{343}{9}\)
∴ Thus the equation of the hyperbola is \(\frac{x^2}{49}-\frac{y^2}{\left(\frac{343}{9}\right)}\) = 1
⇒ \(\frac{x^2}{49}-\frac{9 y^2}{343}\) = 1

Question 15.
Foci (0, ±\(\sqrt{10}\)), passing through (2, 3).
Solution:
Foci (0, ±\(\sqrt{10}\)), passing through (2, 3). Here foci are on the Y-axis.
∴ The equation of the hyperbola is of the form – – 1
Here, the foci are (0, ± \(\sqrt{10}\)) ⇒ c = \(\sqrt{10}\)
We know that a² + b² = c²
⇒ a² + b² = 10
⇒ b² = 10 – a² …………….. (1)
Since the hyperbola passing through point (2, 3)
∴ \(\frac{9}{a^2}-\frac{4}{\left(10-a^2\right)}\) = 1 ………. (2)

From (1) and (2) we have
⇒ \(\frac{9}{a^2}-\frac{4}{\left(10-a^2\right)}\) = 1
9(10 – a²) – 4a² = a²(10 – a²)
90 – 9a² – 4a = 10a² – a⁴
⇒ a⁴ – 23a² + 90 = 0
⇒ a⁴ – 18a² – 5a² + 90 = 0
⇒ a²(a² – 18) – 5(a² – 18) = 0
⇒ (a² -18)(a² – 5) = 0
⇒ a² = 18 or 5
In hyperbola c > a, i.e., c² > a²
⇒ ∴ a² = 5
⇒ b² = 10 – a²
⇒ b² = 10 – 5
⇒ b² = 5
∴ Thus the equation of hyperbola is \(\frac{y^2}{5}-\frac{x^2}{5}\) = 1

Question 16.
Find the equation of the hyperbola of given length of transverse axis 6 whose vertex bisect., the distance between the centre and the focus.
Solution:
Given 2a – 6 ⇒ a = 3 ⇒ a² = 9.
Since the vertex bisects the distance between the
centre and focus. So c = 2a
⇒ c = 2(3) = 6
⇒ c = 6
∴ c² = a² + b²
⇒ b² = c² – a²
⇒ b² = 6² – 3²
⇒ b² = 27
⇒ b = \(\sqrt{27}\)
⇒ b = 3√3
The equation of the hyperbola is \(\frac{x^2}{9}-\frac{y^2}{27}\) = 1.

For systematic learning, students can include AP Inter 1st Year English Grammar with Answers and Expansion of Ideas in English Inter 1st Year in their daily study routine.

AP Inter 1st Year English Expansion of Ideas with Answers

Expansion of ideas is the method by which you go into further depth of your ideas. You must first understand the meaning and then try to restate the given phrase or statement. The best technique to expand a sentence or a proverb is to write synonyms of the original keywords and then expand it using those synonyms. To further expand your topic, you should end your paragraph with either a moral or a conclusion to leave a lasting impression.

For a good expansion of ideas, the theme must have the following features.

• Organise your thoughts in a logical sequence of priority.
• Avoid using cliches, cliched language, or extraneous material to the topic at hand.
• It is important to use basic and brief sentences in your writing.
• Use colloquial and figurative language when appropriate to use your expansion.
• Make sure your writing is flawless in grammar, spelling, and punctuation.

Exercises

From Unit – 1

Question 1.
Using the hints provided, write at least ten lines, expanding the ideas into complete sentences. Be creative!
If I Could Fly…
Idea : Imagine you could fly for one day. Write a paragraph about where you would go, what you would do, and how you would feel.
Hints:
If I could fly, I would __________
I would see __________
Along with me, I would take __________
I would try __________
At the end of the day, I would feel __________
Answer:

If I Could Fly …………..

If I could fly, I would soar high above the clouds like a free bird and explore the world from the sky. I would see the towering mountains, sparkling rivers, and bustling cities far below. Along with me, I would take my camera to capture the magical view from the skies. I would try flying over the Eiffel Tower in Paris and circle around the Taj Mahal in India. I would glide through rainbows and dance with the wind. I would wave at people walking on the streets, wondering if they could see me. I would visit places I have only read about in books, like the Great Wall of China and the Grand Canyon. I would fly to a peaceful island and rest on a cloud, feeling the warmth of the sun on my face. At the end of the day, I would feel happy, free, and grateful for the amazing journey I had through the sky.

Question 2.
Using the hints provided, write at least ten lines, expanding the ideas into complete sentences. Be creative!

A Day in a World without Darkness
Idea : Imagine you are living in a world where doesn’t exist; where every moment is filled with unending light. Write a diary entry describing one day in such a world.

Hints:

• The sun always shines; there’s no night to rest.
• Cities are bustling, even at midnight-time feels endless.
• Plants grow rapidly, but some animals struggle without darkness.
• Can we appreciate light if there’s no darkness?

Instruction: Write at least 10 lines describing what life would feel like in a world without darkness. Reflect on how it changes routines, nature, and the way you view light itself.
Answer:

Diary Entry: A Day in a World without Darkness

June 15, 2042
Dear Diary,

The sun never sets here-it hovers like a watchful eye, painting the sky in eternal gold. I woke up at what my clock claims is 3 AM, but the streets below are alive with chatter, as if it’s noon. Without night, time has lost its rhythm; work, meals, and sleep blur into one endless cycle. Cafes serve coffee at midnight, and children play in parks under the unyielding light.

The plants are thriving, stretching taller each day, but the owls have vanished, and the bats cling to man-made caves. I miss the stars. Last week, scientists projected artificial constellations onto the sky, but it’s not the same. Mom says we’ve forgotten how to rest – our bodies ache for shadows.

I tried closing my blackout curtains today, but the glow seeped in like a persistent whisper. It’s strange: in a world drowned in light, we crave darkness just to remember what brightness means. Would a rainbow matter if the sky never wept? I wonder if we’ve traded mystery for convenience. Tomorrow, I’ll plant a moonflower – just to pretend.

x x x x

Question 3.
Using the hints provided, write atleast ten lines, expanding the ideas into complete sentences. Be creative !

A Day in Their Shoes
Idea: Imagine you are a sanitary worker at your college. Write a journal entry describing your day – your work, thoughts, and feelings. Reflect on the challenges you face and what keeps you motivated.

Hints:

• Morning: Start early, clean classrooms, empty trash bins, pick up leftover snacks and papers.
• School Activity: Teachers spill coffee, students litter, run on freshly mopped floors.
• Evening : Tidy up the lunch area, pick up forgotten belongings, sanitize and mop the whole college.
• Challenges : Repetitive tasks, feeling unacknowledged, exhaustion by the end of the day.
• Joy: A simple “thank you,” or seeing the college spotless brings quite satisfaction.

Answer:

Journal Entry : A Day in My Shoes

5:30 AM: My alarm buzzes-another day begins before the sun rises. I tie my worn-out shoes, knowing the college won’t clean itself. The empty corridors echo as I sweep classrooms, collecting yesterday’s forgotten notes and crumpled snack wrappers.

8:00 AM : Students trickle in, their laughter bouncing off freshly mopped floors. A teacher rushes past. Coffee sloshing onto the tiles I just cleaned. I sigh but wipe it up – no time to dwell.

12:30 PM : The lunch area is a battlefield of spilled rice and juice boxes. I pick up a lost ID card, wondering if its owner will ever notice it’s gone.

3:00 PM : My back aches from bending, but I keep scrubbing. A student nods at me-just a glance, but it feels like a tiny spark of warmth.

6:00 PM : As I empty the last trash bin, I pause. The college gleams, silent and orderly. It’s not much, but it’s mine. Someone left a note on the staff board: “Thanks for keeping our home clean.” I smile. Maybe tomorrow, I’ll start again.

Question 4.
Using the hints provided, write at least ten lines, expanding the ideas into complete sentences. Be creative !

The Lost Animal
Idea : Imagine an animal that got separated from its home. Write a story about its journey to find its way back. Describe its emotions, the challenges it faces, and how it eventually gets home – or perhaps find a new place to call home.

Hints :

• First – Getting lost – fear, confusion, and loneliness – what the animal feels after realizing it’s lost.
• But – Finding strength – the animal uses its instincts, learns to survive, and meets new creatures along the way.
• Then – Facing obstacles – hunger, wild animals, harsh weather, or dangerous forests – how the animal overcomes these challenges.
• Later – Meeting helpful friends – other animals who either guide or help the lost animal.

Answer:

The Lost Animal

A little red fox named Rani lived in a quiet forest near the hills. One sunny afternoon, while chasing a bright blue butterfly, Rani ran farther than he ever had before. When the butterfly vanished, Rani looked around and realized, he didn’t recognize a single tree. The forest looked different. The smells were unfamiliar. Fear wrapped around him like a cold wind. He whimpered softly, his tiny paws trembling. ‘Where is home?’ he thought. The silence around him felt heavier than ever. The once-exciting world now seemed too big, too wild and too lonely.

But after a while, Rani sat down and took a deep breath. I can’t stay scared forever,’ he whispered. He remembered how his mother once told him, Trust your nose, little one.’ So he sniffed the air and began walking. Slowly, he learned to find berries to eat and dug little hole to sleep in at night. Along the way, Rani faced many troubles. One night, heavy rain poured down, soaking his fur and turning the ground into slippery mud. Another time, he was chased by a wild dog and had to hide in a hollow log until it left. Hunger often made him weak, and some days he found no food at all. But he kept going.

One morning, while crossing a shallow stream, Rani heard a croak. A friendly old frog named Tinku greeted him. ‘You look far from home, young fox,’ said Tinku. Rani nodded sadly, ‘I’m trying to find it again.’ Tinku smiled and said, ‘Follow the sound of the river. It flows near the hills.’ Later, Runi met a family of squirrels who shared their nuts and told him which ways to avoid. A wise owl gave him directions from the treetops. Step by step, paw by paw, Rani got closer. One evening, as the sun disappeared behind the hills, Rani smelled something; familiar leaves, warm grass, and… his family. He ran as fast as he could. And there they were: his mother, his brothers, all waiting and searching. Tears of joy filled his eyes as he leapt into his mother’s paws. Rani was finally home. And though the journey had been hard, he was braver, wiser, and stronger than before.

From Unit 2:

Question 5.
Using the hints provided, write at least ten lines, expanding the ideas into complete sentences. Be creative!

Letter to the Future
Idea : Write a letter to future generations describing the current state of the environment. Think about issues like pollution, climate change, and the loss of natural habitat. What kind of changes do you hope for by the time they are alive? What steps should they take to protect the planet? How can they help reduce waste, save water, and protect animals?

Hints:

• First – Describe the current environmental problems: pollution, deforestation, melting ice,etc.
• But – Express hope for change and the power of future actions.
• Then – Share your hopes for a cleaner, healthier world.
• Later – Give advice on reducing waste, conserving water, and supporting sustainable practices.
• Finally – Encourage future generations to stay strong and keep protecting for the Earth.

Answer:

Letter to the Future

Dear Future Generations,

Today, our beautiful Earth is suffering-the skies are filled with smoke, forests are vanishing, ice is meiting, and animals are losing their homes. But we believe in the power of change, and in your ability to make things better than they are now. I dream of a future where rivers flow clean, skies are blue, and trees cover every hill and field. By the time you read this, I hope the world is healing, thanks to choices you and others have made.

Please use less plastic, recycle what you can, and turn off lights when not needed. Save water – every drop counts – and plant trees wherever possible. Support farmers who grow food naturally and companies that care about the planet. Protect every living creature, big or small, because they all have a role in Earth’s balance. Never forget: the Earth is not a gift from our ancestors, but a loan from our children. Stay strong, stay kind, and always stand up for our one and only home – the Earth.

With hope and faith,
A Friend from the Past

Question 6.
Using the hints provided, write at least ten lines, expanding the ideas into complete sentences. Be creative !

Invincible Hope
Idea: Imagine that someone you care about has failed in his exams, and is feeling sad and hopeless. Encourage and inspire him to stay strong and to have hope.

Hints:

• Start with : ‘Dear friend, I know things are hard, but you are strong.’
• Say : ‘Remember when you ?’
• Give advice : ‘Take one step at a time, and I’m here for you.’
• End with : ‘I believe in you!”

Answer:

Invincible Hope

Dear friend,

I know things are hard, but you are strong.

Remember when you practised day and night for that speech and amazed everyone with your words?

That same determination still lives inside you. Failing an exam doesn’t define who you are – it just means you have another chance to grow. Take one step at a time, and I’m here for you, every step of the way.

You’re not alone, and this moment of failure is not the end-just a bend in the road.

Cry if you need to, but then stand up and try again, braver than before.

Your dreams are still waiting for you, and they are worth every effort.

I believe in you!
x x x x

Question 7.
Using the hints provided, write at least ten lines, expanding the ideas into complete sentences. Be creative!

An Apology Letter to Yourself
Idea : Write a letter to yourself, apologise for something you regret, like not studying properly for an exam.

In the letter, explain why you are sorry, and promise to do better in future. Reflect on how you can improve and what steps you’ll take to make things right.

Hints :

• Dear Me, I’m writing to apologize for …………….
• I feel I didn’t give my best effort, and …………
• I will make a plan to study …………..
• I will ask teachers for help …………..
• I will try harder and …………

Answer:

An Apology Letter to Myself

Dear Me,

I’m writing to apologize for not preparing properly for the exam and wasting time when I should have been focused.

I feel I didn’t give my best effort, and now I understand how important it is to stay disciplined and organized.

I let distractions win, and I regret not respecting my own potential.

I will make a plan to study every day, even if it’s just for an hour, and stick to it with honesty.

I will ask teachers for help when I’m stuck, instead of pretending I understand everything.

I will try harder and keep track of my progress, no matter how small.

I’ll avoid procrastination by setting short, realistic goals and rewarding myself for completing them.

This is not the end-just a reminder that I can rise stronger if I take responsibility.

I believe I can do better – and this time, I will.

With determination,
Me

Question 8.
Using the hints provided, write at least ten lines, expanding the ideas into complete sentences. Be creative !

The Magic Door
Idea: Imagine you find a magic door in your home. This door leads to an amazing world filled with magical creatures, strange landscapes, or fantastic objects. Write a story about what you find behind the door. What do you see, hear, and feel? Is there anything that surprises you? How does this adventure change you?

Hints:

• I accidentally found a magical door in my grandparents’ house ………….
• As the door opened, I saw ……….. (describe creatures, objects, or landscapes)
• I could hear ……………. (describe sounds like birds singing, wind blowing, or magical noises)
• I found several creatures and animals facing a problem …………. (describe a problem that is faced by many creatures and animals in that world)
• I wanted to help them by ……………. (describe how you helped them and made it a better world)

Answer:

The Magic Door

I accidentally found a magical door in my grandparents’ house, hidden behind an old bookshelf in the attic. As the door opened, I saw glowing trees with silver leaves, rivers that floated in the sky, and talking animals with shining eyes. A giant owl greeted me and told me I was in the Land of Everlight. I could hear the soft humming of crystal winds, the cheerful chatter of fairies, and the gentle roar of a distant waterfall. But the magical world wasn’t happy-its sun had stopped rising, and the flowers were losing their colors.

I found several creatures and animals gathered in sadness because their magical gem, the Heart of Everlight, had been stolen. I wanted to help them by solving the riddles of the Forest of Whispers and recovering the gem from the Cave of Shadows. With courage and kindness, I brought the light back to their world. As I returned through the magic door, I felt stronger, braver, and full of wonder. That adventure taught me that even small hearts can make a big difference.

From Unit – 3

Question 9.
Using the hints provided, write at least ten lines, expanding the ideas into complete sentences. Be creative!

Interviewing Inspiration
Idea : Imagine interviewing someone who inspires you – a famous figure, a family member, or a fictional character. Write a creative conversation exploring their journey to success. Include details about the setting to make it vivid and engaging.

Hints:

• What inspired you to follow your passion?
• Who supported you the most on your journey?
• Were there tough moments in your journey? How did you overcome them?
• What is the best piece of advice you’ve ever received?
• What advice would you give to someone who’s still searching for their passion?

Answer:

An Imaginary Interview with Dr. A.P.J. Abdul Kalam

It was a quiet morning at the Science Centre in Rameswaram, where I had the rare honour of sitting down with Dr. A.RJ. Abdul Kalam, India’s “Missile Man” and former President. The air was filled with the salty breeze of the nearby sea, and the room echoed with curiosity.

I : Sir, what inspired you to follow your passion in science and service?
Dr. Kalam : As a boy, I was fascinated by the skies. Watching birds fly made me dream of flying machines. That curiosity turned into passion.
I : Who stood by you during this journey?
Dr. Kalam : My parents, especially my father, supported me with values of hard work and honesty. Later, my teachers shaped my scientific thinking.
I : Were there any difficult moments?
Dr. Kalam : Many. Failures in rocket launches tested my patience. But we learned from every mistake and moved forward.
I : What’s the best advice you ever received?
Dr. Kalam : Dream big, but more importantly, work hard to make those dreams real.
I : What advice would you give to today’s youth?
Dr. Kalam : Never give up. Keep searching. Your passion may be hidden, but once you find it, give it your everything.

Question 10.
Using the hints provided, write at least ten lines, expanding the ideas into complete sentences. Be creative!

Refugee’s Journal Entry (One Day)
Idea: Write a journal entry from the perspective of a refugee. Imagine their struggles, hopes, and emotions as they experience a day in their uncertain journey.

Hints:

• ” Today I woke up in a crowded shelter……..”
• “We have very little food today, and I feel so weak………..”
• “I can’t stop thinking about my home…”
• “In this refugee camp, we are all afraid about…….”
• “But, I try to stay hopeful, hoping that one day we will find a place where……..”

Answer:

Refugee’s Journal Entry (One Day)

Date: 05.09.20xx

Today, I woke up in a crowded shelter with dozens of other families, all huddled together beneath torn blankets. The air was damp, and the sound of crying children echoed around us. We have very little food today, and I feel so weak that I can barely stand. My younger brother gave me his piece of bread, even though he was hungry too. I can’t stop thinking about my home – the smell of my mother’s cooking, the laughter in the courtyard, the trees that once bloomed outside our window. It feels like another lifetime.

In this refugee camp, we are all afraid about what tomorrow may bring – disease, hunger, or the sound of gunfire returning. But I try to stay hopeful, hoping that one day we will find a place where we are not seen as outsiders but welcomed as human beings. Until then, I hold on to dreams – and to each other.

Question 11.
Using the hints provided, write at least ten lines, expanding the ideas into complete sentences. Be creative!

One Day with Tomorrow’s Technology
Idea: Imagine waking up one day in the future, surrounded by incredible new technologies that make life exciting, fun, and easier. Think about how your day would look with these inventions. Write a simple story about one day of your life with these futuristic technologies.

Hints : Gadgets – Pet robot to clean your room – A shirt that changes colour with mood – Flying backpacks to go to college – A pen that takes notes by itself – A fridge that prepares the best food – A holo-playground – At home, an A.I robot to explain tricky concepts – A pillow that tracks your dreams
Answer:

One Day with Tomorrow’s Technology

Today I woke up feeling fresh, thanks to my smart pillow that tracked my dreams and played calming music all night. As I stretched, my pet robot Sparky rolled in and started cleaning my room, even organizing my books in alphabetical order!

I slipped on my mood shirt, which turned bright green- my “excited” color – because I had a science quiz today. No worries though, because my A.I. learning robot had already explained the tough concepts while I ate breakfast yesterday.

Downstairs, the smart fridge had prepared my favorite breakfast – hot idlis with chutney – after scanning my energy levels. With my flying backpack, I zoomed to college in minutes, waving to friends mid-air.
During class, I didn’t need to write a thing. My smart pen floated in the air and took neat notes all by itself. Later, I played with friends in a holo-playground – a magical space where we battled dragons and ran on moon surfaces.

The future is amazing – and this was just one day!

Question 12.
Using the hints provided, write at least ten lines, expanding the ideas into complete sentences. Be creative!

Overcoming-Failure
Idea : Imagine you are a student who failed an important exam. Write a paragraph about how you would feel, what you would do to overcome the failure, and how you would regain confidence. Use the hints below to guide your writing.

Hints : Feeling shocked – fear of disappointing parents – comparing yourself to others – realizing everyone fails sometimes – finding encouragement – reflecting on mistakes – identifying weak areas – creating a new study plan – setting achievable goals – seeking help from teachers – attending extra classes – asking questions – learning from failure – passing the next exam – feeling proud understanding failure is a stepping stone – motivated and confident – sharing your experience to inspire others.
Answer:

Overcoming-Failure

When I saw my exam results and realized I had failed, I was completely shocked. I felt scared and disappointed, especially thinking about how my parents would feel. I began comparing myself to my classmates and felt even worse. But after some time, I reminded myself that everyone fails at some point in life. I found encouragement from my friends and teachers who told me not to give up. I took time to reflect on my mistakes and identified the subjects and topics I was weak in. Then I created a new study plan with achievable goals. I started attending extra classes, asking more questions, and seeking help from my teachers whenever I had doubts. Slowly, I began to improve and gain confidence. In the next exam, I passed with good marks and felt very proud. I now understand that failure is not the end, but a stepping stone to success. This experience has motivated me, and I even share my story with others to inspire them not to give up.

Question 13.
Using the hints provided, write at least ten lines, expanding the ideas into complete sentences.

Imagine you went to a picnic with your family in a garden near your village or town. Play a game asking all the members to keep their mobile phones in a box, lock them up for one or two hours, and spend time talking to each other. Write down all that might happen.

Hints: Initially everyone is irritated – slowly get used – first father starts to speak of his school days – mother recollects your (her children’s) childhood memories – sister speaks of how she felt when she was lost and found at a fair – you speak of your favourite memory with each of your family member.
Answer:

A Peaceful Picnic without Mobile Phones

Last Sunday, my family and I went on a picnic to a beautiful garden near our village. As we spread the mat under a shady tree, I proposed a small challenge to everyone: to put their mobile phones in a box and lock it up for two hours. At first, everyone, especially my sister and uncle, frowned and complained – it felt like I had taken away their lifeline. However, as the minutes passed, we began to enjoy the calm, natural surroundings and each other’s company.

Surprisingly, it was my father who broke the silence by sharing stories from his school days, how he was the best cricket player in his class and once got punished for sneaking mangoes from the headmaster’s garden! Mother laughed and soon began telling us about our childhood; how I once tried to feed a chapati to a toy duck, and how my sister used to cry when our neighbour’s dog barked.

Then my sister opened up about the time she got lost at a fair when she was five, how scared she was, and how relieved she felt when Dad finally found her near the candy stall. I felt a deep sense of connection as I shared my favourite memories with each of them: how I love the quiet moments with Dad during our evening walks, how Mom’s bedtime stories always calmed me, and how my sister and I used to make up silly songs on rainy days.

Laughter, surprise, and even a few tears filled those two hours. By the time we opened the box and gave back the phones, no one rushed to check messages. We just sat there, smiling, realizing that real connection lives outside the screen.

For systematic learning, students can include AP Inter 1st Year English Grammar with Answers and Non-Verbal Information in English Inter 1st Year in their daily study routine.

AP Inter 1st Year English Non-Verbal Information with Answers

Non-verbal communication has two key aspects. The first includes the elements we use while speaking – such as facial expressions, hand gestures, way of standing and sitting, and body movements. The second involves conveying information through visual formats like tables, pie charts, bar graphs, flow charts, root maps, and tree diagrams. These visual tools are highly effective in presenting information clearly, often making it easy for even common persons to understand complex data. Therefore, students should develop the ability to present information both visually and verbally for effective and comprehensive communication.

1. pie charts / diagrams

A pie chart is a type of graph that displays data in the form of a circular diagram. Each slice of the pie represents a portion of the total, showing the relative size of each category. It is a form of pictorial representation that helps illustrate how a whole is divided into different parts. The slices are often labeled with details to indicate each category’s contribution to the overall matter. To create a pie chart, both categorical variables and numerical variables are required. In this chart, the term ‘pie’ symbolizes the entire dataset, while the slices represent the individual parts that make up the whole.

Example:

Question 1.
Study the following table and draw a pie diagram based on the information given in the table.

Answer:

Try This:

A summer survey conducted among 500 people in an area revealed their favourite beverages to beat the heat:

• Cool drinks: 148 people preferred them over natural drinks.
• Buttermilk: 102 people chose this as a healthy option.
• Sugandhi soda (Nannari): 90 people opted for it.
• Coconut water: 80 people preferred it.
• Iced tea: Chosen by 20 people.
• Cold coffee: Chosen by 15 people.
• Local drinks: The remaining participants preferred local beverages over all the above options.

Based on this data, create a pie chart to visually represent the distribution of preferences.

2. Tree diagrams

A tree diagram is a visual tool used to represent the hierarchical structure of information in a clear, pictorial form. It is called a tree diagram because its layout resembles the shape of a tree, though it is typically drawn upside down or left to right with the root at the top and branches extending downward. Tree diagrams are useful for breaking down complex information into simpler, organized parts. Here’s how you can create one:

1. Choose your main concept, idea, or topic.
   This should be the broadest or most general idea you want to explore.
2. Place the main concept at the top of the diagram.
   Since tree diagrams are hierarchical, always start with the central theme at the top.
3. Create the first level of branches.
   These branches represent the key ideas or steps directly related to main concept.
4. Add additional branches.
   From each of the first-level branches, add more branches to represent supporting details, sub-ideas, or specific outcomes.
5. Continue branching until completion.
   Keep expanding until all paths lead to conclusions, decisions, or final outcomes.

Tree diagrams are simple to draw and extremely helpful for planning, organizing thoughts, decision-making, and understanding complex systems.

Example

Things in our surroundings form two groups namely living things and non-livings. Living things are of two kinds, plants and animals. There are two types of plants – flowering and non-flowering plants. If we consider animals, some animals lay eggs which are called oviparous animals. Some animals give birth to their young ones. They are viviparous animals.

Read the above paragraph and draw a tree diagram based on the information.
Answer:

Try This

The Indian government is structured into three main branches: Legislative, Executive, and Judiciary. The Legisla¬tive branch is responsible for making laws. Further it is divided into Parliament and State legislatures. Parliament is further divided into Lok Sabha and Rajya Sabha. State legislature has Vidhan Sabha and Vidhan Parishad. The executive branch is responsible for implementing and enforcing laws. This branch is classified into Central Executive and State Executive. Central Executive has President, Prime Minister and Council of Ministers and Bureaucracy. State Executive has Governor, Chief Minister and Council of Ministers and State Bureaucracy. The judiciary interprets laws, resolves disputes, and ensures justice. Judiciary is further classified into Supreme Court, High Courts and Subordinate courts. Subordinate courts are divided into District and Sessions Courts and Lower courts.

Show the above information in a tree diagram.

Answer:

3. Tables:

A table is a systematic arrangement of information or data, typically organized in rows and columns. Tables are widely used in research, data analysis, and communication because they allow for easy comparison and interpretation of data.

Creating a Table: A Step-by-Step Guide

Drawing a table is a simple and effective way to organize data.

Follow these steps :

1. Determine the purpose of the table. Identify what infor-mation you want to compare, categorize, or summarize. Knowing the purpose helps define the structure.
2. Figure out the number of columns and rows needed. De¬cide how much data you have and how it should be grouped.
3. Represent individual items, records, or entities in rows. Each row should contain data for a single unique obser-vation or record. This ensures clarity and avoids confusion.
4. Represent attributes, categories, or variables in columns. Each column should describe a specific aspect of the data (e.g., name, age, score), and the data within each column should be of the same type.
5. Label rows and columns with meaningful headers. Use clear, descriptive headings to ensure the table is easy to read and understand.

Tip : When designing a data table, always organize it so that each row represents a unique observation, and each column represents a specific variable. This makes the data easier to analyze, compare, and interpret.

Example

Four friends decided to compare their monthly expenses on different categories. Srikar spends Rs. 2000 on food. Rs. 1500 on entertainment, Rs. 1000 on travel, and Rs. 500 on skill development. Giridhar spends Rs. 2500 on food, Rs. 1000 on entertainment, Rs. 1500 on travel, and Rs. 750 on skill development. Prabhakar allocates Rs. 1800 for food, Rs. 1200 for entertainment, Rs. 800 for travel, and Rs. 600 for skill development. Siddhu spends Rs. 2200 on food, Rs. 1300 on entertainment, Rs. 900 on travel, and Rs. 700 on skill development.

Read the above passage and draw a table based on the given information.
Answer:

Try This

Convert the information given below into a table.

The University Wits were a group of influential playwrights and poets of the late 16th century who significantly shaped English literature. Among them, Christopher Marlowe (1564-1593) stood out with his masterpieces Doctor Faustus and Tamburlaine the Great known for their powerful use of blank verse and ambitious characters. Thomas Kyd (1558-1594), a pioneer of the revenge tragedy genre, gained fame with The Spanish Tragedy and Cornelia. Robert Greene (1558-1592), known for his prose romances and pastoral comedies, authored Friar Bacon and Friar Bungay and Pandosto. John Lyly (1554-1606) popularized the Euphuistic style in his works Euphues: The Anatomy of Wit and Endymion.

Thomas Nashe (1567-1601), celebrated for his satirical prose, is remembered for The Unfortunate Traveller and Pierce Penniless: His Supplication to the Devil. George Peele (1556-1596) brought lyrical and pastoral drama to the forefront with The Arraignment of Paris and David and Bethsabe. Finally, Thomas Lodge (1558¬1625), known for his romantic and pastoral literature, authored Rosalynde and A Looking Glass for London and En¬gland. Together, these writers revolutionized English drama and prose, paving the way for the golden age of Elizabethan literature.
Answer:
Here’s the information about the University Wits presented in a table:

4. Bar-Graphs

A bar graph (or bar chart) is a type of graph that uses rect-angular bars to visually represent and compare data. The height or length of each bar corresponds to the value or the measure it represents, making it easy to compare different quantities at a glance.

Bar graphs are particularly useful for:

Comparing different categories of information.
Showing how values change over time.
Explaining complex data in a clear and visual format.

They are especially effective when one of the variables is categorical (such as types of fruits, countries, or age groups).

Structure of Bar Graph :

X-axis (horizontal): Usually represents the categories or time periods being measured.
Y-axis (vertical) : Represents the values or quantities corre¬sponding to each category.

Each bar is drawn with a height (or length) proportional to the value it represents, allowing for easy visual comparison.

Example

Read the following passage and draw a bar graph based on the given information.

In a parking area there are a number of vehicles being parked every day. As per the data given by the watchman of that parking area, on Monday, 60 cars, 92 bikes and 23 bicycles were parked. On Tuesday, 57 cars, 101 bikes and 20 bicycles were parked. On Wednesday, 62 cars, 105 bikes and 25 bicycles were parked.
Answer :

Try This

At a sports complex, the number of participants in different activities was recorded over four days. The data is as follows:

• Monday: 35 played basketball, 50 played tennis, and 25 used the gym.
• Tuesday: 40 played basketball, 55 played tennis, and 30 used the gym.
• Wednesday : 45 played basketball, 60 played tennis, and 35 used the gym.
• Thursday : 50 played basketball, 65 played tennis, and 40 used the gym.

Based on this data, draw a bar graph representing the number of participants in each activity over the four days. Use different colors for each activity to ensure clarity.
Answer:

5 . Route Maps

A route map is a detailed diagram or plan that shows the way from a starting point to a specific destination. It typi¬cally includes important information such as the starting location, destination, and the route to follow. A route map often highlights roads, streets, and key landmarks, and may also provide estimated distances or travel times. These maps are designed to help users navigate efficiently between two or more locations, making them especially useful for travel, transportation, and planning purposes.

Example

Anjali got down at city bus-stop. She should go to her friend’s house which is nearby. She called her friend Malathi and took directions to reach Malathi’s home.

Here are the directions :

1. “After getting down at the city bus stop you will see a medical shop right opposite the bus-stop.”
2. “Take the only road beside the medical shop.”
3. “If you walk for two minutes you will see a super bazar on your left.”
4. “As you cross the super bazar, you arrive at a Y-junction.”
5. “There take left. After walking a few steps you will see Sweta Padma Apartments. Come into the apartment and our flat number is 402.”

Answer:

Try This

Route Map Directions : From Vijayawada to Tirupati:

1. Start: Begin at Vijayawada. Take the NH16 highway heading south.
2. After 30 km, you’ll pass through Guntur. Continue on NH16.
3. Drive 150 km further to reach Ongole.
4. Continue on NH16 for another 150 km to reach Nellore.
5. Turn Off : At Nellore, take a left turn onto NH716 towards Tirupati.
6. Follow NH716 for 130 km, passing through smaller towns like Naidupeta.
7. You will arrive at Tirupati, where you can visit the famous Sri Venkateswara Temple.

Draw a route map from Vijayawada to Tirupati with clear labels for each landmark and a line representing the route. Use symbols for landmarks to make your map visually engaging!
Answer:

Exercises

Question 1.
In 2024 Summer Olympics in Paris, 184 countries sent their athletes to compete in various games. With 594 athletes, the United States had the largest delegation at the Olympics and won 40 gold medals, 44 silver and 42 bronze. Following the U.S, France with 572 athletes and won 16 gold, 26 silver and 22 bronze medals. Australia with 460 athletes won 18 gold, 19 silver and 16 bronze medals. Germany with 427 athletes won 12 gold, 13 silver and 8 bronze medals and India with 110 athletes won 1 silver and 5 bronze.
Answer:

Question 2.
Topic : Modes of Transportation

A city survey recorded the preferred modes of daily transportation among its residents:

• Public Transport: 40% of the residents.
• Private Vehicles – cars : 20% of the residents.
• Private Vehicles – bikes : 25% of the residents.
• Cycling : 5% of the residents.
• Walking : 10% of the residents.

Represent this data as a pie chart.
Answer:

Question 3.
A well-planned vegan diet focuses on whole, plant-based foods and excludes all animal-derived products. A raw High- Carb, Low-Fat (HCLF) vegan diet typically includes carbohydrates, with about 80% of the total calories coming from fruits, vegetables, and other carbohydrate-rich foods. Proteins, often sourced from nuts, seeds, leafy greens, and legumes, make up approximately 10% of the calorie intake. Similarly, fats, derived from sources like avocados, nuts, and seeds, also account for 10% of total calories. This proportion supports a high-energy, nutrient-dense eating style while minimizing fat intake. Represent this data by a pie chart.
Answer:

Question 4.
A student’s performance in different subjects is as given in the table. Convert the following table into a bar graph.

Answer:

Question 5.
The human nervous system is classified into Central Nervous System (CNS) and Peripheral Nervous System (PNS). Brain is the centre of the CNS consisting of Cerebrum (responsible for voluntary actions, emotions, and sensory perception) and Cerebellum (controls balance and coordination). The Brainstem regulates heart rate, breathing, and reflexes and the Spinal Cord connects the brain to the rest of the body. The Peripheral Nervous System (PNS) has the following: Somatic Nervous System which controls voluntary movements (e.g., skeletal muscles) and Autonomic Nervous System (ANS). It is further divided into Sympathetic Nervous System (“Fight or flight” response) and Parasympathetic Nervous System (“Rest and digest” response).

Represent the above classification of the human nervous system in the form of a tree diagram.
Answer:

Question 6.
India experiences six primary seasons, each with unique characteristics. Spring, known as Vasanta in Hindi, occurs from March to May (corresponding to the months Chait and Baisakh in the Hindi calendar). Summer, or Grishma, spans May to July, aligning with the Hindi months Jeth and Asarh. The rainy season, referred to as Varsha, lasts from July to September and corresponds to the Hindi months Sawan and Bhadon. Autumn, called Sharada, extends from September to November, represented by the Hindi months Ashwin and Kartik. Winter, or Hemanta, is observed from November to January, covering the Hindi months Aghan and Pausha. Lastly, the cold season, known as Shishira, occurs from January to March, coinciding with the Hindi months Magha and Phalgun. Convert this information into a table diagram.
Answer:

Question 7.
You are planning a trip from your house to the local park. Along the way, you will pass several landmarks, including a grocery store, a library, a bus stop, and a cafe. Use the following instructions to draw a route map showing your journey:

Start from your house. Turn to your left and walk till you reach a grocery shop on your right. Once you cross the grocery shop you will come to a T- junction. There take left and walk till you reach Public Library on your right. Just beside by the library there’s a pathway that leads to the bus stop. On crossing the bus-stop take left turn at the cafe. In two minutes you will reach the park.

Draw the map with clear labels for each landmark and a line to represent the route you will take. Include symbols for the landmarks to make your map visually engaging !
Answer:

Question 8.
In a public library, visitors are categorized based on their purpose of visit. The librarian recorded the following data over four days:

• Monday : 45 visitors for borrowing books, 78 for reading, 32 for internet use.
• Tuesday : 50 for borrowing books, 85 for reading, 40 for internet use.
• Wednesday : 55 for borrowing books, 92 for reading, 35 for internet use.

Answer:

Question 9.
The Ramayana, divided into seven parts or kandas, narrates the life and adventures of Lord Rama and his companions. Bala Kanda details the origins of Rama and his brothers, Lakshmana, Bharata, and Shatrughna.

Ayodhya Kanda depicts the events leading to Rama’s 14-year exile. Aranya Kanda chronicles the final year of exile, featuring Rama defeating demons, Sita’s abduction by Ravana. In Kishkindha Kanda, Rama aids Sugriva in defeating Vali, and Hanuman’s southern army discovers Sita’s location in Lanka. Sundara Kanda focuses on Hanuman’s adventures, while Yuddha Kanda recounts the epic war between Rama and Ravana, culminating in Ravana’s defeat and Sita’s Agni Pareeksha. Rama’s exile ends with his celebrated return to Ayodhya. Finally, Uttara Kanda describes the later years, including Sita’s banishment, the birth of Lava and Kusha, and their upbringing by Valmiki.
Answer:

Question 10.
You are a student at Government Junior College, Guntur. You have won first prize in an elocution competition and need to visit the Collectorate, Guntur, to receive your award from the District Collector. Since you do not know the way to Collectorate, your friend provides you with the following directions.

Directions:
Start at Government Junior College near Naaz Centre in Sambasiva Pet, Guntur. Exit the college and turn left to reach a T-junction. At the junction, turn left and continue until you reach the traffic signals at the rossroads. There, turn right and go straight for approximately 800 metres. You will see Police Parade Ground (PPG) on your left. The Collectorate entrance gate is directly opposite the PPG.

Task: Based on the directions, draw a route map with clear labels for each landmark. Use a line to represent the path you will take and include symbols for the landmarks to make your map visually engaging.
Answer:

Question 11.
The field of science has been studied in many historic civilizations. It is broadly classified into three main branches : Formal Sciences, Natural Sciences and Applied Sciences. Formal Sciences include the subjects of Mathematics, Statistics and Logic. Natural Sciences are separated into Life Sciences and Physical Sciences. Between these, Life Sciences comprise the studies of Biology and Medicine, while Physical Sciences consist of Chemistry and Physics. Applied Sciences cover the disciplines of Engineering and Medical Science.
Draw a tree diagram to show the above data.

Answer:

Question 12.
Read the following passage and draw a table based on the information.

Silpa Electronic Store sells home appliances like Coffee Makers, Grinders, Mixers, Dish Washers, Ovens and Rice Cookers. Revenue earned by this store in the year 2024 is as follows :

210 Coffee Makers were sold for Rs. 30,000. 260 Grinders were sold for Rs. 96,000. 180 Mixers were sold for Rs. 45,000. 128 Dish Washers were sold for Rs. 3,20,000. 115 Ovens were sold for Rs. 80,000. The highest sold appliance was Rice Cookers, 460 sold for Rs. 48,000. Show the above data in a table form.

Answer:
Sales of Home Appliances

Referring to the AP Inter 1st Year Maths Study Material Chapter 10 Conic Sections Exercise 10e Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Conic Sections Solutions Exercise 10e

I.

Question 1.
If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.
Solution:
Taking the origin as the vertex and the axis of the reflector as positive X-axis, the equation of the parabolic section is rightward parabola y² = 4ax
Depth OA = 5 cm, diameter L

∴ Coordinates of point L are (OA, AL)= (5, 10).
Since point L lies on parabola, we have
(10)² = 4a(5)
∴ a = \(\frac{100}{20}\) = 5
Focus is (a, 0) = (5, 0) i.e., at A.
Hence, focus is at the mid-point of the given diameter.

Question 2.
An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.
Solution:
Let ABA’ be the given semi-elliptic arch such that width AA’ = 8 m and OB = 2 m.
The arch is a part of ellipse whose semi-major axis is \(\frac{8}{2}\) = 4 m and
semi-minor axis is OB = 2 m.
i.e., a = 4 m, b = 2 m.

∴ Equation of the ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1.
Putting values of a and b, we have
\(\frac{x^2}{16}+\frac{y^2}{4}\) = 1 ………… (i)
We have, height PR at P where AP (distance from one end say A of the arch APA’) = 1.5 m.
Now OP = OA – AP = 4 – 1.5 = 2.5 m.
For the point R, x = OP = 2.5 = \(\frac{25}{10}=\frac{5}{2}\)
Putting x = \(\frac{5}{2}\) in (i), we get \(\frac{x^2}{16}+\frac{y^2}{4}\) = 1
\(\frac{y^2}{4}\) = 1 – \(\frac{25}{64}\)
y² = \(\frac{64-25}{16}=\frac{39}{16}\)
y = \(\frac{\sqrt{39}}{4}=\frac{6.24}{4}\) = 1.56
Hence, the required height of arch = PR = ordinate (i.e., y) of R = 1.56 m (approx.)

II.

Question 1.
An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?
Solution:
Taking the vertex of the parabola as the origin O and axis along OY’, the equation of the parabolic arch is downward parabola.
x² = – 4ay [∵ upward parabola can’t have a base] …….(i)
Since MB = \(\frac{1}{2}\)base; AB = \(\frac{5}{2}\) = 2.5 m
(From the figure)
and OM = 10 m, the coordinates of B are (2.5, -10).
B lies on (i) ⇒ (2.5)² = -4a(-10)
⇒ 6.25 = 40 a
⇒ a = \(\frac{6.25}{40}\)

Putting this value of a in (i),
equation (i) becomes
x² = –\(-\frac{6.25}{10}\)y
⇒ x² = -0.625y ……..(ii)

Let the width of the parabola at a distance
2m [= ON (say)] from the vertex of a parabola be CD.
If CD = 2x, then coordinates of D are (x, – 2). Since D(x, – 2) lies on (ii).
∴ x² = -0.625 × (-2) = 1.250
= \(\frac{1250}{1000}=\frac{125}{100}=\frac{5}{4}\)
⇒ x = \(\frac{\sqrt{5}}{2}\)
Hence, CD = width at depth 2m = 2x = 2 × \(\frac{\sqrt{5}}{2}\) = √5 = 2.33 m (approximately)

Question 2.
The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle.
Solution:
Let the cable of suspension bridge (hanging in the form of a parabola) be the parabolic arc COB with vertex at the lowest point O and axis vertical.
Let LM – 100 m be the horizontal roadway such that CL = BM = (longest wire) 30 m and OA = (shortest wire) 6 m. [∵ O, the vertex of the parabola is the lowest point of this upward parabola].
We have to find the length of supporting wire NP attached to the roadway at N where AN = 18 m. Now AL = AM = 50 m. (∵ LM = 100m )
Let the coordinate axes be chosen as shown in the figure. We know that equation of the upward parabola with vertex at O(0, 0) and axis along positive Y-axis is
x² = 4ay …….(i)

Since it passes through the point B whose co-ordinates are
x = OH = AM = 50 and
y = BH = BM – HM = BM – OA = 30 – 6 = 24
Putting x = 50 and y = 24 in (i), we have
(50)² = 4a(24)
⇒ 2500 = 96a
⇒ a = \(\frac{2500}{96}=\frac{625}{24}\)
Putting the value of a in (i), the equation of the parabola is
x² = 4 × \(\frac{625 y}{24}\)
or
x² = \(\frac{625 y}{6}\) …………..(ii)
Through the point N draw a vertical line to meet the parabola at point P. Then co¬ordinates of point P are
x = OK = AN = 18 and y = PK = PN – KN = PN – OA = PN – 6.
Putting these values of x and y in (ii),
(18)² = \(\frac{625}{6}\)(PN – 6)
⇒ 324 × 6 = 625(PN – 6)
Length of the supporting wire attached to the roadway at point N which is 18 m from the middle point A = PN = 9.11 m.

Question 3.
A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the X-axis.
Solution:
Let AB be the rod of length 12 cm and P(x, y) be the point on it such that AP = 3 cm. (given)
∴ PB = AB – AP = 12 – 3 = 9 cm .
From P, draw PL and PM perpendiculars on Y-axis and X-axis respectively.
∴ PL = x and PM = y.
Let ∠OAB = θ. ∠LPB is also = θ [Corresponding angles]

In right angled ΔAMP, sin θ = \(\frac{M P}{A P}=\frac{y}{3}\) …….(i)
In right angled ΔPLB, cos θ = \(\frac{P L}{P B}=\frac{x}{9}\) ……(ii)
Putting these values of cos θ and sin θ
from (ii) and (i) in the relation cos²θ + sin²θ = 1
We have, \(\frac{x^2}{81}+\frac{y^2}{9}\) = 1
⇒ \(\frac{x^2}{81}+\frac{y^2}{9}\) = 1
Which is the required locus of point P.

III.

Question 1.
Find the area of the triangle formed by the lines joining the vertex of the parabola x² = 12y to the ends of its latus rectum.
Solution:
Equation of the parabola is x² = 12 y.
Comparing with upward parabola x² = 4ay, we have 4a = 12. ∴ a = 3.
In the figure, S is the focus, O is the vertex and AB (⊥ OS) is the latus rectum.
(By definition of latus rectum)

Latus rectum AB = 4a = 12.
Since the focus S is (0, a) = (0, 3). ∴ OS = 3 [
Area of OAB = \(\frac{1}{2}\)AB × OS [∵ Area of the ∆^le = \(\frac{1}{2}\) base × height]
= \(\frac{1}{2}\) × 12 × 3 = 18 sq. units.

Question 2.
A man running a race-course notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the path traced by the man.
Solution:
Let S and S’ be the two flag posts and P be the position of the man at any instant. Since PS + PS’ = 10 (given),
By definition of an ellipse, the path traced by the man is an ellipse having its foci at the two flag posts and length of the major axis = 10.

Because length of the major axis = 2a
∴ 2a =10
⇒ a = 5.

Also SS’ = distance between the foci = (2c) = 8 (given)
2c = 8
⇒ c = 4
Using the relation c² = a² – b², we have 4² = 52 – b²
⇒ b² = 25 – 16 = 9.
Putting the values of a² and b² in (i), the equation of the (ellipse) path is \(\frac{x^2}{25}+\frac{y^2}{9}\) = 1

Question 3.
An equilateral triangle is inscribed in the parabola y² = 4ax. where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.
Solution:
The given parabola is y² = 4ax. ………. (i)
Its vertex is the origin O. Let OAB be the equilateral triangle inscribed in the parabola. Let each side of triangle be of length l.
By symmetry, AB ⊥ OX and ∠BOC = ∠AOC = 30°.
In ΔOCA, cos 30° = \(\frac{\mathrm{OC}}{\mathrm{OA}}=\frac{\mathrm{x}}{l} \Rightarrow \frac{\sqrt{3}}{2}=\frac{\mathrm{x}}{l}\)
⇒ x = l\(\frac{\sqrt{3}}{2}\)
and sin 30 = \(\frac{\mathrm{AC}}{\mathrm{OA}}=\frac{\mathrm{y}}{l} \Rightarrow \frac{1}{2}=\frac{\mathrm{y}}{l}\)
⇒ y = \(\frac{l}{2}\)

∴ Coordinates of point A are (x, y) = \(\left(\frac{l \sqrt{3}}{2}, \frac{l}{2}\right)\)
Since A lies on parabola (i),
∴ Coordinates of point A satisfy equation (i)
∴ \(\left(\frac{l}{2}\right)^2\) = 4a.\(\frac{\sqrt{3}}{2}\)l
\(\frac{l^2}{4}\) = 2a√3l
l² = 8a√3l
Dividing both sides by l, ⇒ l = 8√3a
Hence, the length of the side of an equilateral triangle = 8√3a.

Practice AP Inter 1st Year Maths Study Material Chapter 10 Conic Sections MCQ to identify your strengths and weak areas.

AP Inter 1st Year Maths Conic Sections MCQ

Circle

Question 1.
The diametes of a circle are along 2x + y – 7 = 0 and x + 3y – 11 = 0.Then the equation of this circle, which also passes through (5, 7) is ( 3 )
1) x² + y² – 4x – 6y – 16 = 0
2) x² + y² – 4x – 6y – 20 = 0
3) x² + y² – 4x – 6y – 12 = 0
4) x² + y² + 4x + 6y – 12 = 0
Answer:
3) x² + y² – 4x – 6y – 12 = 0

Explanation:
Point of intersection of 2x + y – 7 = 0, x + 3y – 11 = 0 is (2, 3).
2x + y – 7 = 0, x + 3y – 11 = 0 are diameters ⇒ Centre = C(2, 3)
Circle passes through A(5, 7) ⇒ Radius = CA = \(\sqrt{9+16}=\sqrt{25}\) = 5.
Equation of the circle is (x – 2)² + (y – 3)² = 5²
∴ x² + y² – 4x – 6y – 12 = 0

Question 2.
Consider the circle x² + y² – 4x – 2y + c = 0 whose centre is A(2, 1). If the point P(10, 7) is such that the line segment PA meets the circle in 9 with PQ = 5, then c = ( 4 )
1) -15
2) 20
3) 30
4) -20
Answer:
4) -20

Explanation:
Given centre of the circle is x² + y² – 4x – 2y + c = 0 and centre A(2, 1) and P = (10, 7) PA meets the circle in Q with PQ = 5.
PA = PQ + QA
⇒ \(\frac{1}{2}\) = 5 + r
⇒ 10 = 5 + r
⇒ r = 5.
⇒ \(\frac{1}{2}\) = 5
⇒ \(\frac{1}{2}\) = 5
⇒ 5 – c = 25
⇒ c = -20

Question 3.
The abscissae of two points A and B are the roots of the equation x² + 2ax – b² = 0 and their ordinates are the roots of the equation y² + 2py – q² = 0. The radius of the circle with AB as diameter is
1) \(\sqrt{a^2+b^2+p^2+q^2}\)
2) \(\sqrt{a^2+p^2}\)
3) \(\sqrt{b^2+q^2}\)
4) \(\sqrt{a^2+b^2-p^2+q^2}\)
Answer:
1) \(\sqrt{a^2+b^2+p^2+q^2}\)

Explanation:
Let A = (x₁, y₁), B = (x₂, y₂)
x₁, x₂ are the roots of x² + 2ax – b² = 0
⇒ (x – x₁) (x – x₁) = 0
y₁, y₂ are the roots of y² + 2py – q² = 0
⇒ (x – y₁) (x – y₂) = 0
Equation of the circle with AB as diameter is
x² + 2ax – b² + y² + 2py – q² = 0
x² + y² + 2ax + 2py – b² – q² = 0
Radius of the circle = \(\\sqrt{a^2+p^2+b^2+q^2}=\sqrt{a^2+b^2+p^2+q^2}\)

Question 4.
The equation of the circle concentric with the circle x² + y² – 6x + 12y + 15 = 0 and of double its area is
1) x² + y² – 6x + 12y – 15 = 0
2) x² + y² – 6x + 12y – 30 = 0
3) x² + y² – 6x + 12y – 25 = 0
4) x² + y² – 6x + 12y – 20 = 0
Answer:
1) x² + y² – 6x + 12y – 15 = 0

Explanation:
Equation of the given circle is x² + y² – 6x + 12y + 15 = 0.
Radius r₁ = \(\frac{1}{2}\)
Area of the given circle = 30π
Equation of the required circle is x² + y² – 6x + 12y + k = 0.
Radius r₂ = \(\frac{1}{2}\)
Area of the required circle = (45 – k)π
Given that (45 – k)π = 2(30π) ⇒ k = – 15
Required circle is x² + y² – 6x + 12y – 15 = 0.

Question 5.
A square is inscribed in the circle x² + y² – 2x + 4y + 3 = 0 and its sides are parallel to the co-ordinate axes then one vertex of the square is
1) (1 + √2, -2)
2) (1 – √2, -2)
3) (1,- 2 + √2)
4) None
Answer:
4) None

Explanation:
Given x² + y² – 2x + 4y + 3 = 0
(x – 1)² – 1 + (y + 2)² – 4 + 3 = 0
⇒ (x – 1)² + (y + 2)² = 2
Centre C = (1, -2), Radius r = √2
Since diameter d = 2r = 2√2
Let ‘s’ is length of square, then d = s√2 ⇒ s√2 = 2√2 ⇒ s = 2
Half of side length is \(\frac{2}{2}\) = 1.
Since possible vertices are (1 + 1, -2 + 1) = (2, -1) ; (1 + 1, -2 – 1) = (2, -3)
(1 – 1, -2 + 1) = (0, -1) ; (1 – 1, -2 – 1) = (0, -3)
None of the option is correct.

Question 6.
The equation of the circle with centre (cos a, sin 1) and radius 1 is
1) x² + y² = 1
2) x² + y² – 2x cos a – 2y sin a = 0
3) x² + y² + 2x cos a + 2y sin a = 0
4) x² + y² – 2x cos a – 1 = 0
Answer:
2) x² + y² – 2x cos a – 2y sin a = 0

Explanation:
Equation of the circle with centre (cos a, sin a) and radius 1 is (x – cos a)² + (y – sin a)² = (1)²
⇒ x² + y² – 2x cos a – 2y sin a + cos²a + sin²a = 1
⇒ x² + y² – 2x cos a – 2y sin a = 0

Question 7.
If r1 and r2 are radii of the circles x² + y² – 4x – 8y – 44 = 0 and x² + y² + 6x + 8y – 96 = 0 respectively then r₁ : r₂ is
1) 8 : 11
2) 9 : 11
3) 16 : 21
4) 64 : 123
Answer:
1) 8 : 11

Explanation:
r₁ = radius of x² + y² – 4x – 8y – 44 = 0
⇒ r₁ = \(\sqrt{4+16+44}=\sqrt{64}\) = 8.
r₂ = radius of x² + y² + 6x + 8y – 96 = 0
⇒ r₂ = \(\sqrt{9+16+96}=\sqrt{121}\) = 11
∴ r₁ : r₂ = 8 : 11.

Question 8.
The radius of the circle touching the lines 3x – 4y + 5 = 0; 6x – 8y – 9 = 0 is
1) 1
2) \(\frac{23}{15}\)
3) \(\frac{20}{19}\)
4) \(\frac{19}{20}\)
Answer:
4) \(\frac{19}{20}\)

Explanation:
If r is the radius of the circle touching the lines 3x – 4y + 5 = 0; 6x – 8y – 9 = 0, then
2r = \(\frac{10+9}{\sqrt{36+64}}=\frac{19}{10}\)
⇒ r = \(\frac{19}{20}\)

Question 9.
The perimeter of the circle 3(x² + y²) = 16 is
1) \(\frac{4}{\sqrt{3}}\)π
2) \(\frac{16}{\sqrt{3}}\)π
3) \(\frac{8}{\sqrt{3}}\)π
4) \(\frac{1}{\sqrt{3}}\)π
Answer:
3) \(\frac{8}{\sqrt{3}}\)π

Explanation:
Equation of the given circle is 3(x² + y²) = 16
⇒ x² + y² = \(\frac{16}{3}\)
Perimeter of the circle = 2πr = 2π × \(\frac{4}{\sqrt{3}}=\frac{8}{\sqrt{3}}\)π

Question 10.
The radius of circle 3(x² + y²) – 3x + 9y + 10 = 0
1) \(\sqrt{\frac{15}{2}}\)
2) \(\frac{5}{\sqrt{2}}\)
3) \(\sqrt{\frac{15}{3}}\)
4) Not defined
Answer:
4) Not defined

Explanation:
Equation of the given circle is 3(x² + y²) – 3x + 9y + 10 = 0
⇒ x² + y² – x + 3y + \(\frac{10}{3}\) = 0
Radius of the circle = \(\frac{1}{2}\) Which is not defined.

Parabola

Question 1.
Parabola has the origin as its focus and the line x = 2 as the directrix. Then the vertex of the parabola is at
1) (1, 0)
2) (0, 1)
3) (2, 0)
4) (0, 2)
Answer:
1) (1, 0)

Explanation:
S = (0, 0), Z = (2, 0)
⇒ A = \(\frac{S+Z}{2}\) = (1, 0)

Question 2.
If the focus of a parabola is (0, -3) and its directrix is y = 3 then its equation is
1) x² = -12y
2) x² = 12y
3) y² =-12x
4) y² = 12x
Answer:
1) x² = -12y

Explanation:
Focus of the parabola is S = (0, -3)
Equation of the directrix is y = 3 ⇒ y – 3 = 0.
If P(x, y) is a point on the parabola, then PS = distance of P from the directrix.
⇒ \(\sqrt{(x-0)^2+(y+3)^2}\) = (y -3)²
⇒ x² + y² + 6y + 9 = y² – 6y + 9
⇒ x² = -12y.
Equation of the parabola is x² = -12y.

Question 3.
If the parabola y² = 4ax passes through the point (3, 2) then the length of its latusrectum is
1) \(\frac{2}{3}\)
2) \(\frac{4}{3}\)
3) \(\frac{1}{3}\)
4) 4
Answer:
2) \(\frac{4}{3}\)

Explanation:
Parabola y² = 4ax passes through the point (3, 2)
⇒ (2)² = 4a(3)
⇒ 4a = \(\frac{4}{3}\)

Length of latusrectum = 4a = \(\frac{4}{3}\)

Question 4.
If the vertex of the parabola is the point (-3, 0) and the directrix is the line x + 5 = 0 then its equation is
1) y² = 8(x + 3)
2) x² = 8(y + 3)
3) y² = -8(x + 3)
4) y² = 8(x + 5)
Answer:
1) y² = 8(x + 3)

Explanation:
Given equation of the directrix is x = – 5 which is parallel to Y-axis.
∴ Principal axis of the parabola is parallel to X-axis. Vertex = (-3, 0).
Equation of the parabola is (y – 0)² = 4a(x + 3) ⇒ y² = 4a(x + 3)
We know the equation of the directrix is x = h – a ⇒ x = – 3 – a, but given as x = – 5.
∴ – 5 = -3 – a ⇒ a = 5 – 3 = 2
Equation of the parabola is y² = 8(x + 3).

Ellipse

Question 1.
The eccentricity of an ellipse, when its centre at the origin is \(\frac{1}{2}\). If one of the directrices is x = 4, then the equation of the ellipse is ( 4 )
1) 3x² + 4y² = 1
2) 4x² + 3y² = 1
3) 4x² + 3y² = 12
4) 3x² + 4y² = 12
Answer:
4) 3x² + 4y² = 12

Explanation:
Centre = (0, 0), Eccentricity e = \(\frac{1}{2}\) and directrix is x = 4
⇒ \(\frac{a}{e}\) = 4
⇒ 2a = 4
⇒ a = 2
b² = a²(1 – e²) = 4(1 – \(\frac{1}{4}\)) = 3 ;
Equation of the ellipse is \(\frac{x^2}{4}+\frac{y^2}{3}\) = 1
⇒ 3x² + 4y² = 12

Question 2.
The equation of the ellipse whose focus is at (4, 0) and whose eccentricity \(\frac{4}{5}\) is
1) \(\frac{x^2}{3^2}+\frac{y^2}{5^2}\) = 1
2) \(\frac{x^2}{5^2}+\frac{y^2}{3^2}\) = 1
3) \(\frac{x^2}{5^2}+\frac{y^2}{4^2}\) = 1
4) \(\frac{x^2}{4^2}+\frac{y^2}{5^2}\) = 1
Answer:
2) \(\frac{x^2}{5^2}+\frac{y^2}{3^2}\) = 1

Explanation:
e = \(\frac{4}{5}\), (ae, 0) = (4, 0)
⇒ ae = 4
⇒ a = 5 ; b2 = 25(1 – \(\frac{16}{25}\)) = 9
∴ Equation of the ellipse is \(\frac{x^2}{25}+\frac{y^2}{9}\) = 1
⇒ \(\frac{x^2}{5^2}+\frac{y^2}{3^2}\) = 1

Question 3.
The ellipse x² + 4y² = 4 is inscribed is a rectangular (rectangle) aligned with the co-ordinate axes which in turn is inscribed in another. The equation of the ellipse is
1) x² + 12y² = 16
2) 4x² – 8y² = 48
3) 4x² + 64y² = 48
4) x² + 16y² = 16
Answer:
1) x² + 12y² = 16

Explanation:
The given ellipse is x² + 4y² = 4 ⇒ \(\frac{x^2}{4}+\frac{y^2}{1}\) = 1.
The vertex of the rectangle (2, 1) is in I quadrant. Again, the ellipse circumscribing the rectangle passes through the point (4, 0). So its equation is \(\frac{x^2}{16}+\frac{y^2}{b^2}\) = 1

A(2, 1) lies on the above ellipse, \(\frac{4}{16}+\frac{1}{b^2}\) = 1
⇒ \(\frac{1}{\mathrm{~b}^2}\) = 1 – \(\frac{1}{4}\)
⇒ b² = \(\frac{4}{3}\)
Thus the equation of the desired ellipse is \(\frac{x^2}{16}+\frac{3}{4}\)y² = 1
x² + 12y² = 16

Question 4.
If focus of an ellipse is at the origin the directrix is the line x = 4 and the eccentricity is \(\frac{1}{2}\). Then the length of the semi-major axis is
1) \(\frac{2}{3}\)
2) \(\frac{4}{3}\)
3) \(\frac{5}{3}\)
4) \(\frac{8}{3}\)
Answer:
4) \(\frac{8}{3}\)

Explanation:
Given S = (0, 0), e = \(\frac{1}{2}\), equation to directrix is x – 4 = 0.
We have CZ = \(\frac{a}{e}\)
⇒ CS + SZ = \(\frac{a}{e}\)
⇒ ae + SZ = \(\frac{a}{e}\)
⇒ SZ = \(\frac{a}{e}\) – ae.

Distance from focus to directrix = a(\(\frac{1}{e}\) – e)
|0 – 4| = a(2 – \(\frac{1}{2}\)) ⇒ a\(\left(\frac{3}{2}\right)\) = 4
⇒ a = \(\frac{8}{3}\).
∴ Length of the semi major axis is a = \(\frac{8}{3}\).

Question 5.
The major axes of an ellipse is three times minor axis, then the eccentricity is
1) \(\frac{2 \sqrt{2}}{3}\)
2) \(\frac{2}{3}\)
3) \(\frac{\sqrt{2}}{3}\)
4) \(\frac{1}{3}\)
Answer:
1) \(\frac{2 \sqrt{2}}{3}\)

Explanation:
2a = 3 . 2b ⇒ a = 3b
∴ e² = \(\frac{a^2-b^2}{a^2}=\frac{8 b^2}{9 b^2}=\frac{8}{9}\)
e = \(\frac{2 \sqrt{2}}{3}\)

Question 6.
The eccentricity of ellipse x² + 3y² = 6 is
1) \(\frac{2}{3}\)
2) \(\sqrt{\frac{2}{3}}\)
3) \(\frac{\sqrt{3}}{2}\)
4) \(\frac{2}{\sqrt{3}}\)
Answer:
2) \(\sqrt{\frac{2}{3}}\)

Explanation:
Given ellipse is x² + 3y² = 6
⇒ \(\frac{x^2}{6}+\frac{y^2}{2}\) = 1 (a > b).
Eccentricity of the ellipse = \(\sqrt{\frac{a^2-b^2}{a^2}}=\sqrt{\frac{6-2}{6}}=\sqrt{\frac{4}{6}}=\sqrt{\frac{2}{3}}\)

Hyperbola

Question 1.
The equation of the conic with focus at (1, – 1), directrix along x – y + 1 = O and with eccentricity √2 is
1) x² – y² = 1
2) xy = 1
3) 2xy – 4x + 4y + 1 = 0
4) 2xy + 4x – 4y + 1 = 0
Answer:
3) 2xy – 4x + 4y + 1 = 0

Explanation:
Focus S = (1, – 1), eccentricity e = √2, directrix along x – y + 1 = 0.
Let (x, y) be a point on the hyperbola ⇒ SP = e PM
⇒ SP² = e² PM²
⇒ (x – 1)² + (y + 1)² = 2\(\frac{1}{2}\)
⇒ x² + y² – 2x + 2y + 2 = x² + y² + 1 – 2xy – 2y + 2x
⇒ 2xy – 4x + 4y + 1 = 0, which is the equation of the conic.

Question 2.
If one of the foci of the hyperbola is origin the corresponding directrix is 3x + 4y + 1 = 0 and the eccentricity of the hyperbola is √5, then equation of the hyperbola is
1) 4x² + 11y² + 24xy + 6x + 8y + 1 = 0
2) 8x² + 9y² + 24xy + 6x + 6y + 1 = 0
3) 8x² + 9y² + 24xy + 6x + 8y + 1 = 0
4) 8x² + 9y² + 24xy + 6x + 8y + 1 = 0
Answer:
1) 4x² + 11y² + 24xy + 6x + 8y + 1 = 0

Explanation:
Let P(x₁, y₁) be a point in the hyperbola. The distance from P to origin is \(\sqrt{x_1^2+y_1^2}\).
The distance from P to the d.irectrix is \(\left|\frac{3 x_1+4 y_1+1}{5}\right|\)
∴ \(\sqrt{x_1^2+y_1^2}=\sqrt{5}\left|\frac{3 x_1+4 y_1+1}{5}\right|\)
⇒ 5(x₁² + y₁²) = (3x₁ + 4y₁ + 1)
⇒ 5x₁² + 5y₁² = 9x₁² + 16y₁ + 1 + 24x₁y₁ + 6x₁ + 8y₁
⇒ 4x₁² + 24x1y1 + 11y₁² + 6x₁ + 8y₁ + 1—O
∴ The equation of the hyperbola is 4x² + 11y² + 24xy + 6x + 8y + 1 – 0

Question 3.
The vertices of the hyperbola are (2, 0), (-2, 0) and the foci are (3, 0), (-3, 0) the equation of the hyperbola is
1) \(\frac{x^2}{5}-\frac{y^2}{4}\) = 1
2) \(\frac{x^2}{4}-\frac{y^2}{5}\) = 1
3) \(\frac{x^2}{5}-\frac{y^2}{2}\) = 1
4) \(\frac{x^2}{2}-\frac{y^2}{5}\) = 1
Answer:
2) \(\frac{x^2}{4}-\frac{y^2}{5}\) = 1

Explanation:
Vertices of the hyperbola = (±2, 0) ⇒ a = 3
Foci of the hyperbola = (±3, 0) ⇒ ae – 3 ⇒ 2e = 3 ⇒ e = \(\frac{3}{2}\)
b² – a²(e² – 1)
⇒ b² – 4(\(\frac{9}{4}\) – 1)
⇒ b² = 5.
Equation of the hyperbola is \(\frac{x^2}{4}-\frac{y^2}{5}\) = 1

Question 4.
The foci of the hyperbola 9y² – 4x² = 36 are the points ( 1 )
1) (0, ±√13)
2) (±3, 0)
3) (0, ±3)
4) (±13, 0)
Answer:
1) (0, ±√3)

Explanation:
Given hyperbola is 9y² – 4x² = 36
Eccentricity, e = \(\sqrt{\frac{a^2+b^2}{a^2}}=\sqrt{\frac{9+4}{4}}=\frac{\sqrt{13}}{2}\)
Foci of the hyperbola = (0, ±be) = (0, ±2 × \(\frac{\sqrt{13}}{2}\)) = (0, ±\(\sqrt{13}\))

Question 5.
The distance between the foci of the hyperbola x² – 3y² – 4x – 6y – 11 = o is
1) 4
2) 6
3) 8
4) 10
Answer:
3) 8

Explanation:
x² – 3y² – 4x – 6y – 11 = o
⇒ (x² – 4x + 4) – 3(y² + 2y + 1) = 12
⇒ (x – 2)² – 3(y + 1)² = 12
⇒ \(\frac{(x-2)^2}{12}-\frac{(y+1)^2}{4}\) = 1
⇒ a² = 12, b² = 4
e = \(\sqrt{\frac{a^2+b^2}{a^2}}=\frac{\sqrt{12+4}}{\sqrt{12}}=\frac{4}{2 \sqrt{3}}=\frac{2}{\sqrt{3}}\)
Distance between foci = 2ae = 2\(\sqrt{12}\left(\frac{2}{\sqrt{3}}\right)\) = 8.

Question 6.
The equation \(\frac{1}{2}\) = 1 represents a hyperbola if c ( 1 )
1) lies between 5 and 9
2) does not lie between 5 and 9
3) lies between 0 and 5
4) lies between 0 and 9
Answer:
1) lies between 5 and 9

Explanation:
\(\frac{x^2}{9-c}+\frac{y^2}{5-c}\) = 1 represents a hyperbola if
9 – c > 0 ⇒ 9 > c; 5 – c < 0 ⇒ 5 < c
∴ 5 < c < 9.

Question 7.
If e and e’ are the eccentricities of the ellipse 5x² + 9y² = 45 and the hyperbola 5x² – 4y² = 45 respectively then ee’
1) 9
2) 5
3) 4
4) 1
Answer:
4) 1

Explanation:
5x² + 9y² = 45
⇒ \(\frac{x^2}{9}+\frac{y^2}{5}\) = 1
⇒ e = \(\frac{\sqrt{9-5}}{3}=\frac{2}{3}\)

5x² – 4y² = 45
⇒ \(\frac{x^2}{9}-\frac{y^2}{\frac{45}{4}}\) = 1
e’ = \(\frac{\sqrt{9+\frac{45}{4}}}{3}=\frac{9}{6}=\frac{3}{2}\)
∴ ee’ = \(\frac{2}{3} \times \frac{3}{2}\) = 1

Question 8.
If the foci of the ellipse \(\frac{1}{2}\) = 1 and the hyperbola \(\frac{1}{2}\) = 1 coincide then b² =
1) 4
2) 5
3) 8
4) 9
Answer:
2) 5

Explanation:
Eccentricity of the ellipse \(\frac{x^2}{25}+\frac{y^2}{16}\) = 1 is \(\sqrt{\frac{a^2-b^2}{a^2}}=\sqrt{\frac{25-16}{25}}=\frac{3}{5}\). Foci are (3, 0), (-3, 0).
Eccentricity of the hyperbola \(\frac{x^2}{4}-\frac{y^2}{b^2}\) = 1 is . Foci are (\(\sqrt{\frac{4+b^2}{4}}\), 0). (-\(\sqrt{\frac{4+b^2}{4}}\), 0).
Given that S = \(\sqrt{4+b^2}\)
⇒ 9 = 4 + b²
⇒ b² = 5.

Question 9.
For the hyperbola \(\frac{x^2}{\cos ^2 \alpha}+\frac{y^2}{\sin ^2 \alpha}\) = 1 which of the following remains constant when a varies?
1) Eccentricity
2) Directrix
3) Abscissae of vertices
4) Abscissae of foci
Answer:
4) Abscissae of foci

Explanation:
\(\frac{x^2}{\cos ^2 \alpha}+\frac{y^2}{\sin ^2 \alpha}\) = 1. We have sin²α = cos²α (e² – 1)
tan²α = e² – 1
⇒ e² = sec²α
⇒ e = sec α ……….(1)
Equation of the directrix is x = a/e.
⇒ x = \(\frac{\cos \alpha}{\sec \alpha}\)
⇒ x = cos²α ……….(2)
Vertices = (± cos α, 0) …………. (3)
Foci = (±ae, 0) = (± cos α sec α, 0) = (±1, 0) …………(4)
Abscissae of foci are constant.

Referring to the AP Inter 1st Year Maths Study Material Chapter 9 Straight Lines Exercise 9d Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Straight Lines Solutions Exercise 9d

I.

Question 1.
Find the values of k for which the line (k – 3)x – (4 – k²)y + k² – 7k + 6 = 0 is
a) Parallel to X-axis,
b) Parallel to Y-axis.
c) Passing through the origin.
Solution:
The given equation of the line is (k – 3) x – (4 – k²) y + k² – 7k + 6 = 0 …………. (1)
a) If the given line is parallel to the X – axis then,
Slope of the given line = Slope of the X – axis
Then given line can be written as (k – 3) x + k² – 7k + 6 (4 – k²)y
y = \(\frac{(k-3)}{\left(4-k^2\right)} x+\frac{k^2-7 k+6}{\left(4-k^2\right)}\) Which is of the form y = mx + c
Slope of the given line = \(\frac{1}{2}\), Slope of the X-axis = 0
⇒ \(\frac{(k-3)}{\left(4-k^2\right)}\) = 0
⇒ k – 3 = 0
⇒ k = 3
Thus, the given line is parallel to X-axis, then the value of k = 3.

b) If the given line is parallel to the Y-axis, it is vertical. Hence, its slope will be undefined.
The slope of the given line is = \(\frac{(k-3)}{\left(4-k^2\right)}\)
Now, \(\frac{(k-3)}{\left(4-k^2\right)}\) is undefined at k = 4 ⇒ k = ±2
Thus, if the given line is parallel to the Y-axis, then the value of k = ± 2.

c) If the given line is passing through the origin, then point (0, 0) satisfies the given equation of the line.
(k – 3) (0) – (4 – k²) (0) + k² – 7k = 6 = 0
⇒ k² – 7k + 6 = 0
⇒ k² – 6k – k + 6 = 0
⇒ (k – 6)(k – 1) = 0
⇒ k = 6 or k = 1
Thus, if the given line is passing through the origin, then the value of k is either 1 or 6.

Question 2.
Find the equations of the line, which cut – off intercepts on the axes whose sum and product are 1 and -6 respectively.
Solution:
Let the intercepts cut by the given lines on the axes be a and b. It is given that
a + b = 1 …….. (1); ab = -6 ….(2)
On solving equation (1) and (2), we obtain a = 3 and b = -2 or a = -2 and b = 3 It is known that the equation of the line whose intercepts on the axes are a and b is x y
\(\frac{x}{a}+\frac{y}{b}\) = 1 ⇒ bx + ay – ab = 0
Case I : a = 3 and b = -2
In this case, the equation of the line is -2x + 3y + 6 = 0 ⇒ 2x – 3y = 6

Case II : a = -2 and b = 3
In this case, the equation of the line is 3x – 2y + 6 = 0 ⇒ -3x + 2y = 6
Thus, the required equations of the lines are 2x – 3y = 6 and -3x + 2y = 6.

Question 3.
What are the points on the Y-axis whose distance from the line \(\frac{1}{2}\) = 1 is 4 units.
Solution:
Let (0, b) be the point on Y-axis whose distance from line \(\frac{1}{2}\) = 1 is 4 units.
The given line can be written as 4x + 3y – 12 = 0 ……… (1)
On comparing equation (1) to the general equation of line Ax + By + C = 0, we obtain A = 4, B = 3 and C = -12.
It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by d = \(\frac{1}{2}\)
Therefore, if (0, b) is the point on the Y-axis whose distance from line \(\frac{1}{2}\) = 1 is 4 units. Then,
⇒ 4 = \(\frac{1}{2}\)
⇒ 4 = \(\frac{1}{2}\)
⇒ 20 = |3b – 12|
⇒ 20 = (3b – 12)
⇒ (3b – 12) = 20 or (3b – 12) = -20
⇒ 3b = 32 ⇒ b = \(\frac{32}{3}\) or 3b = -8 ⇒ b = \(\frac{-8}{3}\)
Thus, the required points are (0, \(\frac{32}{3}\)) and (0, \(\frac{-8}{3}\))

Question 4.
Find the perpendicular distance from the origin to the line joining the points (cos θ, sin θ) and (cos Φ, sin Φ).
Solution:
The equation of the line joining the points (cos0, sin0) and (cos Φ, sin Φ) is given by
y – sin θ = \(\frac{\sin \phi-\sin \theta}{\cos \phi-\cos \theta}\) (x – cos θ)
⇒ y(cos Φ – cos θ) – sin0(cos Φ – cos θ) = x (sin Φ – sin θ) – cos θ(sin Φ – sin θ)
⇒ x (sin Φ -sin θ) + y (cos Φ – cos θ) + cos θ sin Φ – cos θ sin θ – sin θ cos Φ + sin θ cos θ = 0
⇒ x(sin Φ – sin0) + y (cos Φ cos θ) + sin Φ – θ) = 0
⇒ Ax + By + C = 0, where A = sin θ – sin Φ, B = cos Φ – cos θ and C = sin (Φ – θ)
It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by d = \(\frac{\left|\mathrm{Ax}_1+\mathrm{By}_1+\mathrm{C}\right|}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}\)
Therefore, the perpendicular distance (d) of the given line from point (x₁, y₁) = (0, 0) is

But we know that sin θ = 2(cos\(\frac{\theta}{2}\).sin\(\frac{\theta}{2}\))
Then, \(\frac{\left|2 \cos \left(\frac{\phi-\theta}{2}\right) \sin \left(\frac{\phi-\theta}{2}\right)\right|}{\left|2 \sin \left(\frac{\phi-\theta}{2}\right)\right|}=\left|\cos \left(\frac{\phi-\theta}{2}\right)\right|\)

Question 5.
Find the equation of the line parallel to Y-axis and draw through the point of intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.
Solution:
The equation of any line parallel to the Y-axis is of the form x = a ………. (1)
The two given lines are x – 7y + 5 = 0 ……………(2)
3x + y = 0 ………. (3)
On solving equation (2) and (3), we obtain x = –\(\frac{5}{22}\) and y = \(\frac{15}{22}\)
Therefore, \(\left(-\frac{5}{22}, \frac{15}{22}\right)\) is the point of intersection of lines (2) and (3)
Since, line x = a passes through the point \(\left(-\frac{5}{22}, \frac{15}{22}\right)\) ⇒ a = \(-\frac{5}{22}\)
Thus, the required equation of the line is x = \(-\frac{5}{22}\)

Question 6.
Find the equation of a line drawn perpendicular to the line \(\frac{1}{2}\) = 1 through the point, where it meets the Y-axis.
Solution:
The equation of the given line is \(\frac{1}{2}\) = 1.
This equation can also be written as 3x + 2y – 12 = 0.
y = –\(\frac{3}{2}\)x + 6 which is of the form y = mx + c.
Hence, the slope of the given line is
On substituting x = 0 in the equation of the given line, we obtain \(\frac{y}{26}\) = 1 ⇒ y = 6
The given line intersects the Y-axis at (0, 6).
The equation of the line that has a slope of \(\frac{2}{3}\) and passes through the point (0, 6) is
⇒ (y – 6) = \(\frac{2}{3}\)(x – 0)
⇒ 3y – 18 = 2x
⇒ 2x – 3y + 18 = 0
Thus, the required equation of the line is 2x – 3y + 18 = 0

Question 7.
Find the length of the perpendicular drawn from the point (3, 4) to the straight line 3x – 4y + 10 = 0.
Solution:
Length of the perpendicular = \(\frac{\left|a x_1+b y_1+c\right|}{\sqrt{a^2+b^2}}=\frac{|3(3)-4(4)+10|}{\sqrt{9+16}}=\frac{3}{5}\)

Question 8.
Find the distance between the following parallel lines,
i) 3x – 4y = 12, 3x – 4y = 7
Solution:
Equations of the given lines are 3x – 4y – 12 = 0, 3x – 4y – 7 = 0
Distance between parallel lines = \(\frac{\left|c_1-c_2\right|}{\sqrt{a^2+b^2}}=\frac{|-12+7|}{\sqrt{9+16}}=\frac{5}{5}\) = 1

ii) 5x – 3y – 4 = 0, 10x – 6y – 9 = 0
Solution:
Equations of the given lines can be taken as 10x – 6y – 8 = 0, 10x – 6y – 9 = 0
Distance between parallel lines = \(\frac{\left|c_1-c_2\right|}{\sqrt{a^2+b^2}}=\frac{|-8+9|}{\sqrt{100+36}}=\frac{1}{2 \sqrt{34}}\)

Question 9.
Find the value of k, if the straight lines y – 3kx + 4 = 0 and (2k – 1) x – (8k – 1)y -6 = 0 are perpendicular.
Solution:
Equations of the given lines are -3kx + y + 4 = 0, (2k – 1) x – (8k – 1) y – 6 = 0
The lines are perpendicular.
Here ax = -3k, b1 =1, a2 = 2k – 1, b2 = -(8k – 1)
a₁a₂ + b₁b₂ = 0
⇒ -3k(2k – 1) -1 (8k -1) = 0
⇒ -6k² + 3k – 8k + 1 = 0
⇒ 6k² + 5k – 1 = 0
⇒ (k + 1) (6k – 1) = 0
⇒ k = -1 or k = \(\frac{1}{6}\)

Question 10.
(- 4, 5) is a vertex of a square and one of its diagonals is 7x – y + 8 = 0. Find the equation of the other diagonal.
Solution:
ABCD is a square.

Equation of the diagonal AC is 7x – y + 8 = 0
The diagonal BD can be taken as x + 7y + k = 0
BD passes through D (- 4, 5)
-4 + 35 + k = 0 ⇒ k = -31
Equation of BD is x + 7y – 31 = 0.

II.

Question 1.
Find the area of the triangle formed by the lines y-x = 0, x + y = 0 and x – k = 0.
Solution:
The equations of the given lines are, y – x = 0 …………(1)
x + y = 0 ……….(2)
x – k = 0 ………..(3)
The point of intersecction of lines (1) and (2) is given by x = 0 and y = 0
The point of intersection of lines (2) and (3) is given by x = k and y = – k
The point of intersecction of lines (3) and (1) is given by x = k and y = k
Thus, the vertices of the triangle formed by the three given lines are (0, 0), (k, -k) and (k, k).
We know that the area of a triangle whose vertices are (x₁, y₁), (x₂, y₂) and (x₃, y₃) is
½|x₁(y₂ – y₃)+ x₂ (y₃ – y₁) + x₃(y₁ – y₂) |
Therefore, area of the triangle formed by three given lines
= ½|0(-k – k) + k (k – 0) + k (0 + k)|
= ½|k² + k²| = ½(2k)² = k²
Hence, the area of the triangle is k2 square units.

Question 2.
Find the value of p so that the three lines 3x + y – 2 = 0, px + 2y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point.
Solution:
The equations of the given lines are 3x + y – 2 = 0 ………(1)
px + 2y – 3 = 0 ………. (2)
2x – y – 3 = 0 ……… (3)
On solving equations (1) and (3), we obtain x = 1 and y = -1
Since these three lines may intersect at one point, the point of intersection of lines (1) and (3) will also satisfy line (2)
⇒ p (1) + 2 (-1) – 3 = 0
⇒ p – 2 – 3 = 0
⇒ p = 5
Thus, the required value of p = 5.

Question 3.
If three lines whose equations are y = m₁x + c₁, y = m₂x + c₂ and y = m₃x + c₃ are concurrent, then show that m₁ (c₂ – c₃) + m₂ (c₃ – c₁) + m₃ (c₁ – c₂) = 0.
Solution:
The equations of the given lines are y = m₁x + c₁ …….. (1)
y = m₂x + c₂ ………. (2)
y = m₃x + c₃ ………. (3)

On subtracting equation (1) from (2) we obtain
(m₂ – m₁)x + (c₂ – c₁) = 0
⇒ (m₁ – m₂)x = (c₂ – c₁)
⇒ x = \(\frac{\mathrm{c}_2-\mathrm{c}_1}{\mathrm{~m}_1-\mathrm{m}_2}\)

On substituting this value of x in (1), we obtain y = m₁\(\left(\frac{\mathrm{c}_2-\mathrm{c}_1}{\mathrm{~m}_1-\mathrm{m}_2}\right)\) + c₁
= \(\frac{m_1 c_2-m_1 c_1}{m_1-m_2}\) + c₁

= \(\frac{m_1 c_2-m_1 c_1+c_1\left(m_1-m_2\right)}{m_1-m_2}\)

= \(\frac{m_1 c_2-m_1 c_1+m_1 c_1-m_2 c_1}{m_1-m_2}\)

y = \(\frac{\mathrm{m}_1 \mathrm{c}_2-\mathrm{m}_2 \mathrm{c}_1}{\mathrm{~m}_1-\mathrm{m}_2}\)

Therefore, \(\left(\frac{\mathrm{c}_2-\mathrm{c}_1}{\mathrm{~m}_1-\mathrm{m}_2}, \frac{\mathrm{~m}_1 \mathrm{c}_2-\mathrm{m}_2 \mathrm{c}_1}{\mathrm{~m}_1-\mathrm{m}_2}\right)\) is the point of intersection of the lines (1) and (2).
It is given that lines (1), (2) and (3) are concurrent.
Hence the point of intersection of lines (1) and (2) will also satisfy equation (3).
On substituting the values of x and y in (3), we obtain
⇒ \(\frac{m_1 c_2-m_2 c_1}{m_1-m_2}\) = m₃\(\left(\frac{\mathrm{c}_2-\mathrm{c}_1}{\mathrm{~m}_1-\mathrm{m}_2}\right)\) + c₃
⇒ \(\frac{m_1 c_2-m_2 c_1}{m_1-m_2}=\frac{m_3 c_2-m_3 c_1+c_3 m_1-c_3 m_2}{m_1-m_2}\)
⇒ m₁c₂ – m₂c₁ – m₃c₁ + m₃c₁ – m₁c₃ + m₂c₃ = 0
⇒ m₁c₂ – m₂c₁ – m₃c₂ + m₃c₁ – m₁c₃ + m₂c₃ = 0
⇒ m₁(c₂ – c₃) + m₂ (c₃ – c₁) + m₃ (c₁ – c₂) = 0
Hence, m₁(c₂ – c₃) + m₂ (c₃ – c₁) + m₃ (c₁ – c₂) = 0 proved.

Question 4.
Find the equation of the lines through the points (3, 2) which make an angle of 45° with the line x – 2y = 3.
Solution:
Let the slope of the required line be m₁.
The given line can be represented as y = \(\frac{1}{2}\)x – \(\frac{3}{2}\), which is of the form y = mx + c
Slope of the given line = m₂ = ½
It is given that the angle between the required line and line x – 2y = 3 is 45°.
We know that if θ is the acute angle between lines l₁ and l₂ with the slopes m₁ and m₂ respectively.
Then, tan θ = \(=\left|\frac{\mathrm{m}_2-\mathrm{m}_1}{1+\mathrm{m}_1 \mathrm{~m}_2}\right|\)

Case I: m₁ = 3
The equation of the line passing through (3, 2) and having a slope of 3 is :
y – 2 = 3(x – 3)
⇒ y – 2 = 3x – 9
⇒ 3x – y = 7

Case II: m₁ = –\(\frac{1}{3}\)
The equation of the line passing through (3, 2) and having a slope of –\(\frac{1}{3}\) is :
y – 2 = –\(\frac{1}{3}\)(x – 3)
⇒ 3y – 6 = -x + 3
⇒ x + 3y = 9

Question 5.
Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.
Solution:
Let the equation of the line having equal intercepts on the axes be
\(\frac{1}{2}\) = 1
⇒ x + y = a ……….(1)
On solving equations 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0
we obtain x = \(\frac{1}{13}\) and y = \(\frac{5}{13}\)
Therefore, \(\left(\frac{1}{13}, \frac{5}{13}\right)\) is the point of intersection of the two given lines.

Since equation (1) passes through the point \(\left(\frac{1}{13}, \frac{5}{13}\right)\)
\(\frac{1}{13}+\frac{5}{13}\) = a
a = \(\frac{6}{13}\)
Equation (1) becomes x + y = \(\frac{6}{13}\)
13x + 13y = 6
Thus, the required equation of the line id 13x + 13y = 6

Question 6.
Show that the equation of the line passing through the origin and making an angle θ with the line y = mx + c, is \(\frac{y}{x}=\frac{m \pm \tan \theta}{1 \mp m \tan \theta}\)
Solution:
Let the equation of the line passing through the origin be y = m₁x
If this line makes an angle of 9 with line y = mx + c, then angle is given by

Question 7.
In what ratio, the line segment joining (-1, 1) and (5, 7) is divided by the line x + y = 4 ?
Solution:
The equation of the line joining the points (-1, 1) and (5, 7) is
y – 1 = \(\frac{7-1}{5+1}\)(x + 1)
⇒ y – 1 = \(\frac{6}{6}\)(x + 1)
⇒ x – y + 2 = 0 …………….(1)
The equation of the given lines is x + y = 4 ………….(2)
The point of intersection of line (1) and (2) is x = 1 and y = 3
Let point (1, 3) divides the line segment joining (-1, 1) (5, 7) in the ratio 1 : k
Accordingly, by section formula (1, 3) = \(\left(\frac{k(-1)+1(5)}{1+k}, \frac{k(1)+1(7)}{1+k}\right)\)
⇒ (1, 3) = \(\left(\frac{-k+5}{1+k}, \frac{k+7}{1+k}\right)\)
⇒ \(\frac{-k+5}{1+k}\) = 1, \(\frac{k+7}{1+k}\) = 3
\(\frac{-k+5}{1+k}\) = 1 ⇒ -k + 5 = 1 + k
⇒ 2k = 4 ⇒ k = 2
Thus, the line joining the points (-1, 1) and (5, 7) is divided by line x + y = 4 in the ratio 1:2.

Question 8.
Find the equations of the straight lines passing through (1, 3) and (i) parallel to (ii) perpendicular to the line passing through the points (3, -5) and (-6, 1).
Solution:
A (3, -5), B(-6, 1) are the given points. mmn
Slope of the AB = \(\frac{y_1-y_2}{x_1-x_2}=\frac{-5-1}{3+6}=\frac{-6}{9}=\frac{-2}{3}\)

i) The required line is parallel to AB and passes through (1, 3).
Equation of the required line is
y – 3 = \(\frac{-2}{3}\)(x – 1)
⇒ 3y – 9 = -2x + 2
⇒ 2x + 3y – 11 = 0

ii) The required line is perpendicular to AB. Equation of required line can be taken as 3x – 2y + k = 0
This line passes through A(1, 3) ⇒ 3 – 6 + k = 0 ⇒ k = 6 – 3 = 3
∴ The required line is 3x – 2y + 3 = 0

Question 9.
Find the equation of the line perpendicular to the line 3x + 4y + 6 = 0 and making an intercept -4 on X -axis.
Solution:
Equation of the given line is 3x + 4y + 6 = 0
Equation of the line perpendicular to the given line is
4x – 3y = k ⇒ \(\frac{4 x}{k}-\frac{3 y}{k}\) = 1 ⇒ \(\frac{1}{2}\) + y = 1; x- intercept = \(\frac{k}{4}\) = -4 ⇒ k = -16
Equation of the required line is 4x – 3y = -16 ⇒ 4x – 3y + 16 = 0

Question 10.
A (-1, 1), B (5, 3) are opposite vertices of a square in the XY-plane. Find the equation of the other diagonal (not passing through A, B) of the square.
Solution:
Let ABCD be a square. A (-1, 1), B (5, 3) are opposite vertices of the square.
Slope of AB = \(\frac{1-3}{-1-5}=\frac{-2}{-6}=\frac{1}{3}\)

The diagonal CD is perpendicular to AB
Slope of CD = –\(\frac{1}{m}\) = -3
‘O’ is the mid point of the diagonals.
Coordinates of O are \(\left(\frac{-1+5}{2}, \frac{1+3}{2}\right)\) = (2, 2)
CD passes through O(2, 2)
Equation of CD is y – 2 = -3(x – 2)
⇒ y – 2 = -3x + 6
⇒ 3x + y – 8 = 0

Question 11.
Find the foot of the perpendicular drawn from (3, 0) upon the straight line 5x + 12y – 41 = 0.
Solution:
Equation of the line is 5x + 12y – 41 = 0, (x₁, y₁) = (3, 0)
If(x₂, y₂) is the foot of the perpendicular from (x₁, y₁) on the line ax + by + c = O, then
\(\frac{x_2-x_1}{a}=\frac{y_2-y_1}{b}=\frac{-\left(a x_1+b y_1+c\right)}{a^2+b^2}\)

⇒ \(\frac{x_2-3}{5}=\frac{y_2-0}{12}=-\frac{(15+0-41)}{25+144}\)

⇒ \(\frac{x_2-3}{5}=\frac{y_2}{12}=\frac{26}{169}=\frac{2}{13}\)

⇒ x₂ – 3 = \(\frac{10}{13}\)

x₂ = 3 + \(\frac{10}{13}=\frac{39+10}{13}=\frac{49}{13}\) and y₂ = \(\frac{1}{2}\)

∴ Foot of the perpendicular = (x₂, y₂) = \(\left(\frac{49}{13}, \frac{24}{13}\right)\)

Question 12.
Show that the distance of the point (6, -2) from the line 4x + 3y – 12 is half the distance of the poInt (3, 4) from the line 4x – 3y = 12.
Solution:
Equation of AB is 4x + 3y – 12 = 0.
PQ Length of the perpendicular from P = \(\frac{|24-6-12|}{\sqrt{16+9}}=\frac{6}{5}\)

Equation of CD is 4x – 3y – 12 = 0.
RS Length of the perpendicular from R = \(\frac{|12-12-12|}{\sqrt{16+9}}=\frac{12}{5}\)
∴ PQ = \(\frac{1}{2}\)RS

III.

Question 1.
Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2x – y = 0.
Solution:
The given lines are 2x – y = 0 ……….. (1); 4x + 7y + 5 = 0 ……….. (2)
Let A (1, 2) is a point on the line (1) and B be the point of intersection of lines (1) and (2).
On solving equations (1) and (2), we obtain x = \(-\frac{5}{18}\) and y = \(-\frac{5}{9}\)
Coordinates of point B are \(\left(-\frac{5}{18},-\frac{5}{9}\right)\)
By using distance formula, the distance between points A and B can be obtained as
AB = \(\sqrt{\left(1+\frac{5}{18}\right)^2+\left(2+\frac{5}{9}\right)^2}=\sqrt{\left(\frac{23}{18}\right)^2+\left(\frac{23}{9}\right)^2}=\sqrt{\left(\frac{23}{9 \times 2}\right)^2+\left(\frac{23}{9}\right)^2}\)

= \(\sqrt{\left(\frac{23}{9}\right)^2\left(\frac{1}{2}\right)^2+\left(\frac{23}{9}\right)^2}=\sqrt{\left(\frac{23}{9}\right)^2\left(\frac{1}{4}+1\right)}=\frac{23}{9} \sqrt{\left(\frac{5}{4}\right)}\)

= \(\frac{23}{9} \times \frac{\sqrt{5}}{2}=\frac{23 \sqrt{5}}{18}\)
Thus, the required distance is \(\frac{23 \sqrt{5}}{18}\) units.

Question 2.
Find the direction in which a straight line must be drawn through the point (-1, 2) so that its point of intersection with the line x + y = 4 may be at a distance of 3 units from this point.
Solution:
Let y = mx + c be the line through the point (-1, 2).
Accordingly, 2 = m (-1) + c
⇒ 2 = -m + c
⇒ c = m + 2
⇒ y = mx + m + 2 …….. (1)
The given line is x + y = 4 ……… (2)
On solving equations (1) and (2), we obtain x = \(\frac{2-\mathrm{m}}{\mathrm{~m}+1}\) and y = \(\frac{5 m+2}{m+1}\)

Therefore, \(\left(\frac{2-\mathrm{m}}{\mathrm{~m}+1}, \frac{5 \mathrm{~m}+2}{\mathrm{~m}+1}\right)\) is the point of intersection of lines (1) and (2)
Since this point is at a distance of 3.units from point (-1, 2), accordingly to distance formula,
⇒ \(\sqrt{\left(\frac{2-m}{m+1}+1\right)^2+\left(\frac{5 m+2}{m+1}-2\right)^2}\) = 3

⇒ \(\left(\frac{2-m+m+1}{m+1}\right)^2+\left(\frac{5 m+2-2 m-2}{m+1}\right)^2\) = 3

⇒ \(\frac{9}{(\mathrm{~m}+1)^2}+\frac{9 \mathrm{~m}^2}{(\mathrm{~m}+1)^2}\) = 9

⇒ \(\frac{1+m^2}{(m+1)^2}\) = 1
⇒ 1 + m² = m² + 1 + 2m
⇒ 2m = 0
⇒ m = 0
Thus, the slope of the required line must be zero i.e., it must be parallel to the X-axis.

Question 3.
The hypotenuse of a right – angled triangle has its ends at the points (1,3) and (-4, 1). Find the equation of the legs (perpendicular sides ) of the triangle, which are parallel to the axes.
Solution:
Let ABC be the right – angled triangle, where C = 90°. Let the slope of AC = m
Hence, the slope of BC = –\(\frac{1}{m}\)
Equation of AC : y – 3 = m (x – 1) ⇒ (x – 1) = \(\frac{\mathrm{y}-3}{\mathrm{~m}}\)
Equation of BC : x + 4 = -m(y – 1)
For a given value of m, we can get these equations
For, m = 0, y – 3 = 0; x + 4 = 0,
For m = ∞, x – 1 = 0; y – 1 = 0

Question 4.
Find the image of the given points with respect to the line
(i) point (3, 8) w.r.t line x + 3y = 7
(ii) Point (1,2) w.r.t line 3x + 4y – 1 = 0
Solution:
(i) The equation of the given line is x + 3y = 7 …………(1)
Let point B (a, b) be the image of point A (3, 8).
Accordingly, line (1) is the perpendicular bisector of AB.
Slope of AB = \(\frac{b-8}{a-3}\), while the slope of the line (1) is –\(\frac{1}{3}\)

Since line (1) is perpendicular to AB
⇒ \(\left(\frac{b-8}{a-3}\right)\left(-\frac{1}{3}\right)\) = -1
⇒ \(\frac{b-8}{3 a-9}\) = 1
⇒ b – 8 = 3a – 9
⇒ 3a – b – 1 = 0
⇒ 3a – b = 1 ………….. (2)
Mid point of AB will also satisfy line (1).
Hence, from equation (1), we have
⇒ \(\frac{1}{2}\) + 3\(\frac{1}{2}\) = 7
⇒ a + 3 + 3b + 24 = 14
⇒ a + 3b = -13 ………… (3)

On solving equations (2) and (3), we obtain a = -1 and b = -4
Thus, the image of the given point with respect to the given line is (-1, -4).

ii) Let (x₁, y₁) = (1, 2) and line in 3x + 4y – 1 = 0
Image (h, k) of the point (x₂, y₂) w.r.t line ax + by + c = 0
\(\frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{-2\left(a x_1+b y_1+c\right)}{a^2+b^2}\)

⇒ \(\frac{h-1}{3}=\frac{k-2}{4}=\frac{-2(3+8-1)}{9+16}=\frac{-20}{25}=\frac{-4}{5}\)

⇒ \(\frac{h-1}{3}=\frac{-4}{5}\)

⇒ h = \(\frac{-7}{5} ; \frac{k-2}{4}=\frac{-4}{5}\)

⇒ k = \(=\frac{-6}{5}\)
∴ Image = \(\left(-\frac{7}{5}, \frac{-6}{5}\right)\)

Question 5.
If the lines y = 3x + 1 and 2y = x + 3 are equally indicated to the line y = mx + 4, find the value of m.
Solution:
The equations of the given lines are,
y = 3x + 1 ……….. (1), 2y = x + 3 …………. (2), y = mx + 4 ……….. (3)
Slope of line (1), m₁ = 3;
Slope of line (2), m₂ = ½;
Slope of line (3), m₃ = m
It is given that lines (1) and (2) are equally inclined to line (3). This means that the given angle between lines (1) and (3) equals the angle between lines (2) and (3).
∴ \(\left|\frac{m_1-m_3}{1+m_1 m_3}\right|=\left|\frac{m_2-m_3}{1+m_2 m_3}\right|\)

⇒ \(\left|\frac{3-\mathrm{m}}{1+3 \mathrm{~m}}\right|=\left|\frac{\frac{1}{2}-\mathrm{m}}{1+\frac{1}{2} \mathrm{~m}}\right|\)

⇒ \(\left|\frac{3-\mathrm{m}}{1+3 \mathrm{~m}}\right|=\left|\frac{1-2 \mathrm{~m}}{\mathrm{~m}+2}\right|\)

⇒ \(\frac{3-m}{1+3 m}= \pm\left(\frac{1-2 m}{m+2}\right)\)

Case I :
If \(\frac{3-\mathrm{m}}{1+3 \mathrm{~m}}=\frac{1-2 \mathrm{~m}}{\mathrm{~m}+2}\)
⇒ (3 – m) (m + 2) = (1 – 2m) (1 + 3m)
⇒ -m² + m + 6 = 1 + m – 6m²
⇒ 5m² + 5 = 0
⇒ m² + 1 = 0
⇒ m² = -1
Here, m = √-1, which is not real. Hence, this case is not possible.

Case II :
If \(\frac{3-\mathrm{m}}{1+3 \mathrm{~m}}=-\left(\frac{1-2 \mathrm{~m}}{\mathrm{~m}+2}\right)\)
⇒ (3 – m)(m + 2) = -(1 – 2m)(1 + 3m)
⇒ -m² + m + 6 = – (1 + m – 6m²)
⇒ 7m² – 2m – 7 = 0
⇒ m = \(\frac{-2 \pm \sqrt{4-4(7)(-7)}}{2(7)}\)

⇒ m = \(\frac{-2 \pm 2 \sqrt{1+49}}{14}\)

⇒ m = \(\frac{1 \pm 5 \sqrt{2}}{7}\)
Thus, the required value of m = \(\frac{1 \pm 5 \sqrt{2}}{7}\)

Question 6.
If sum of the perpendicular distances of a variable point P (z, y) from the lines x + y – 5 = 0 and 3x – 2y + 7 = 0 is always 10. Show that P must move on a line.
Solution:
The equations of the lines are x + y – 5 = 0…………… (1)
3x – 2y + 7 = 0 …………. (2)
The perpendicular distance of P (x, y) from lines (1) and (2) are respectively given by
d₁ = \(\frac{|x+y-5|}{\sqrt{(1)^2+(1)^2}}\) and d₂ = \(\frac{|3 x-2 y+7|}{\sqrt{(3)^2+(-2)^2}}\)
⇒ d₁ = \(\frac{|x+y-5|}{\sqrt{2}}\) and d₂ = \(\frac{|3 x-2 y+7|}{\sqrt{13}}\)
It is given that d₁ + d₂ = 10
Therefore, \(\frac{|x+y-5|}{\sqrt{2}}+\frac{|3 x-2 y+7|}{\sqrt{13}}\) = 10
⇒ √13|x + y – 5| + √2|3x – 2y + 7| – 10\(\sqrt{26}\) = 0
Assuming x + y – 5 and 3x – 2y + 7 are positive.
⇒ √13(x + y – 5) + √2(3x – 2y + 7) – 10\(\sqrt{26}\) = 0
⇒ √13x + √13y – 5√13 + 3√2x – 2√2y + 7√2 – 10√26= 0
⇒ (√15 + 3√2)x + (√13 – 2√2)y + (7√2 -5√13 – 10√26) = 0
Since, (√13 + 3√2)x + (√13 – 2√2)y + (7√2 – 5√13 – 10√26) = 0 is the equation of line.
Similarly, we can obtain the equation of line for any signs of x + y- 5 = 0 and 3x – 2y + 7 = 0.
Thus, point P must move on a line.

Question 7.
Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.
Solution:
The equations of the given lines are 9x + 6y – 7 = 0 …….. (1)
3x + 2y + 6 = 0 ………. (2)
Let P(h, k) be the arbitrary point, which is equidistant from lines (1) and (2). Then the perpendicular distance of P(h, k) from the line (1) is given by
d₁ = \(\frac{|9 \mathrm{~h}+6 \mathrm{k}-7|}{\sqrt{(9)^2+(6)^2}}=\frac{|9 \mathrm{~h}+6 \mathrm{k}-7|}{\sqrt{117}}=\frac{|9 \mathrm{~h}+6 \mathrm{k}-7|}{3 \sqrt{13}}\)
And the perpendicular distance of P(h, k) from line (2) is given by
d₂ = \(\frac{|3 \mathrm{~h}+2 \mathrm{k}+6|}{\sqrt{(3)^2+(2)^2}}=\frac{|3 \mathrm{~h}+2 \mathrm{k}+6|}{\sqrt{13}}\)

Since P(h, k) is equidistant from lines (1) and (2)
⇒ d₁ = d₂
\(\frac{1}{2}\) ⇒ |9h + 6k – 7| = 3|3h + 2k + 6|
⇒ 9h + 6k – 7 = ± 3(3h + 2k + 6)

Case I : 9h + 6k – 7 = 3(3h + 2k + 6)
⇒ 9h + 6k – 7 = 9h + 6k + 18
⇒ 25 = 0
Which is an absurd, hence this case is not possible.

Case II : 9h + 6k – 7 = -3 (3h + 2k + 6)
⇒ 9h + 6k – 7 = -9h – 6k – 18
⇒ 18h + 12k + 11 = 0
Thus, the required equation of the line is 18x + 12y + 11 = 0

Question 8.
A ray of light passing through the point (1, 2) reflects on the X-axis at point A and the reflected ray passes through the point (5, 3). FInd the coordinates of A.
Solution:
Let the coordinates of point A be (a, 0).
Draw a line, AL perpendicular to the X-axis.

We know that angle of incidence is equal to angle of reflection.
Hence, let ∠BAL = ∠CAL = Φ and ∠CAX = θ
Now, ∠OAB = 180° – (θ + 2Φ)
= 180° [θ + 2(90° – θ)]
= 180° – (θ + 180° – 2θ)
= 180° – 180° + θ = θ.
Therefore, ∠BAX = 180° – θ
Now, Slope of line AC = \(\frac{3-0}{5-a}\)
⇒ tan θ = \(\frac{3}{5-a}\) ………. (1)

Slope of line AB = \(\frac{2-0}{1-a}\)

tan(180° – θ) = \(\frac{2}{1-a}\)

-tan θ = \(\frac{2}{1-a}\)

tan θ = \(\frac{-2}{1-a}\) ….. (2)

From equations (1) and (2), we obtain
⇒ \(\frac{1}{2}\)
⇒ 3a = -3 = -10 + 2a
⇒ a = \(\frac{13}{5}\)
Thus, the coordinates of point A are (\(\frac{13}{5}\), 0)

Question 9.
Prove that the product of the lengths of the perpendiculars drawn from the points (\(\sqrt{a^2-b^2}\), 0) and (\(-\sqrt{a^2-b^2}\), 0) to the line \(\frac{x}{a}\) cos θ + \(\frac{y}{b}\) sin θ = 1 is b².
Solution:
The equation of the given line is \(\frac{x}{a}\) cos θ + \(\frac{y}{b}\) sin θ = 1
bx cos θ + ay sin θ – ab = 0 …………… (1)

Length of the perpendicular from point (\(\sqrt{a^2-b^2}\), 0) to the line (1) is
p₁ = \(\frac{\left|b \cos \theta\left(\sqrt{a^2-b^2}\right)+a \sin \theta(0)-a b\right|}{\sqrt{b^2 \cos ^2 \theta+a^2 \sin ^2 \theta}}=\frac{\left|b \cos \theta\left(\sqrt{a^2-b^2}\right)-a b\right|}{\sqrt{b^2 \cos ^2 \theta+a^2 \sin ^2 \theta}}\) ……………… (2)

Length of the perpendicular from point (\(-\sqrt{a^2-b^2}\), 0) to the line (2) is
p₂ = \(\frac{\left|b \cos \theta\left(-\sqrt{a^2-b^2}\right)+a \sin \theta(0)-a b\right|}{\sqrt{b^2 \cos ^2 \theta+a^2 \sin ^2 \theta}}=\frac{\left|b \cos \theta\left(-\sqrt{a^2-b^2}\right)-a b\right|}{\sqrt{b^2 \cos ^2 \theta+a^2 \sin ^2 \theta}}\)

On multiplying equation (2) and (3), we obtain

Question 10.
A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.
Solution:
The equations of the given lines are 2x – 3y + 4 = 0 ………… (1)
3x + 4y – 5 = 0 …………….. (2)
6x – 7y + 8 = 0 …………. (3)
The person is standing at the junction of the paths represented by lines (1) and (2).
On solving equations (1) and (2), we obtain x = –\(\frac{1}{17}\) and y = \(\frac{22}{17}\).

Thus, the person is standing at point \(\left(-\frac{1}{17}, \frac{22}{17}\right)\).

The person can reach path (3) in the least time if he walks along the perpendicular line to (3) from point \(\left(-\frac{1}{17}, \frac{22}{17}\right)\)

Now, slope of the line (3) = \(\frac{6}{7}\)
Slope of the line perpendicular to line (3) = \(-\left(\frac{1}{\frac{6}{7}}\right)=-\frac{7}{6}\) and having a slope of \(\frac{-7}{6}\) is given by
(y – \(\frac{22}{17}\)) = –\(\frac{7}{6}\) (x + \(\frac{1}{17}\))
⇒ 6(17y – 22) = -7(17x + 1)
⇒ 102y – 132 = -119x – 7
⇒ 119x + 102y = 125
Hence, the path that the person should follow us 119x + 102y = 125.

Question 11.
Show that the lines x – 7y – 22 = 0, 3x + 4y + 9 = 0 and 7x + y – 54 = 0 form a right angled isosceles triangle.
Solution:
The equations of the straight lines are
x – 7y – 22 = 0 ………… (1)
3x + 4y + 9 = 0 ……….. (2)
7x + y – 54 = 0 ………. (3)

Let A be the angle between (1) & (2)
cos A = \(\frac{\left|a_1 a_2+b_1 b_2\right|}{\sqrt{a_1^2+b_1^2} \sqrt{a_2^2+b_2^2}}=\frac{|3-28|}{\sqrt{1+49} \sqrt{9+16}}\)

= \(\frac{25}{5 \sqrt{2} .5}=\frac{25}{25 \sqrt{2}}=\frac{1}{\sqrt{2}}\)
⇒ A = 45°

Let B be the angle between (2) & (3)
cos B = \(\frac{21+4}{\sqrt{9+16} \sqrt{49+1}}=\frac{25}{25 \sqrt{2}}=\frac{1}{\sqrt{2}}\)
⇒ B = 45°

Let C be the angle between (3) & (1)
cos C = \(\frac{7-7}{\sqrt{1+49} \sqrt{49+1}}\) = 0
⇒ C = 90°
Since ∠A = ∠B = 45° and ∠C = 90°
Therefore, given lines form a right angled isosceles triangle.

Question 12.
Find the equation of the straight line passing through the point (-3, 2) and making an angle of 45° with the straight line 3x – y + 4 = 0.
Solution:
Let the slope of the required line be m₁.
Given line is 3x – y + 4 = 0, slope m₂ = 3.
If 0 is the angle between them ⇒ tan θ = \(\left|\frac{\mathrm{m}_1-\mathrm{m}_2}{1+\mathrm{m}_1 \mathrm{~m}_2}\right|\)
⇒ tan 45 = \(\left|\frac{m_1-3}{1+3 m_1}\right|\)
⇒ 1 = \(\frac{m_1-3}{1+3 m_1}\)
⇒ ± 1 = \(\frac{m_1-3}{1+3 m_1}\)
⇒ 1 + 3m₁ = m₁ – 3
⇒ 2m₁ = -4
m₁ = -2 (or)
⇒ -(1 + 3m₁) = m₁ – 3
⇒ -1 – 3m₁ = m₁ – 3
⇒ 2 = 4m₁
⇒ m₁ = ½
The equation of line passing through (-3, 2) with slope m₁ = -2 is
y – y₁ = m(x – x₁)
⇒ y – 2 = -2(x + 3)
⇒ y – 2 = -2x – 6
⇒ 2x + y + 4 = 0
The equation of line passing through (-3, 2) with slope m = -½ is
y – 2 = \(\frac{1}{2}\) (x + 3)
⇒ 2y – 4 = x+ 3
⇒ x – 2y + 7 = 0

For systematic learning, students can include AP Inter 1st Year English Grammar with Answers and Reading Comprehension in English Inter 1st Year in their daily study routine.

AP Inter 1st Year English Reading Comprehension with Answers

Comprehension is a very important skill. It refers to the process of understanding the information presented in a written or oral text. In the Intermediate Public Examination (English) the students’ ability to understand a printed text is tested. This is, therefore, classified as reading comprehension’. Understanding a spoken text is known as listening comprehension’.

Comprehension needs a careful study of the given text. Apart from the explicit meaning there may be implicit mean¬ings and inferences.

In the Public examination, questions of various types follow the given text. Understanding the questions properly is as or rather more, important as understanding the text. Reading the questions first, therefore, will be of some help in improving one’s comprehension skills.

Careful observation of the way the given language is used and regular practice will ensure mastering this all important skill – comprehension.

Examples

Read the following passages and answer the questions that follow.

1. THE TALE OF FARMER

A farmer’s horse ran away, and his neighbors said, “What bad luck!”

The farmer replied, “Maybe.”

The next day, the horse returned with two wild horses. The neighbors exclaimed, “What good luck!”

The farmer replied, “Maybe.”

Later, the farmer’s son tried to ride one of the wild horses but fell and broke his leg. The neighbors lamented, “What bad luck!”

The farmer replied, “Maybe.”

Soon after, soldiers came to recruit young men for war, but they left the farmer’s son because of his broken leg. The neighbors celebrated, “What good luck !”

The farmer simply said, “Maybe.”

This parable shows that life is unpredictable and that it’s wise to reserve judgment. What seems like bad luck may turn into good fortune, and vice versa.

Questions:

Question 1.
What does the parable show ?
Answer:
The parable shows that life is unpredictable and that it’s wise to reserve judgment.

Question 2.
What is a synonym for the word ‘fortune’ ?
Answer:
‘luck’

Question 3.
What was the consequence of the farmer’s son trying to ride one of the wild horses ?
Answer:
He fell and broke his leg.

Question 4.
What part of speech is the word ‘reserve’ in the passage ?
Answer:
‘verb’

Question 5.
Why did the soldiers leave the farmer’s son when they came to recruit young men for war ?
Answer:
Because of his broken leg.

2. ECHO CHAMBER

Do you know how your social media feed often shows things that match your interests or opinions ? Did you ever wonder why that happens ? You may think it’s just showing you what you like. Correct! Algorithms are designed to show you content that you’re likely to agree with or enjoy, based on what you’ve liked or shared in the past. This can create an echo chamber, where you mostly see opinions that match yours and don’t get exposed to other points of view. Do you think that could be a problem ? It might make you think everyone agrees with you, even if they don’t. It can make it harder to understand other people’s opinions or even realize that there are other perspectives. It’s important to step outside that echo chamber by seeking out different ideas, even if you disagree with them. It helps you think critically and grow as a person. How do you think we could do that together ?

Maybe we could follow some different kinds of accounts or read news from various sources ?

That’s a great idea. Let’s explore different viewpoints so we can talk about them. It’ll make all of us better at understanding the world!

Questions:

Question 1.
What is an echo chamber ?
Answer:
An echo chamber is where you mostly see opinons that match yours.

Question 2.
Why does social media often show content that matches our interests ?
Answer:
Algorithms are designed to show you content that you’re likely to agree with or enjoy, based on what you’ve liked or shared in the past.

Question 3.
State True or palse: Understanding different viewpoints helps us understand the world better.
Answer:
True.

Question 4.
What effect can echo chambers have on our thinking ?
a) Narrow-minded
b) Broad-minded
c) Open to new ideas
Answer:
(a) Narrow-minded

Question 5.
Identify the part of speech of the word ‘various’ in the phrase ‘various sources’.
Answer:
Adjective

Try the Following

1. HERO WORSHIPPING

Hero-worshipping is part of societies. Those heroes might be film stars, political leaders, sports stars or other persons. Hero-worshippers display passionate admiration and imitation of their ‘heroes’. Hero-worshipping is a human quality that shows that a person is overawed by the personality cult. A hero becomes a demigod for him. His hero’s positive attributes become his guiding and inspiring principles and he becomes blind to the hero’s negative traits and weaknesses.

The instinct of hero worship springs from an internal inferiority complex. The result is that they are distanced from reality and live in a dream world of their own. It thus becomes the biggest stumbling block in the progress of persons indulging in it. This plain passion often lands us in a cloud of illusion.

However, with rational thinking and perseverance we can produce more heroes rather than more hero-worshippers.

Questions:

Question 1.
What do hero-worshippers typically display towards their heroes ?
Answer:
They display passionate admiration and imitation of their heroes.

Question 2.
State True or False: A hero-worshipper becomes blind to the hero’s negative traits and weaknesses.
Answer:
True.

Question 3.
According to the passage, from where does the instinct of hero worship spring ?
Answer:
It springs from an internal inferiority complex.

Question 4.
What is the ultimate result of indulging in hero-worshipping ?
Answer:
It distances people from reality, causing them to live in a dream world, and becomes a stumbling block in their progress.

Question 5.
What two qualities do we need to cultivate in order to produce more heroes rather than more hero-worshippers ?
Answer:
Rational thinking and perseverance.

2. DISABILITY AND PITY

As someone with a disability, I’d like people to know there’s a line between kindness and pity. It’s a realization I’ve come to after a lifetime of struggling with my disability. There are a few of the barriers I face every single day. I will continue to face them for the rest of my life. This is my reality. But it’s not these barriers that get under my skin. I’m a smart, educated, ambitious person, yet many people can’t see past my disability. Don’t get me wrong: I’ve never been bullied. It’s actually the opposite – I’m repeatedly pitied. At least once a day, I can see pity in the eyes or voice of someone I interact with.

The same can’t be said for a woman who insisted on praying over my head to God. She prayed that God shouid take away my disability and fix me. That situation was embarrassing and uncalled for. I didn’t ask for her pity, nor did I ask for her prayers to “fix” me. I don’t pity myself and neither should you. We all love to be shown kindness, but unless someone asks for pity, don’t give it. Pity makes us feel less than others or not enough.

Questions:

Question 1.
How does pitying make someone feel ?
Answer:
Pitying makes someone feel less than others or not enough.

Question 2.
What can people see if they look past the writer’s disability ?
Answer:
They can see that the writer is a smart, educated, and ambitious person.

Question 3.
State True or False: Disabled people are often bullied, and it is problematic for them.
Answer:
False. The writer specifically says, ‘I’ve never been bullied.’

Question 4.
According to the passage, what is more meaningful than pity ?
Answer:
Kindness is more meaningful than pity.

Question 5.
As per the passage, what do disabled people want ?
a) Pity
b) Dignity and respect
c) Charity
Answer:
(b) Dignity and respect

3. CHANGE STARTS FROM HOME

Why do we expect girls to come home early? Because we let boys roam around till late in the evening. Why do we expect girls to learn cooking? Because we never bother to teach our boys how to cook. Why do we expect our girls to be humble and gentle? Because we tell boys to be strong and bold. Discrimination starts at home. It begins in the way we raise our children, in the expectations we set, and the roles we assign. A million homes make up a city, and a thousand cities make up a nation. But as we move from home to society, the discrimination multiplies, spreading far and wide.

it wears many faces — catcalling, eve-teasing, harassment, stalking, abuse, domestic violence, and worse. The unfortunate reality is that these forms of discrimination are not just limited to one part of society; they touch lives everywhere, often in ways we don’t even notice.

The journey of change has to start with us. If we want to create a world where girls aren’t held back by outdated expectations, we need to start questioning and reshaping the norms in our own homes. It’s not just about raising girls right—it’s about raising boys to respect, appreciate, and support them as equals. Only then can we hope for a society that truly stands for fairness, respect, and equality.

Questions:

Question 1.
What role can we play to reduce discrimination, according to the passage ?
Answer:
We can reduce discrimination by questioning and reshaping the norms in our own homes, and by raising boys to respect, appreciate, and support girls as equals.

Question 2.
State True or False: Raising our girls in the right way is enough to change society.
Answer:
False. The passage says it’ not just about raising girls right—it’s also about how we raise boys.

Question 3.
Where does discrimination often begin, as suggested in the passage ?
Answer:
Discrimination often begins at home.

Question 4.
How does the passage suggest that discrimination spreads from one place to another ?
Answer:
It spreads as we move from home to society, multiplying and affecting more people along the way.

Question 5.
What are some examples of discrimination mentioned in the passage?
Answer:
Catcalling, eve-teasing, harassment, stalking, abuse, and domestic violence.

Exercises

1. Read the following passage carefully and answer the questions that follow.

Social Media

Instagram is the worst social media network for mental health and wellbeing, according to a recent survey. It was associated with high levels of anxiety, depression, bullying, sleep quality and body image and FOMO, or the “fear of missing out.”

There were certainly some benefits associated with social networking for example, self-expression, community building and emotional support.

However, young people who spend more than two hours a day on social networking sites are more likely to report psychological distress. “Seeing friends constantly on holiday or enjoying nights out can make young people feel like they are missing out while others enjoy life,” the #Status Of Mind report states. “These feelings can promote a ‘compare and despair’ attitude.”

Social media posts can also set unrealistic expectations and create feelings of inadequacy and low self-esteem. This may explain why Instagram, where personal photos take centre stage, received the worst scores lor body image and anxiety. As one survey respondent wrote, “Instagram easily makes girls and women feel as if their bodies aren’t good enough as people add filters and edit their pictures in order for them to look ‘perfect’.” The more social networks a young adult uses, the more likely he or she is to report depression and anxiety.

Tips to reduce social media consumption :

1. Set time limits
2. Turn off notifications
3. Replace social media with hobbies like reading, playing, exercising

Questions:

Question 1.
According to the passage, what is the likely outcome for a young adult who uses multiple social networks ?
Answer:
They are more likely to report depression and anxiety.

Question 2.
List two tips to reduce social media consumption mentioned in the passage.
Answer:

1. Turn off notifications,
2. Replace the social media with hobbies like reading, playing, exercising

Question 3.
According to a recent survey, which social media network is worst for mental health and wellbeing ?
Answer:
Instagram

Question 4.
What does the term FOMO stand for ?
Answer:
Fear of Missing Out

Question 5.
State True or False: Young people who spend more than two hours a day on social networking sites are less likely to report psychological distress.
Answer:
False.

2. Read the following passage carefully and answer the questions that follow.

Atomic Habits

A habit is an act that you perform automatically by instinct. Like when you walk into a dark room, you instinctively turn on a light switch, right ? Habits are actions you don’t even have to think about, which is why you might not realize how a small daily action can have a powerful effect on your life.

We don’t typically notice tiny changes because the impact is not seen immediately. The impact may not be seen for months, or even years, but the results can be enormous. For instance, eating a plate of panipuri certainly won’t make you unhealthy; however, if you eat a plate of panipuri every week, then you’ll likely see a change in weight over the course of a year. Similarly, going to gym one day won’t immediately make you in shape. Instead, its small habits over time that compound into major results.

While many people believe that making big changes is the key to success, it’s the small things we do each day that have the biggest impact on our lives. Our small daily habits will affect the person we become in the future, so it’s critical that we adopt healthy habits to ensure we live long, successful lives. Luckily, there are many tips and tricks to help you implement new habits today. When you can make your habits an attractive, easy, and satisfying experience, you’ll be more likely to keep up with them.

Questions:

Question 1.
What makes it more likely for someone to keep up with new habits ?
Answer:
When habits are made attractive, easy, and satisfying, people are more likely to keep up with them.

Question 2.
According to many people, what is believed to be the key to success ?
Answer:
Making big changes is believed to be the key to success.

Question 3.
To live and long successful lives, what do we need to adopt ?
Answer:
We need to adopt healthy habits.

Question 4.
Why don’t we typically notice the tiny changes ?
Answer:
Because the impact is not seen immediately; it may take months or even years to become noticeable.

Question 5.
State TRUE or FALSE : Eating a plate of panipuri certainly won’t make you unhealthy; however, if you eat a plate of panipuri every week, then you’ll likely see a change in weight over the course of a year.
Answer:
TRUE.

3. Read the following passage carefully and answer the questions that follow.

Cyber Safety

More than 175,000 children go online for the first time every day – a new child every half second – UNICEF said today. Digital access exposes these children to a wealth of benefits and opportunities, but also to a host of risks and harms, including access to harmful content, sexual exploitation and abuse, cyberbullying, and misuse of their private information.

UNICEF says, the children should keep things like your full name, age, address, phone number, college name, and even small details about your family private. It’s exciting to meet new people online, but always remember: if you haven’t met them in real life, don’t share personal details, photos, or your location with them. And if anyone ever makes you feel uncomfortable, tell your parent/teacher right away. If you see a link, pop-up, or message from someone you don’t know, don’t click on it. Sometimes, these things can take you to places that can harm your device or steal your information. It’s fun to share photos, but remember: once you post something online, it’s hard to take it back. Only share things you’d be comfortable with everyone (including strangers) seeing.

Questions :

Question 1.
Whom can you tell if anyone makes you feel uncomfortable ?
Answer:
My parent or teacher

Question 2.
What is the name of the organization that cautions children to be safe online ?
Answer:
UNICEF

Question 3.
What might happen if you click on a link or message from someone you don’t know ?
Answer:
It can harm your device or steal your information.

Question 4.
According to UNICEF, how many children go online every day ?
Answer:
More than 175,000 children

Question 5.
State True or False: Once you post something online, it’s easy to take it back.
Answer:
False.

4. Read the following passage carefully and answer the questions that follow.

Social Democracy

We must not to be content with mere political democracy. We must make our political democracy a social democracy as well. Political democracy cannot last unless there lies at the base of it social democracy. What does social democracy mean ? It means a way of life which recognizes liberty, equality and fraternity as the principles of life. These principles of liberty, equality and fraternity are not to be treated as separate items in a trinity. They form a union of trinity in the sense that to divorce one from the other is to defeat the very purpose of democracy. Without equality, liberty would produce the supremacy of the few over the many. Equality without liberty would kill individual initiative. Without fraternity equality and liberty will be no deeper than coats of paint.

Questions :

Question 1.
How can political democracy last, according to the passage ?
Answer:
Political democracy can last only if it is supported by social democracy.

Question 2.
State True or False : The principles of liberty, equality, and fraternity are to be treated as separate items in a trinity.
Answer:
False.

Question 3.
What does social democracy mean ?
Answer:
Social democracy means a way of life that recognizes liberty, equality, and fraternity as the principles of life.

Question 4.
Without fraternity, what would be the outcome of liberty and equality?
Answer:
Liberty and equality would be no deeper than coats of paint.

Question 5.
What would happen if equality existed without liberty ?
Answer:
It would kill individual initiative.

5. Read the following passage carefully and answer the questions that follow.

Peer Pressure and Substance Abuse

Peer pressure is a risk factor for drug use, including alcohol use or smoking, among both children and adults. A teen is vulnerable to peer pressure When he feels that peer acceptance is important to them, or if he is sensitive to rejection. Teens may willingly choose to use drugs/alcohol to fit in and avoid rejection or look ‘cool’.

How to resist peer pressure: When feeling pushed to participate in something you don’t want, employ these tips to avoid the activity or leave the situation altogether. Such as making eye contact, and refusing to participate in a polite but firm voice. This should be enough to cause a real friend to back off. Or say you can’t participate because of responsibilities you need to attend to later or the next day. If all fail leave the situation if their pressure continues. If this group of friends continues to force you to engage in the behaviour you aren’t interested in, then they may not be the friends you want to spend time with. These types of social situations can be draining at best and dangerous at worst, leading to unwanted drinking or drug use.

Questions:

Question 1.
State True or False: If someone is forcing you to try drugs/ alcohol, they are not your friends.
Answer:
True.

Question 2.
What should one do if peer pressure continues even after saying no ?
Answer:
Leave the situation altogether.

Question 3.
Why may teens willingly choose to use drugs/alcohol ?
Answer:
To fit in, avoid rejection, or look ‘cool’.

Question 4.
Write one tip to avoid peer pressure.
Answer:
Make eye contact and refuse to participate in a polite but firm voice.

Question 5.
If you say no to drug use, who will back off ?
Answer:
A real friend will back off.

6. Read the following passage carefully and answer the questions that follow.

Physical Activity

Physical inactivity is the fourth leading cause of death worldwide. It’s also associated with chronic illness and disability.

Recent research estimates that the world could see close to half a billion new cases of major chronic diseases by 2030 if people don’t get more active.

Regular physical activity helps to prevent and manage many chronic diseases. Popular ways to be physically active include walking, running, cycling, and playing sports.

The World Health Organization (WHO) recommends that children and adolescents (5-17 years old) get an average of at least 60 minutes per day of moderate to vigorous intense physical activity.

This should incorporate vigorous aerobic activities, as well as those that strengthen muscle and bone, at least three days a week. It’s also recommended that children spend no more than two hours a day on recreational screen time.

Physical activity can also boost mental health and academic performance.

Questions:

Question 1.
What does the World Health Organization recommend for children and adolescents ?
Answer:
They should get an average of at least 60 minutes per day of moderate to vigorous intense physical activity, including vigorous aerobic activities and muscle- and bone¬strengthening activities at least three days a week.

Question 2.
What is the fourth leading cause of death worldwide ?
Answer:
Physical inactivity

Question 3.
How many hours a day should children spend on recreational screen time ?
Answer:
No more than two hours a day

Question 4.
What does recent research suggest if people don’t get more active ?
Answer:
The world could see close to half a billion new cases of major chronic diseases by 2030.

Question 5.
According to the paragraph, what is physical inactivity associated with ?
Answer:
Chronic illness and disability.

7. Read the following passage carefully and answer the questions that follow.

Unity in Diversity

Indian National Army (INA) was founded by Subhas Chandra Bose. It was an organization formed with the aim of securing India’s independence, with the spirit of unity across diverse backgrounds. This chapter in Indian history highlights the power of unity in diversity as a defence against divisive external pressures; however the British captured INA and conducted trials for the officers.

The British prosecuted three INA officers; Shah Nawaz Khan, a Muslim; Gurbaksh Singh Dhillon, a Sikh; and Prem S&hgal, a Hindu; reflecting the natural diversity of the INA itself, which embraced people from all regions and faiths. This diversity, rather than dividing public opinion, became a powerful symbol of unity. Indians across religious and regional lines rallied in solidarity with the officers, viewing their struggle as a representation of the nation’s dream for freedom and dignity. The trials sparked mass protests and reignited the collective drive toward a shared national identity, highlighting the strength of unity in the face of external challenges, reinforcing unity as an instrument for change.

Questions

Question 1.
Who were the three INA officers prosecuted by the British ?
Answer:
Shah Nawaz Khan, Gurbaksh Singh Dhillon, and Prem Sahgal.

Question 2.
What was the aim of the Indian National Army ?
Answer:
To secure India’s independence

Question 3.
Who founded the Indian National Army ?
Answer:
Subhas Chandra Bose

Question 4.
State True or False: The INA officers were from only one religious community.
Answer:
False.

Question 5.
According to the paragraph, what is the best defence against divisive external pressures ?
Answer:
Unity in diversity

8. Read the following passage carefully and answer the questions that follow.

The Stonecutter’s Wish

Once, there was a humble stonecutter who spent his days carving stones from a mountain. One day, he envied a wealthy merchant and wished to be rich. Instantly, he became a merchant, surrounded by wealth.
Soon, he noticed a king commanding respect and power. He wished to be a king. His wish was granted, and he felt powerful until the sun’s heat made him uncomfortable, “The sun is stronger than a king,” he thought, and he became the sun.

As the sun, he noticed a thick cloud blocking his rays. “The cloud is mightier than the sun,” he decided and became a cloud. But then the wind blew him apart, so he wished to be the wind.

The wind rushed mightily across the land until it came upon an immovable mountain. “The mountain is stronger than I am!” he realized, and he became a mighty, unyielding stone.

One day, a stonecutter began chipping away at him. Only then did he understand: he had always been the most powerful thing, just as he was. This parable teaches us to appreciate ourselves and recognize our unique value without envy.

Questions :

Question 1.
Why did the stonecutter wish to be the wind ?
Answer:
Because the wind blew apart the cloud, and he believed the wind was stronger than the cloud.

Question 2.
What is the moral of the parable ?
Answer:
To appreciate ourselves and recognize our unique value without envy.

Question 3.
State True or False: We have always been the most powerful thing, just as we are.
Answer:
True.

Question 4.
Why did the stonecutter want to be a king ?
Answer:
Because he saw that the king commanded respect and power.

Question 5.
What was the stonecutter’s first wish ?
Answer:
To be rich like the wealthy merchant.

9. Read the following passage carefully and answer the questions that follow.

Khejri Massacre

In 1730, the desert of Rajasthan turned red when Amrita Devi, a Bishnoi woman, bravely stood in front of the king’s soldiers to protect the sacred Khejri trees. She hugged a tree and said, “If saving this tree costs my life, it is worth it.” With those words, her head fell to the grouhd.

But her sacrifice didn’t stop there. Ten, twenty, thirty – more people stepped forward, and their heads fell too. Yet the Bishnoi villagers didn’t stop. Fifty, ninety, one hundred – they kept coming, standing strong. Men, women, children, and elders all hugged the trees, refusing to give up. One hundred and fifty, two hundred, two hundred and fifty-five – still, the soldiers kept swinging their swords.

Three hundred, three hundred and fifty-yet the Bishnois didn’t back down. Their courage was stronger than their fear. When the king heard what was happening, he rushed to the scene. Seeing the villagers’ bravery, he was moved.

The king stopped the destruction and promised to protect the Khejri trees forever. Their sacrifice inspired the modern Chipko movement.

Questions:

Question 1.
Which modern movement in India was inspired by the sacrifice of the Bishnois ?
Answer:
The Chipko movement

Question 2.
Who promised to protect the Khejri trees after witnessing the Bishnois’ sacrifice ?
Answer:
The king

Question 3.
How many Bishnois sacrificed their lives to protect the trees ?
Answer:
Three hundred and sixty-three Bishnois

Question 4.
Who was the Bishnoi woman who stood against the king’s soldiers to save the trees ?
Answer:
Amrita Devi

Question 5.
What did Amrita Devi say before sacrificing her life ?
Answer:
If saving this tree costs my life, it ik worth it.

10. Read the following passage carefully and answer the questions that follow.

Mountain Man

In the hot hills of Bihar, Dashrath Manjhi faced a terrible loss when his wife fell on a mountain path and couldn’t get to the hospital in time. Her death filled him with a strong determination. He stood before the huge mountain that blocked his village from the nearest town and decided to break it with just a hammer and chisel.

With every strike of his tools, pieces of rock fell. Neighbours laughed and called him crazy, but Dashrath kept going. “This mountain took my wife,” he said, “I won’t let it take, anyone else.”

Day after day, year after year, he worked tirelessly His hands bled, the seasons changed, years rolled on and 22 years passed. Slowly, a path appeared where the mountain once stood. The villagers, who once mocked him, began to admire his incredible effort.

In the end, Dashrath Manjhi succeeded. He carved a 110- metre-long road through the mountain, turning a 55-kilometre journey into just 15 kilometres. Known as the “Mountain Man,” he showed the world that even the toughest challenges can be overcome with determination. His story inspires people everywhere to believe in themselves and their dreams. Next time when you feel tired or have a mountain to move just remember the Mountain Man.

Questions:

Question 1.
What did people call Dashrath Manjhi when he started breaking the mountain ?
Answer:
They called him crazy.

Question 2.
What tools did Dashrath Manjhi use to break the mountain ?
Answer:
A hammer and chisel

Question 3.
How many years did Dashrath Manjhi work hard to carve a path through the mountain ?
Answer:
22 years

Question 4.
What title did Dashrath Manjhi earn for his achievement ?
Answer:
The ‘Mountain Man’

Question 5.
State True or False: Even the toughest challenges can be overcome with determination.
Answer:
True.

11. Read the following passage carefully and answer the questions that follow.

Dolphins are regarded as the friendliest creatures in the sea and stories of them helping drowning sailors have been common since Roman times. The more we learn about dolphins, the more we realize that their society is more complex than people previously imagined. They look after other dolphins when they are ill, care for pregnant mothers and protect the weakest in the community, as we do. Some scientists have suggested that dolphins have a language but it is much more probable that they communicate with each other without needing words. Could any of these mammals be more intelligent than man ? Certainly the most common argument in favour of man’s superiority over them that we can kill them more easily than they can kill us is the least satisfactory. On the contrary, the more we discover about these remarkable creatures, the less we appear superior when we destroy them.

Questions:

Question 1.
Which ancient period is referred to in the passage ?
Answer:
Roman times are referred to in the passage.

Question 2.
What do the Dolphins do as humans do ?
Answer:
Dolphins look after other dolphins when they are ill, care for pregnant mothers, and protect the weakest in their community-just like humans do.

Question 3.
Write the antonym of the word ‘superior’.
Answer:
The antonym of ‘superior’ is ‘inferior’.

Question 4.
What do some scientists suggest ?
Answer:
Some scientists suggest that dolphins have a language.

Question 5.
How are Dolphins regarded ?
Answer:
Dolphins are regarded as the friendliest creatures in the sea.

12. Read the following passage carefully and answer the questions that follow.

The Battle of Chancellorsville, one of the most famous battles of the Civil War, took place in Virginia in the spring of 1863. For months, the two armies had been staked out on opposite banks of a narrow river. The Confederate troops were led by perhaps the most revered military tactician in American history, General Robert E. Lee. The Union soldiers were led by “Fighting” Joe Hooker.

In appearance, personality, and lifestyle, these men were nearly perfect opposites. Lee, an older man in poor health with a gray beard, had a somber, measured demeanor. Hooker was a blond, strapping young man whose vanity over his appearance was but one aspect of his egotism. Whereas Lee was devout and principled, Hooker was known for his rollicking enjoyment of whiskey.

Questions:

Question 1.
Where did the civil war take place ?
Answer:
The Civil War took place in Virginia.

Question 2.
When did the battle take place ?
Answer:
The battle took place in the spring of 1863.

Question 3.
Who led the troops ?
Answer:
The Confederate troops were led by General Robert E. Lee. The Union soldiers were led by “Fighting” Joe Hooker.

Question 4.
‘Somber’ means __________.
Answer:
‘serious’ or ‘solemn’

Question 5.
What is the name of the battle ?
Answer:
The name of the battle is The Battle of Chancellorsville.

Referring to the AP Inter 1st Year Maths Study Material Chapter 9 Straight Lines Exercise 9c Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Straight Lines Solutions Exercise 9c

I.

Question 1.
Reduce the following equations into slope – intercept form and find their slopes and the y-intercepts.
i) x + 7y = 0
Solution:
The given equation is x + 7y = 0. It can be written as y = –\(\frac{1}{7}\)x + 0
Which is in slope-intercept form, where m = –\(\frac{1}{7}\) and c = 0.

ii) 6x + 3y – 5 = 0
Solution:
The given equation is 6x + 3y – 5 = 0. It can be written as y = \(\frac{1}{3}\)(-6x + 5)
y = -2x + \(\frac{5}{3}\)
Which is in slope-intercept form, where m = – 2 and c = \(\frac{5}{3}\).

iii) y = 0
Solution:
The given equation is y = 0.
It can be written as y = 0 . x + 0
Which is in slope-intercept form, where m = 0 and c = 0.

Question 2.
Reduce the following equations into intercept form and find their intercepts on the axes.
(i) 3x + 2y – 12 = 0
Solution:
The equation of a straight line in intercept form is \(\frac{x}{a}+\frac{y}{b}\) = 1.
i) The given equation is 3x + 2y – 12 = 0.
It can be written as 3x + 2y = 12 ⇒ \(\frac{3 x}{12}+\frac{2 y}{12}\) = 1 ⇒ \(\frac{x}{4}+\frac{y}{6}\) = 1.
Which is in intercept form, where x-intercept = a = 4, y-intercept = b = 6.

ii) 4x – 3y = 6
Solution:
The given equation is 4x – 3y = 6.
It can be written as \(\frac{4 x}{6}-\frac{3 y}{6}\) = 1
⇒ \(\frac{2 x}{3}-\frac{y}{2}\) = 1

⇒ \(\frac{x}{\left(\frac{3}{2}\right)}+\frac{y}{(-2)}\) = 1
Which is in intercept form, where x-intercept = a = 3/2, y-intercept = b = -2.

iii) 3y + 2 = 0
Solution:
The given equation is 3y + 2 = 0. It can be written as 3y = – 2 ⇒ \(\frac{y}{\left(\frac{-2}{3}\right)}\) = 1
Which is in intercept form, where the intercept on the Y-axis is \(\frac{-2}{3}\) and it has no intercept on the X-axis.

Question 3.
Find the distance of the point (-1, 1) from the line 12(x + 6) = 5(y – 2)
Solution:
The given equation of the line is 12(x + 6) = 5(y – 2)
⇒ 12x + 72 = 5y – 10 ⇒ 12x— 5y + 82 = 0 ………. (1)
On comparing equation (1) with Ax + By + C = O, we have A = 12, B = -5 and C = 82.
The perpendicular distance (d) of a line Ax + By + C = O from a point (x₁, y₁) is
d = \(\frac{\left|\mathrm{Ax}_1+\mathrm{By}_1+\mathrm{C}\right|}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}\)
The given point (x₁, y₁) = (-1, 1)
Therefore, the distance of point (-1, 1) from the given line
= \(\frac{|12(-1)+(-5)(1)+82|}{\sqrt{(12)^2+(-5)^2}}=\frac{|-12-5+82|}{\sqrt{169}}=\frac{|65|}{13}\) = 5 units

Question 4.
Find the points on the X-axis, whose distances from the line \(\frac{x}{3}+\frac{y}{4}\) = 1 are 4 units.
Solution:
The given equation of the line is \(\frac{x}{3}+\frac{y}{4}\) = 1 ⇒ 4x + 3y – 12 = 0 ………. (1)
On comparing equation (1) with Ax + By + C = 0, we have A = 4, B = 3 and C = -12
Let (a, 0) be the point on the X-axis whose distance from the given line is 4 units. It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is d = \(\frac{\left|\mathrm{Ax}_1+\mathrm{By}_1+\mathrm{C}\right|}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}\)
Substitute the values of A, B, C in the above equation.
Therefore, 4 = \(\frac{|4 a+3 \times 0-12|}{\sqrt{4^2+3^2}}\)
⇒ 4 = \(\frac{|4 a-12|}{5}\)
⇒ |4a – 12| = 20
⇒ ±(4a – 12) = 20
⇒ 4a-12 = 20 (or) -(4a – 12) = 20
⇒ 4a = 32 (or)
⇒ 4a = 12 – 20 = -8
⇒ a = 8 (or)
⇒ a = – 2
Thus the required points on the X-axis are (-2, 0) and (8, 0).

Question 5.
Find the distance between parallel lines
(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0
(ii) I(x + y) + p = 0 and I(x + y) – r = 0
Solution:
The distance (d) between the parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0 is d = \(\frac{\left|\mathrm{C}_1-\mathrm{C}_2\right|}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}\)

i) The given parallel lines are 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0
Here A = 15, B = 8, C₁ = – 34 and C₂ = 31
Therefore, the distance between the parallel lines is
d = \(\frac{\left|C_1-C_2\right|}{\sqrt{A^2+B^2}}=\frac{|-34-31|}{\sqrt{(15)^2+8^2}}=\frac{|-65|}{\sqrt{289}}=\frac{65}{17}\) units

ii) The given parallel lines are l (x + y) + p = 0 and l (x + y) – r = 0
⇒ lx + ly + p = 0 and lx + ly – r = 0
⇒ Here, A = l, B = l, C₁ = p and C₂ = – r
Therefore the distance between the parallel lines is
d = \(\frac{\left|\mathrm{C}_1-\mathrm{C}_2\right|}{\sqrt{\mathrm{A}^2+units\mathrm{B}^2}}=\frac{|\mathrm{p}+\mathrm{r}|}{\sqrt{l^2+l^2}}=\frac{|\mathrm{p}+\mathrm{r}|}{\sqrt{2 l^2}}=\frac{1}{\sqrt{2}}\left|\frac{\mathrm{p}+\mathrm{r}}{l}\right|\)

Question 6.
Find the equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point (-2, 3).
Solution:
The equation of the given line is 3x – 4y + 2 = 0
Slope = \(\frac{-a}{b}=\frac{-3}{-4}=\frac{3}{4}\)
The equation of the line parallel to 3x – 4y + 2 = 0 and passing through (-2, 3) is 3
(y – 3) = \(\frac{3}{4}\) (x – (- 2))
⇒ 4y – 12 = 3x + 6
⇒ 3x – 4y + 18 = 0.

Question 7.
Find the equation of the line perpendicular to the line x – 7y + 5 = 0 and having x-intercept 3.
Solution:
The given equation of the line is x – 7y + 5 = 0 ⇒ y = \(\frac{1}{7}\) x + \(\frac{5}{7}\)
This is of the form y = mx + c ⇒ Slope of the given line =
The slope of the line perpendicular to the given line is M = \(-\frac{1}{\left(\frac{1}{7}\right)}\) = -7
The equation of the line with slope -7 and x-intercept 3 is y = M(x – a)
⇒ y = -7(x – 3)
⇒ 7x + y = 21

Question 8.
Find the angles between the lines √3x + y = 1 and x + √3y = 1
Solution:
The given lines are √3x + y = 1 and x + √3y = 1
y = -√3x + 1 (1)
and y = –\(\frac{1}{\sqrt{3}}\)x + \(\frac{1}{\sqrt{3}}\) …. (2)
The slope of line (1} is m₁ = -√3 , while the slope of line (2) is m₂ = \(-\frac{1}{\sqrt{3}}\).

The acute angle θ between the two lines is tan θ = \(\left|\frac{\mathrm{m}_1-\mathrm{m}_2}{1+\mathrm{m}_1 \mathrm{~m}_2}\right|\)
⇒ tan θ = \(\left|\frac{-\sqrt{3}+\frac{1}{\sqrt{3}}}{1+(-\sqrt{3})\left(-\frac{1}{\sqrt{3}}\right)}\right|\)
⇒ tan θ = \(\frac{\frac{-3+1}{\sqrt{3}}}{1+1}\)
⇒ tan θ = \(\left|\frac{-2}{2 \times \sqrt{3}}\right|\)
θ = 30°
Thus, the angle between the given lines is either 30° or 180° – 30° = 150°

Question 9.
The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y – 19 = 0 at right angle. Find the value of h.
Solution:
The slope of the line passing through the points (h, 3) and (4,1) is m₁ = \(\frac{1-3}{4-h}=\frac{-2}{4-h}\)
Equation of the given line is 7x – 9y – 19 = 0 ⇒ y = \(\frac{7}{9}\)x – \(\frac{19}{9}\)
Slope of given line m₂ = \(\frac{7}{9}\)
It is given that the two lines are perpendicular m₁ × m₂ = -1
⇒ \(\left(\frac{-2}{4-h}\right) \times\left(\frac{7}{9}\right)\) = -1
⇒ \(\frac{-14}{36-9 h}\) = -1
⇒ 14 = 36 – 9h
⇒ 9h = 36 – 14
⇒ h = \(\frac{22}{9}\)
∴ h = \(\frac{22}{9}\)

Question 10.
Find the ratio In which straight line 3x + 4y = 6 divIdes the line segment joining the points (2, -1) and (1, 1).
Solution:
Given points (x₁ , y₁) = (2, —1), (x₂, y₂) = (1, 1)
Equation of the given line is 3x + 4y = 6
Here, a = 3, b = 4, c = -6
\(\frac{l}{m}=\frac{-\left(\mathrm{ax}_1+\mathrm{by}_1+\mathrm{c}\right)}{\left(\mathrm{ax}_2+\mathrm{by}_2+\mathrm{c}\right)}\)
= \(\frac{-[3(2)+4(-1)+(-6)]}{[3(1)+4(1)+(-6)]}=\frac{-(6-4-6)}{3+4-6}=\frac{-(-4)}{1}=\frac{4}{1}\)
The ratio of the straight lines dividing the line segment joining given points is 4 :1 L₁, L₂ are of opposite signs.
∴ The given points lie on opposite sides of the line.

Question 11.
Find the point of Intersection of the lines 7x + y + 3 = 0, x + y = 0.
Solution:
L₁ = 7x + y + 3 = 0, L₂ = x + y = 0, a₁ = 7, b₁ = 1, c₂ = 3 and a₂ = 1, b₂ = 1, c₂ = 0
Point of intersection of the lines L₁ and L₂ is \(\left[\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}, \frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}\right]\)
= \(\left[\frac{1(0)-(1)(3)}{7(1)-1(1)}, \frac{(3)(1)-0(7)}{7(1)-1(1)}\right]=\left[\frac{0-3}{7-1}, \frac{3-0}{7-1}\right]\)
= \(\left[\frac{-3}{6}, \frac{3}{6}\right]=\left(\frac{-1}{2}, \frac{1}{2}\right)\)

Question 12.
Show that the straight lines (a – b)x + (b – c)y = c – a, (b – c)x + (c – a)y = a – b and (c – a) z + (a – b) y = b – c are concurrent.
Solution:
Given equation is of the lines are (a – b)x + (b – c)y = c – a ………… (1)
(b – c)x + (c – a)y = a – b ………… (2)
(c – a) z + (a – b) y = b – c ………… (3)

From (1) and (2)

Point of intersection of equations (1) and (2) is P(-1, -1)
Substitute the value of P in equation (3),
LHS = (c – a) (-1) + (a – b) (-1) = -c + a – a + b = b – c = RHS
∴ P (-1, -1) is a point on the equation (3). So, the given lines are concurrent.

Question 13.
Find the value of p, if the straight lines x + p = 0, y + 2 = 0 and 3x + 2y + 5 = 0 are concurrent.
Solution:
Equation of the given straight lines are, x + p = 0 (1); y + 2 = 0 …………(2)
and 3x + 2y + 5 = 0 …………(3)
From (2), y = -2
Substitute ‘y’ value in the equation (3)
⇒ 3x – 4 + 5 = 0
⇒ 3x = 4- 5 = -1
⇒ x = –\(\frac{1}{3}\)
Point of intersection of the straight lines (2) and (3) is P(-\(\frac{1}{3}\), -2)
The given lines are concurrent.
‘P’ is a point on x + p = 0
⇒ –\(\frac{1}{3}\) + p = 0
⇒ p = –\(\frac{1}{3}\)

Question 14.
Find the area of the triangle formed by the following straight lines and the coordinate axes,
i) x – 4y + 2 = 0,
ii) 3x – 4y + 12 = 0.
Solution:
i) Equation of the straight line AB is x – 4y + 2 = 0
⇒ -x + 4y = 2
⇒ \(\frac{x}{-2}+\frac{y}{\left(\frac{1}{2}\right)}\) = 1, a = 2, b = \(\frac{1}{2}\)
Area of the triangle OAB = \(\frac{1}{2}\)|ab| = \(\frac{1}{2}\)|-2 x \(\frac{1}{2}\)| = \(\frac{1}{2}\) sq. units.

ii) 3x – 4y + 12 = 0 is the equation of the straight line AB
⇒ -3x + 4y = 12
⇒ \(\frac{x}{-4}+\frac{y}{3}\) = 1, here a = – 4, b = 3.
Area of the triangle OAB = \(\frac{1}{2}\) |ab| = \(\frac{1}{2}\)(-4)(3)| = \(\frac{1}{2}\) (12) = 6 sq. units.

II.

Question 1.
Prove that the line through the point (x , yx) and parallel to the line Ax + By + C = 0 is A (x – x₁) + B(y – y₁) = 0.
Solution:
The given line is Ax + By + C = 0 ⇒ y = \(\left(\frac{-\mathrm{A}}{\mathrm{~B}}\right)\)x + \(\left(\frac{-\mathrm{C}}{\mathrm{~B}}\right)\)
⇒ slope, m = \(\frac{-\mathrm{A}}{\mathrm{~B}}\)
It is known that parallel lines have the same slope. Slope of the other line m = \(\frac{-\mathrm{A}}{\mathrm{~B}}\)
The equation of the line passing through point (x₁, y₁) and having a slope \(\frac{-\mathrm{A}}{\mathrm{~B}}\) is

y – y₁ = m(x – x₁)
⇒ (y – y₁) = \(\frac{-\mathrm{A}}{\mathrm{~B}}\)(x – x₁)
⇒ B(y – y₁) = -A(x – x₁)
⇒ A(x – x₁) + B(y – y₁) = 0
Hence Proved.

Question 2.
Two lines passing through the point (2, 3) intersects each other at an angle of 609. If slope of one line is 2, find equation of the other line.
Solution:
It is given that the slope of the first line, m₁ = 2,
Let the slope of the other line be m₂.
The angle between the two lines is 60°.
tan 60° = \(\left|\frac{\mathrm{m}_1-\mathrm{m}_2}{1+\mathrm{m}_1 \mathrm{~m}_2}\right|\)
⇒ √3 = \(\left|\frac{2-\mathrm{m}_2}{1+2 \mathrm{~m}_2}\right|\)
⇒ √3 = \(\pm\left(\frac{2-m_2}{1+2 m_2}\right)\)

⇒ √3 = \(\frac{1}{2}\)
⇒ √3(1 + 2m₂) = 2 – m₂
⇒ √3 + 2√3m₁ + m₂ + m₂ = 2
⇒ m₂ = \(\frac{2-\sqrt{3}}{(2 \sqrt{3}+1)}\)

(or)

√3 = –\(\frac{1}{2}\)
√3(1 + 2m₂) = -(2 – m₂)
√3 + 2√3m₂ – m₂ = -2
m₂ = \(=\frac{-(2+\sqrt{3})}{(2 \sqrt{3}-1)}\)

Case I:
If m₂ = \(\frac{2-\sqrt{3}}{(2 \sqrt{3}+1)}\)
The equation of the line passing through the point (2, 3) and having a slope of
\(\frac{2-\sqrt{3}}{(2 \sqrt{3}+1)}\) is (y – 3) = \(\frac{2-\sqrt{3}}{(2 \sqrt{3}+1)}\)(x – 2)
⇒ (2√3 + 1)(y – 3) = (2 – √3) (x – 2)
⇒ (√3 – 2) x +(2√3 + 1)y = -4 + 2√3 + 6√3 + 3
⇒ (√3 – 2)x + (2√3 + 1)y = -1 + 8√3
In this case, the equation of the other line is (√3 – 2) x + (2√3 + 1) y = -1 + 8√3

Case II :
If m₂ = \(\frac{-(2+\sqrt{3})}{(2 \sqrt{3}-1)}\)
The equation of the line passing through (2, 3) and having slope \(\frac{-(2+\sqrt{3})}{(2 \sqrt{3}-1)}\) is
(y – 3) = \(\frac{-(2+\sqrt{3})}{(2 \sqrt{3}-1)}\) (x – 2)
⇒ (2√3 – 1)(y – 3) = -(2 + √3)(x – 2)
⇒ (√3 + 2)x + (2√3 – 1)y = 8√3 + 1
In this case equation of other line is (√3 + 2)x + (2√3 – 1)y = 8√3 + 1

Question 3.
Find the equation of the right bisector of the line segment joining the points (3, 4) and (-1, 2).
Solution:
The right bisector of a line segment bisects the lines segment at (3, 4) and (-1, 2)
The end points of the line segment are given as A(3, 4) and B(-1, 2)
Accordingly, mid pint of AB = \(\left(\frac{3-1}{2}, \frac{4+2}{2}\right)\) = (1, 3)
Slope of AB = \(\frac{2-4}{-1-3}=\frac{-2}{-4}=\frac{1}{2}\). Slope of the line perpendicular
to AB = \(-\frac{1}{\left(\frac{1}{2}\right)}\) = -2

The equation of the line passing through (1,3) and having a slope of – 2 is (y – 3) = – 2(x – 1)
⇒ y – 3 = -2x + 2
⇒ 2x + y = 5
Thus, the required equation of the line is 2x + y = 5.

Question 4.
Find the co-ordinates of the foot of the perpendicular from the point (-1, 3) to the line 3x – 4y – 16 = 0.
Solution:
Let (a, b) be the coordinates of the perpendicular from (-1, 3) to the line 3x – 4y – 16 = 0
Slope of the line joining (-1, 3) and (a, b), m₁ = \(\frac{b-3}{a+1}\)
Slope of the line 3x – 4y – 16 = 0 is m₂ = \(\frac{-3}{-4}=\frac{3}{4}\)

Since these two lines are perpendicular, m₁ . m₂ = -1
\(\left(\frac{b-3}{a+1}\right) \times\left(\frac{3}{4}\right)\) = -1
⇒ \(\frac{3 b-9}{4 a+4}\) = -1
⇒ 3b -9 = -4a – 4
⇒ 4a + 3b = 5 …………(1)
Point (a, b) lies on line 3x – 4y = 16, therefore 3a – 4b = 15 ……….(2)
On solving equations (1) and (2), we have a = \(\frac{68}{25}\) and b = \(-\frac{49}{25}\)
Thus, the required co-ordinates of the foot of the perpendicular are \(\left(\frac{68}{25}, \frac{-49}{25}\right)\)

Question 5.
The perpendicular from the origin to the line y = mx+c meets it at the point (-1, 2). Find the values of m and c.
Solution:
The given equation of the line is y = mx + c.
It is given that the perpendicular from the origin meets the given line at (-1, 2). Therefore, the line joining the points (0, 0) and (-1, 2) is perpendicular to the given line.
Slope of the line joining (0, 0) and (-1, 2) = \(\frac{2}{-1}\) = -2
The slope of the given line is m.
m × -2 = -1
⇒ m = \(\frac{1}{2}\) [∵ The two lines are perpendicular]
Since point (-1, 2) lies on the given line, it satisfies the equation y = mx + c
⇒ 2 = m(-1) + c
⇒ 2 = \(\frac{1}{2}\)(-1) + c
⇒ c = 2 + \(\frac{1}{2}\)
∴ m= \(\frac{1}{2}\) and c = \(\frac{5}{2}\).

Question 6.
If p and q are the lengths of perpendiculars from the origin to the lines x cos θ – y sin θ = k cos 2θ and x sec θ + y cosec θ = k, respectively, prove that p² + 4q² = k².
Solution:
The equations of given lines are
x cos θ – y sin θ = k cos 20 ………….(1)
x sec θ + y cosec θ = k ……….(2)
The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is
d =
On comparing equation (1) with Ax + By + C = 0, we have A = cos θ, B = – sin θ, and C = – k cos 2θ
It is given that ‘p’ is the length of the perpendicular from (0, 0) to line (1)
p = \(\left|\frac{\mathrm{A}(0)+\mathrm{B}(0)+\mathrm{C}}{\sqrt{\mathrm{~A}^2+\mathrm{B}^2}}\right|=\frac{|\mathrm{C}|}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}=\frac{|-\mathrm{k} \cos 2 \theta|}{\sqrt{\cos ^2 \theta+\sin ^2 \theta}}\) = |-k cos 2θ| …………..(3)

On comparing equation (2) with Ax + By + C = 0, we have A = sec θ, B = cosec θ, and C = -k
It is given that ‘q’ is the length of the perpendicular from (0, 0) to line (2)
q = \(\left|\frac{A(0)+B(0)+C}{\sqrt{A^2+B^2}}\right|=\frac{|C|}{\sqrt{A^2+B^2}}=\frac{|-k|}{\sqrt{\sec ^2 \theta+{cosec}^2 \theta}}\) ………. (4)

From (3) and (4) we have
p² + 4q² = (-k cos 2θ)² + 4\(\left(\frac{|-\mathrm{k}|}{\sqrt{\sec ^2 \theta+{cosec}^2 \theta}}\right)^2\)
= k² cos² 2θ + \(\frac{4 k^2}{\left(\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin ^2 \theta \cdot \cos ^2 \theta}\right)}\) = k² cos² 2θ + \(\frac{4 \mathrm{k}^2}{\left(\frac{1}{\cos ^2 \theta}+\frac{1}{\sin ^2 \theta}\right)}\)
= k² cos² 2θ + \(\frac{4 \mathbf{k}^2}{\left(\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin ^2 \theta \cdot \cos ^2 \theta}\right)}\)
= k² cos² 2θ + \(\frac{4 k^2}{\left(\frac{1}{\sin ^2 \theta \cdot \cos ^2 \theta}\right)}\)
= k² cos² 2θ + 4k² sin² θ.cos² θ
= k² cos² 2θ + k²(2 sin θ cos θ)²
= k² cos² 2θ + k² sin² 2θ
= k² (cos² 2θ + sin² 2θ) = k²
Hence, we proved that p² + 4q² = k²

Question 7.
In the triangle ABC with vertices A(2, 3), B(4, -1) and C(1, 2), find the equation and length of altitude from the vertex A.
Solution:
Let AD be the attitude of triangle ABC from vertex A.
Slope of BC = \(\frac{2-(-1)}{1-4}=-\frac{3}{3}\) = -1
The equation of the line passing through point (2, 3) and having a slope of 1 is (y – 3) = 1(x – 2)
⇒ x – y + 1 = 0
⇒ y – x = 1
Therefore equation of the altitude from vertex A is y – x = 1.
Length of AD = Length of the perpendicular from A(2, 3) to BC.
The equation of BC is (y + 1) = (-1) (x – 4)
⇒ y + 1 = -x + 4
⇒ x + y – 3 = 0 …………. (2)
On comparing the above line with Ax + By + C = 0 we have A = 1, B = 1, and C = -3
Length of AD = \(\frac{|1 \times 2+1 \times 3-3|}{\sqrt{1^2+1^2}}=\frac{|2|}{\sqrt{2}}\) = √2 units
Thus, the equation and the length of the altitude from vertex A are y – x = 1 and 42 units respectively.

Question 8.
If p is the length of the perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that \(\frac{1}{2}\).
Solution:
It is known that the equation of a line whose intercepts on the axes are a and b is
\(\frac{x}{a}+\frac{y}{b}\) = 1
⇒ bx + ay = ab
⇒ bx + ay – ab – 0 …………….(1)
The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is d = \(\left|\frac{\mathrm{Ax}_1+\mathrm{By}_1+\mathrm{C}}{\sqrt{\mathrm{~A}^2+\mathrm{B}^2}}\right|\)
On comparing equation (1) with Ax + By + C = 0, we have A = b, B = a, and C = – ab Therefore, p is the length of the perpendicular from point (0, 0) to line (1), we have
p = \(\left|\frac{\mathrm{A}(0)+\mathrm{B}(0)+(-\mathrm{ab})}{\sqrt{\mathrm{b}^2+\mathrm{a}^2}}\right|=\frac{|-\mathrm{ab}|}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\)
Squaring on both sides we have
p² = \(\frac{(-a b)^2}{a^2+b^2}\)
⇒ p²(a² + b²) = a²b²
⇒ \(\frac{a^2+b^2}{a^2 b^2}=\frac{1}{p^2}\)
⇒ \(\frac{1}{\mathrm{p}^2}=\frac{1}{\mathrm{a}^2}+\frac{1}{\mathrm{~b}^2}\)

Question 9.
A straight line meets the coordinate axes in A and B. Find the equation of the straight line, when
i) AB is divided in the ratio 2 : 3 at (-5, 2).
Solution:
Let a straight line AB meets the coordinate axes at A and B. Let OA = a and OB = b.
Co- ordinates of A are (a, 0) and co- ordinates of B are (0, b).
P divides AB in the ratio 2:3.
Co – ordinates of P are \(\left(\frac{3 \mathrm{a}}{5}, \frac{2 \mathrm{~b}}{5}\right)\) = (-5, 2)
\(\frac{3 a}{5}\) = -5 ⇒ a = \(-\frac{25}{3} ; \frac{2 b}{5}\) = 2 ⇒ b = 5
Equation of AB is \(\frac{x}{a}+\frac{y}{b}\) = 1
\(\frac{x}{\left(-\frac{25}{3}\right)}+\frac{y}{5}\) = 1
⇒ \(\frac{-3 x}{25}+\frac{y}{5}\) = 1
⇒ -3x + 5y = 25
⇒ 3x – 5y + 25 = 0

ii) AB is divided in the ratio 1 : 2 at (-5, 4).
Solution:
Let a straight line AB meets the coordinate axes at A and B.
Let OA = a and OB = b.
Co – ordinates of A are (a, 0) and co-ordinates of B are (0, b).
P divides AB in the ratio 1:2.
Co – ordinates of P are \(\left(\frac{2 \mathrm{a}}{3}, \frac{\mathrm{~b}}{3}\right)\) = (-5, 4) = \(\frac{2 a}{3}\) = -5
⇒ a = \(-\frac{15}{2} ; \frac{\mathrm{b}}{2}\) = 4
⇒ b = 12

Equation of AB is = \(\frac{x}{a}+\frac{y}{b}\) = 1
⇒ \(\frac{x}{\left(-\frac{15}{2}\right)}+\frac{y}{12}\) = 1
⇒ \(\frac{-2 x}{15}+\frac{y}{12}\) = 1
⇒ -8x + 5y = 60
⇒ 8x – 5y + 60 = 0

iii) (p, q) bisects AB
Solution:
Let OA = a and OB = b.
Co – ordinates of A are (a, 0) and co-ordinates of B are (0, b).
P bisects AB.
⇒ Mid-point of AB \(\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{~b}}{2}\right)\) = (p, q)
Therefore, \(\frac{a}{2}\) = p
⇒ a = 2p, \(\frac{b}{2}\) = q
⇒ b = 2q
Equation of AB is \(\frac{x}{a}+\frac{y}{b}\) = 1
⇒ \(\frac{x}{2 p}+\frac{y}{2 q}\) = 1
⇒ \(\frac{x}{p}+\frac{y}{q}\) = 2

Question 10.
A triangle of area 24 sq. units is formed by a straight line and the coordinate axes in the first quadrant. Find the equation of the straight line, if it passes through (3, 4).
Solution:
Area of a triangle = 24 sq. units.
The line AB passes through the point P (3, 4).
Equation of AB in the intercept form is \(\frac{x}{a}+\frac{y}{b}\) = 1
⇒ \(\frac{3}{a}+\frac{4}{b}\) = 1
⇒ \(\frac{4}{b}\) = 1 – \(\frac{3}{a}=\frac{a-3}{a}\)
b = \(\frac{4 a}{a-3}\)

Area of ∆OAB = 24
⇒ \(\frac{1}{2}\)|ab| = 24
⇒ \(\frac{1}{2}\left|\frac{4 a^2}{a-3}\right|\) = 24
⇒ a² = 12(a – 3)
⇒ a² – 12a + 36 = 0
⇒ (a – 6)² = 0
⇒ a = 6

Substitute the value of ‘a’ in b = \(\frac{4 a}{a-3}=\frac{24}{3}\) = 8
Equation of AB is \(\frac{x}{6}+\frac{y}{8}\) = 1
⇒ 4x + 3y = 24
⇒ 4x + 3y – 24 = 0

Question 11.
Find the set of values of ‘a’ if the points (1, 2) and (3, 4) lie to the same side of the straight line 3x – 5y + a = 0.
Solution:
Equation of the given line is 3x – 5y + a = 0
Let P (x₁, y₁) = (1, 2) and Q (x₂, y₂) = (3, 4)
L₁₁ = 3(1) – 5(2) + a = a – 7,
L₂₂ = 3(3) – 5(4) + a = a – 11
Since P and Q lie on same side a – 7 and a – 11 must both be positive or both be negative.
Let a – 7 > 0, a – 11 > 0 ⇒ a > 7 and a > 11 ⇒ a ∈ (11, ∞)
Let a – 7 < 0, a – 11 < 0 ⇒ a < 7, a < 11 ⇒ a ∈ (-∞, 7)
∴ a ∈ (-∞ , 7) ∪ (11, ∞)

Question 12.
Show that the lines 2x + y – 3 = 0, 3x + 2y – 2 = 0 and 2x – 3y – 23 = 0 are concurrent and find the point of concurrency.
Solution:
Equations of the given lines are 2x + y – 3 = 0 ………….(1)
3x + 2y – 2 = 0 ……… (2)
2x – 3y – 23 = 0 ……..(3)

\(\frac{x}{-2+6}=\frac{y}{-9+4}=\frac{1}{4-3}\)
⇒ x = 4, y = -5
Co – ordinates of P are (4, -5)
Substitute ‘P’ value in 2x – 3y – 23 = 2(4) – 3 (-5) – 23 = 8 + 15 – 23 = 0
P lies on (3). So, the given lines are concurrent. Point of concurrency is P (4, -5)

Question 13.
Find the value of p, if 3x + 4y = 5, 2x + 3y = 4, px + 4y = 6 are concurrent.
Solution:
Equations of the lines are 3x + 4y – 5 = 0 ………… (1)
2x + 3y – 4 = 0 ……………. (2)

\(\frac{x}{-16+15}=\frac{y}{-10+12}=\frac{1}{9-8}\)
⇒ x = 4, y = 2
Point of intersection of (1) and (2) is P (-1, 2)
The given lines are concurrent. P lies on px + 4y = 6.
-p + 8 = 6 ⇒ p = 8 – 6 = 2.

Question 14.
If 3a + 2b + 4c = 0, then show that the equation ax + by + c = 0 represents a family of concurrent straight lines and find the point of concurrency.
Solution:
Given condition is 3a + 2b + 4c = 0
Divide the equation with 4 ⇒ \(\left(\frac{3}{4}\right)\)a + \(\left(\frac{1}{2}\right)\)b + c = 0

For all values of a, b the line ax + by + c = 0 passes through the point \(\left(\frac{3}{4}, \frac{1}{2}\right)\)
The equation ax + by + c = 0 represents a family of concurrent lines.
Point of concurrency is \(\left(\frac{3}{4}, \frac{1}{2}\right)\)

III.

Question 1.
Find the point on the straight line 3x + y + 4 = 0 which is equidistant from the points (-5, 6) and (3, 2).
Solution:
Points A (-5, 6) and B (3, 2). P(x₁, y₁) is the point on 3x + y + 4 = 0
3x₁ + y₂ + 4 = 0 …………. (1)
PA = PB => PA² = PB²
⇒ (x₁ + 5)² + (y₁ – 6)² = (x₂ – 3)² + (y₂ – 2)²
⇒ x₁ + 10x₁ + 25 + y₁ – 12y₁ + 36 = x₁ – 6x₁ + 9 + y₁ – 4y₁ + 4
16x₁ – 8y₁ + 48 = 0
⇒ 2x₁ – y₁ + 6 = 0 …….. (2)
Adding (1) and (2)
⇒ 5x₁ + 10 = 0
⇒ x₁ = -2
Substitute x₁ value (1) -6 + y₁ + 4 = 0
⇒ y₁ = 6 – 4 = 2
∴ The required point P is (-2, 2)

Question 2.
A straight line through P(3, 4) makes an angle of 60° with the positive direction of the X -axis. Find the coordinates of the points on the line which are 5 units away from P.
Solution:
Let P(x₁, y₁) = (3, 4), θ = 60°.
Co-ordinates of any point on the line Q are (x₁ + r cos θ, y₁ + r sin θ)
θ = 60° ⇒ cos θ = cos 60° = \(\frac{1}{2}\) and sin θ = sin 60° = \(\frac{\sqrt{3}}{2}\)
Let r = 5
Co-ordinates of Q are [3 + 5(\(\frac{1}{2}\)), 4 + 5(\(\frac{\sqrt{3}}{2}\))] = \(\left(\frac{11}{2}, \frac{8+5 \sqrt{3}}{2}\right)\)
Let r = -5
Co-ordinates of Q are [3 – 5(\(\frac{1}{2}\)), 4 – 5(\(\frac{\sqrt{3}}{2}\))] = \(\left(\frac{1}{2}, \frac{8-5 \sqrt{3}}{2}\right)\)

Question 3.
A straight line through 9 (√3, 2) makes an angle \(\frac{\pi}{6}\) with the positive direction of the X – axis. If the straight line intersects the line √3 x – 4y + 8 = 0 at P, find the distance PQ.
Solution:
Equation of the straight line AB = √3x – 4y + 8 = 0.
By the problem, AB and PQ are not perpendicular.
PQ makes an angle \(\frac{\pi}{6}\) with the positive direction of X – axis. 6
m = Slope of PQ = tan \(\frac{\pi}{6}=\frac{1}{\sqrt{3}}\) ⇒ PQ passes through Q (√3, 2)
Equation of PQ is y – 2 = (x – √3) ⇒ √3y – 2√3 = x – √3x – √3y = -√3 …….. (1)
Equation of AB is √3x – 4y + 8 = 0 ⇒ √3x – 4y = -8 (2)
Equation (1) x √3 ⇒ √3x – 3y = -3 (3)
Subtract equation (3) from (2) ⇒ -y = -5 ⇒ y = 5
Substitute y value in (1), gives x – √3(5) = – √3 ⇒ x = 4√3
Coordinates of P are (4√3, 5), Coordinates of Q are (√3, 2)
PQ² = (4√3 – √3 )² + (5 – 2)²
= 27 + 9 = 36
⇒ PQ = \(\sqrt{36}\) = 6 units.

Question 4.
A straight line through Q (2, 3) makes an angle \(\frac{3 \pi}{4}\) with the negative direction of the X – axis. If the straight line intersects the line x + y – 7 = 0 at P, find the distance PQ.
Solution:
Equation of the straight line AB is x + y – 7 = 0
The line PQ makes an angle \(\frac{3 \pi}{4}\) with the negative direction of X – axis.
i.e., PQ makes an angle π – \(-\frac{3 \pi}{4}=\frac{\pi}{4}\) with the positive direction of X – axis.
Coordinates of Q are (2, 3)
Coordinates of P are (x₁ + r cos θ, y₁ + r sin θ)
(2 + r.cos\(\frac{\pi}{4}\), 3 + r sin\(\frac{\pi}{4}\)) = (2 + \(\frac{\mathrm{r}}{\sqrt{2}}\), 3 + \(\frac{\mathrm{r}}{\sqrt{2}}\))
P is a point on the line x + y – 7 = 0
⇒ 2 + \(\frac{\mathrm{r}}{\sqrt{2}}\)), 3 + \(\frac{\mathrm{r}}{\sqrt{2}}\)) – 7 = 0
⇒ 2. \(\frac{\mathrm{r}}{\sqrt{2}}\)) = 2
⇒ r = √2

Question 5.
Show that the straight lines x + y = 0, 3x + y- 4 = 0 and x + 3y – 4 = 0 form an isosceles triangle.
Solution:
The equations of the given straight lines are
x + y = 0 ………….. (1)
3x + y – 4 = 0 …………… (2)
x + 3y – 4 = 0 ………… (3)
Let θ₁ be the angle between the equations (1) and (2)
cos θ₁ = \(\frac{a_1 a_2+b_1 b_2}{\sqrt{a_1^2+b_1^2} \sqrt{a_2^2+b_2^2}}=\frac{3+1}{\sqrt{1+1} \sqrt{9+1}}=\frac{4}{\sqrt{20}}=\frac{2}{\sqrt{5}}\)
θ₁ = cos^-1\(\left(\frac{2}{\sqrt{5}}\right)\)

Let θ₂ be the angle between the equations (2) and (3)
cos θ₁ = \(\frac{a_1 a_2+b_1 b_2}{\sqrt{a_1^2+b_1^2} \sqrt{a_2^2+b_2^2}}=\frac{3+3}{\sqrt{9+1} \sqrt{1+9}}=\frac{6}{10}=\frac{3}{5}\)
θ₁ = cos^-1\(\left(\frac{3}{5}\right)\)

Let θ₃ be the angle between the equations (3) and (1)
cos θ₁ = \(\frac{a_1 a_2+b_1 b_2}{\sqrt{a_1^2+b_1^2} \sqrt{a_2^2+b_2^2}}=\frac{1+3}{\sqrt{1+1} \sqrt{1+9}}=\frac{4}{\sqrt{20}}=\frac{2}{\sqrt{5}}\)
θ₁ = cos^-1\(\left(\frac{2}{\sqrt{5}}\right)\)
Since θ₁ = θ₃
∴ The triangle formed by the given straight lines is an isosceles triangle.

Referring to the AP Inter 1st Year Maths Study Material Chapter 9 Straight Lines Exercise 9b Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Straight Lines Solutions Exercise 9b

I. In Exercises 1 to 7, find the equation of the line which satisfy the given conditions.

Question 1.
Write the equations for the X and Y – axes.
Solution:
The y-coordinate of every point on the X-axis is 0. Therefore, the equation of the X-axis is y = 0. The ‘x’ coordinate of every point on the Y-axis is zero.
Therefore, the equation of Y-axis is x = 0.

Question 2.
Find the equation of line passing through the point (-4, 3) with slope ½.
Solution:
We know that the equation of the line passing through the point (x₀, y₀) whose slope is m is (y – y₀) = m(x – x₀).
Thus, the equation of the line passing through the point (- 4, 3), whose slope is ½
(y – 3) = ½(x + 4)
⇒ 2(y -3) = x + 4
⇒ 2y – 6 = x + 4
⇒ x – 2y + 10 = 0

Question 3.
Find the equation of the line passing through (0, 0) with slope m.
Solution:
We know that the equation of the line passing through the point (x₀, y₀) whose slope’m’ is (y – y₀) = m(x – x₀).
Thus, the equation of the line passing through the point (0, 0) whose slope’m’ is
(y – 0) = m(x – 0)
⇒ y = mx.

Question 4.
Find the equation of the line passing through (2, 2√3) and inclined with X-axis at an angle of 75°.
Solution:
The slope of the line that inclines with the X-axis at an angle of 75° is m = tan 75°
⇒ m = tan(45° + 30°) = \(\frac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 45^{\circ} \tan 30^{\circ}}\)

= \(\frac{1+\frac{1}{\sqrt{3}}}{1-1 \cdot \frac{1}{\sqrt{3}}}=\frac{\frac{\sqrt{3}+1}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}\)
We know that the equation of the line passing through the point (x₀, y₀) whose slope m is (y – y₀) = m(x – x₀)
Thus, if a line passes through (2, 2√3) and inclined with the X-axis at an angle of
75°, then the equation of the line is (y – 2√3) = \(=\frac{\sqrt{3}+1}{\sqrt{3}-1}\)(x – 2)
⇒ (y – 2√3)(√3 -1) = (√3 + 1)(x – 2)
⇒ y(√3 – 1) – 2√3(√5 – 1) = x (√3 +1) – 2 (√3 + 1)
⇒ (√3 + 1) x – (√3 -1) y = 2√3 + 2 – 6 + 2√3
⇒ (√3 +1) x – (√3 – 1) y = 4√3 – 4
⇒ (√3 +1) x – (√3 -1) y = 4(√3 – 1)

Question 5.
Find the equation of line intersecting the X-axis at a distance of 3 units to the left of origin with slope -2.
Solution:
It is known that, if a line with slope m makes x-intercept a. Then the equation of the line is y = m(x – a).
For the line intersecting the X-axis at a distance of 3 units to the origin, a = – 3
The slope of the line is given as m = – 2
Thus, the required equation of the line is y = – 2[x – (- 3)]
⇒ y = -2x – 6
⇒ 2x + y + 6 = 0

Question 6.
Find the equation of line intersecting the Y-axis at a distance of 2 units above the origin and making an angle of 30° with positive direction of the X-axis.
Solution:
It is known that, if a line with slope m makes y-intercept c, then, the equation of the line is y = mx + c.
Here, c = 2 and m = tan 30° = \(\frac{1}{2}\)
Thus, the required equation of the line is y = \(\frac{1}{\sqrt{3}}\)x + 2
⇒ y = \(\frac{x+2 \sqrt{3}}{\sqrt{3}}\)
⇒ √3y = x + 2√3
∴ x – √3y + 2√3 = 0

Question 7.
Find the equation of the line passing through the points (-1, 1) and (2, -4).
Solution:
It is known that, if the equation of the line passing through the points (x₁; y₁) and (x₂, y₂) is y – y₁ = \(\frac{1}{2}\) (x – x₁)
Therefore, the equation of the line passing through the points (-1, 1) and (2, – 4) is
(y – 1) = \(\frac{-4-1}{2+1}\) (x + 1)
⇒ (y – 1) = \(\frac{-5}{3}\)(x + 1)
⇒ 3(y – 1) = – 5(x + 1)
⇒ 3y – 3 = -5x – 5
∴ 5x + 3y + 2 = 0

Question 8.
The vertices of ΔPQR are P(2, 1), Q(-2, 3) and R(4, 5). Find the equation of the median through the vertex R.
Solution:
Given that the vertices of ΔPQR are P(2, 1), Q(-2, 3) and R(4, 5).
Let RL be the median through vertex R.
Accordingly, L is the mid point of PQ.

By mid point formula, the coordinates of point L is given by \(\left(\frac{2-2}{2}, \frac{1+3}{2}\right)\) = (0, 2)

It is known that, the equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is y – y₁ = \(\frac{y_2-y_1}{x_2-x_1}\)(x – x₁)

Therefore, the equation of RL can be determined by substituting (x₁, y₁) = (4, 5) and (x₂, y₂) = (0, 2)
Hence, y – 5 = \(\frac{2-5}{0-4}\)(x – 4)
⇒ y – 5 = \(\frac{-3}{-4}\) (x – 4)
⇒ 4(y – 5) = 3(x – 4)
⇒ 4y – 20 = 3x -12
⇒ 3x – 4y + 8 = 0
Thus, the required equation of the median through vertex R is 3x – 4y + 8 = 0.

Question 9.
Find the equation of the line passing through (-3, 5) and perpendicular to the line through the points (2, 5) and (-3, 6).
Solution:
The slope of the line joining the points (2, 5) and (-3, 6) is m = \(\frac{6-5}{-3-2}=\frac{1}{-5}\)
We know that, two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.
Therefore, slope of the line perpendicular to the line through the points (2, 5) 1 1
and (- 3, 6) is = \(-\frac{1}{m}=-\frac{1}{\left(-\frac{1}{5}\right)}\) = 5
Now, the equation of the line passing through the point (- 3, 5) whose slope is 5, is
(y – 5) = 5(x + 3)
⇒ (y – 5) = 5x + 15
⇒ 5x – y + 20 = 0

Question 10.
A line perpendicular to the line segment joining the points (1, 0) and (2. 3) divides it in the ratio l:n. Find the equation of the line.
Solution:
According to the section formula, the coordinates of the point that divides the line segment joining the points (1,0) and (2, 3) in the ratio 1 : n is
\(\left(\frac{n(1)+1(2)}{1+

The slope of the line joining the points (1, 0) and (2, 3) is m = [latex]\frac{3-0}{2-1}\) = 3

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other. Therefore, slope of the line that is perpendicular to the line joining the points (1, 0) and (2, 3) is \(-\frac{1}{m}=-\frac{1}{3}\).

Now the equation of the line passing through \(\left(\frac{\mathrm{n}+2}{\mathrm{n}+1}, \frac{3}{\mathrm{n}+1}\right)\) and whose slope is \(-\frac{1}{3}\) is (y – \(\frac{3}{(\mathrm{n}+1)}\)) = \(\frac{-1}{3}\)(x – \(\frac{n+2}{n+1}\))
⇒ 3[(n + 1)y – 3] = -[x(n + 1) – (n + 2)]
⇒ 3(n + 1)y – 9 = – (n + 1)x + n + 2
⇒ (1 + n)x + 3(1 + n)y = n + 11

Question 11.
Find the equation of a line that cuts off equal Intercepts on the coordinate axes and passes through the point (2, 3).
Solution:
The equation of a line in the intercept form is \(\frac{x}{a}+\frac{y}{b}\) = 1 …….(1)
Here, a and b are the intercepts on X and Y axes respectively. It is given that the line cuts off equal intercepts on both the axes. This means that a = b. Therefore, equation (!) reduces to \(\frac{x}{a}+\frac{y}{b}\) = 1
⇒ x + y = a ……….(2)
Since the line passes’through point (2, 3), equation (2) reduces to 2 + 3 = a
a = 5
On substituting the value of ‘a’ in the equation (2), we have x + y = 5
This is the required equation of the line.

Question 12.
If the portion of a straight line intercepted between the axes of coordinates is bisected at (2p, 2q), write the equation of the straight line.
Solution:
Let a,b be the intercepts of the line.
∴ The line cut X – axis at A(a, 0), Y – axis at B (0, b)
Mid point if \(\overline{\mathrm{AB}}\) is \(\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{~b}}{2}\right) \Rightarrow\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{~b}}{2}\right)\) = (2p, 2q)

⇒ \(\frac{a}{2}\) = 2P, \(\frac{b}{2}\) = 2q

⇒ a = 4p, b = 4q
Equation of the line is \(\frac{x}{4 p}+\frac{y}{4 q}\) = 1
⇒ \(\frac{x}{p}+\frac{y}{q}\) = 4

Question 13.
Find the angle made by the straight line y = -√3x + 3 with the positive direction of the X – axis measured in the counter clockwise direction.
Solution:
Let ‘θ’ be the inclination.
Slope = -√3
⇒ tan θ = -√3
⇒ θ = \(\frac{2 \pi}{3}\)

Question 14.
The intercepts of a straight line on the axes of coordinates are a and b. If p is the length of the perpendicular drawn from the origin to this line, write the value of p in terms of a and b.
Solution:
Equation of the line having intercepts a, b is \(\frac{x}{a}+\frac{y}{b}\) = 1 ⇒ bx + ay – ab = 0 …(1)
p = Length of the perpendicular from (0, 0) to line (1).
= \(\frac{|-a b|}{\sqrt{b^2+a^2}}=\frac{|a b|}{\sqrt{a^2+b^2}}\)

II.

Question 1.
Find the equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.
Solution:
The equation of a line in the intercept form is \(\frac{x}{a}+\frac{y}{b}\) = 1 ……. (1)
Here, a and b are the intercepts on X and Y axes respectively, it is given that a + b = 9
⇒ b = 9 – a ……….(2)
From eqns. (1) and (2) we have \(\frac{x}{a}+\frac{y}{9-a}\)
It is given that the line passes through the point (2, 2). Therefore, equation (3) reduces to
\(\frac{2}{a}+\frac{2}{9-a}\) = 1
⇒ 2\(\left(\frac{1}{a}+\frac{1}{9-a}\right)\) = 1

⇒ \(\frac{2}{a}+\frac{2}{9-a}\) = 1
⇒ 18 = 9a – a²
⇒ a² – 9a + 18 = 0
⇒ a² – 6a – 3a + 18 = 0
⇒ a(a – 6) – 3(a – 6) = 0
⇒ (a – 6)(a -3) = 0
⇒ a = 6 or a = 3

If a = 6 then b = 9 – 6 = 3,
Therefore, the equation of the line is \(\frac{x}{6}+\frac{y}{3}\) = 1
⇒ x + 2y – 6 = 0

If a = 3 then b = 9 – 3 = 6, therefore, the equation of the line is
\(\frac{x}{3}+\frac{y}{6}\) = 1
⇒ 2x + y – 6 = 0

Question 2.
Find the equation of the Une through the point (0, 2) making an angle \(\frac{2 \pi}{3}\) with the positive X-axis. Also, find the equation of the line parallel to it and crossing the Y-axis at a distance of 2 units below the origin.
Solution:
The slope of the line making an angle \(\frac{2 \pi}{3}\) with the positive X-axi is m = tan(\(\left(\frac{2 \pi}{3}\right)\)) – √3
Equation of a line passing through (0, 2) and slope √3 is
(y – 2) = -√3(x – 0)
⇒ y – 2 = -√3x
⇒ √3x + y – 2 = 0
The slope of the line parallel to the line √3x + y – 2 = 0 is .4.
It is given that the line parallel to line √3x + y – 2 = 0 crosses the Y – axis 2 units below the origin i.e., it passes through (0, -2).
Hence, the equation of the line passing through the point (0, -2) and having a slope √3 is y – (-2) = -√3(x – 0)
⇒ y + 2 = √3x
⇒ √3x + y + 2 = 0

Question 3.
The perpendicular from the origin to a line meets it at the point (-2, 9). Find the equation of the line.
Solution:
The slope of the line joining the origin (0, 0) and point (-2, 9) is m₁ = \(\frac{9-0}{-2-0}=\frac{-9}{2}\)
Accordingly, the slope of the line perpendicular to the line joining the origin and point (- 2, 9) is m₂ = \(\frac{-1}{m_1}=-\frac{1}{\left(\frac{-9}{2}\right)}=\frac{2}{9}\)
Now, the equation of the line passing through the point (-2, 9) and having a slope ‘2/9’ is
9y – 81 = 2x + 4
⇒ 2x – 9y + 85 = 0

Question 4.
The length L (in centimetres) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L = 125.134 when C = 110, express L in terms of C.
Solution:
It is given that when C = 20, the value of L is 124.942
whereas when C = 110, the value of L is 125.134
Accordingly, points (20, 124.942) and (110, 125.134) satisfy the linear relation between L and C.
Now, assuming C along the X-axis and L along the Y-axis, we have two points, (20, 124.942) and (110, 125.134) in the xy plane.
Therefore, the linear relation between L and C is the equation of the line passing through the points (20, 124.942) and (110, 125.134)
(L – 124.942) = \(\frac{125.134-124.942}{110-20}\)(C – 20)
L – 124.942 = \(\frac{0.192}{90}\)(C – 20)
L = \(\frac{0.192}{90}\)(C – 20) + 124.942.
This is the required linear relations.

Question 5.
The owner of a milk, store finds that, he can sell 980 litres of milk each week at Rs. 14/ litre and 1220 litres of milk each week at Rs. 16/ litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs. 17/litre?
Solution:
The relationship between selling price and demand is linear. Assuming selling price per litre along the X-axis and demand along the Y-axis, we have two points i.e. (14,980) and (16,1220) in the xy-plane that satisfy the linear relationship between selling price and demand.

Therefore, the linear relationship between selling price per litre and demand is the equation of the line passing through the points (14, 980) and (16, 1220).
y – 980 = \(\frac{1220-980}{16-14}\) (x – 14)
⇒ y – 980 = \(\frac{240}{2}\)(x – 14)
⇒ y – 980 = 120(x – 14)
⇒ y = 120(x – 14) + 980
when x = ₹ 17/litre, y = 120(17 – 14) + 980
⇒ y = 120 × 3 + 980 = 360 + 980 = 1340
∴ The owner of the milk store could sell 1340 litres of milk weekly at Rs. 17/ litre.

Question 6.
P(a, b) is the mid-point of a line segment between axes. Show that equation of the line is \(\frac{x}{a}+\frac{y}{b}\) = 2.
Solution:
Let AB be the line segment between the axes and let P(a, b) be its mid-point.
Let the coordinates of A and B are (0, y) and (x, 0) respectively. Since P(a, b) is the mid-point of AB.
\(\left(\frac{0+x}{2}, \frac{y+0}{2}\right)\) = (a, b)
⇒ \(\left(\frac{\mathrm{x}}{2}, \frac{\mathrm{y}}{2}\right)\) = (a, b)
⇒ \(\frac{x}{2}\) = a and \(\frac{y}{2}\) = b
∴ x = 2a and y = 2b

Thus, the respective coordinates of A and B are (0, 2b) and (2a, 0). The equation of the line passing through points (0, 2b) and (2a, 0) is
(y – 2b) = \(\frac{(0-2 b)}{(2 a-0)}\)(x – 0)
⇒ y – 2b = \(\frac{1}{2}\)(x)
⇒ a(y – 2b) = -bx
⇒ ay – 2ab = – bx
⇒ bx + ay = 2ab
On dividing both sides by ab, we have \(\frac{b x}{a b}+\frac{a y}{a b}=\frac{2 a b}{a b}\) ⇒ \(\frac{x}{a}+\frac{y}{b}\) = 2
Thus, the equation of the line is \(\frac{x}{a}+\frac{y}{b}\) = 2

Question 7.
Point R(h, k) divides a line segment between the axes in the ratio 1 : 2. Find the equation of the line.
Solution:
Let AB be the line segment between the axes such that point R(h, k) divides AB in the ratio 1 : 2.
Let the respective coordinates of A and B are (x, 0) and (0, y). Since R(h, k) divides AB in the ratio 1 : 2, according to the section formula
(h, k) = \(\left(\frac{1 \times 0+2 \times x}{1+2}, \frac{1 \times y+2 \times 0}{1+2}\right)\)
⇒ (h, k) = \(\left(\frac{2 \mathrm{x}}{3}, \frac{\mathrm{y}}{3}\right)\)

⇒ h = \(\frac{2 x}{3}\) and k = \(\frac{y}{3}\)

⇒ x = \(\frac{3 h}{2}\) and y = 3k
Therefore, the respective coordinates of A and B are (\(\frac{3 h}{2}\), o) and (0, 3k).
Now, the equation of line AB is
(y – 0) = \(\frac{3 k-0}{0-\frac{3 h}{2}}\) (x – \(\frac{1}{2}\))
⇒ y = \(-\frac{2 \mathrm{k}}{\mathrm{~h}}\)(x – \(\frac{3 \mathrm{h}}{2}\))
⇒ hy = -2kx + 3hk
⇒ 2kx + hy = 3hk
Thus, the required equation of the line is 2kx + hy = 3hk.

Question 8.
By using the concept of equation of a line, prove that the three points are collinear. C
i) (3, 0), (-2, -2) and (8, 2)
Solution:
Let A(3, 0), B(-2, -2), C(8, 2) be the given points.
Equation of the line \(\frac{1}{2}\) is y – y₁ = \(\frac{y_2-y_1}{x_2-x_1}\) (x – x₁)
⇒ y – 0 = \(\frac{-2-0}{-2-3}\)(x – 3)
y = \(\frac{2}{5}\)(x – 3)
⇒ 5y = 2x – 6
⇒ 2x – 5y – 6 = 0 …(1)
Put C(8, 2) in (1) ⇒ 2(8) – 5(2) – 6 = 0
Hence given points A, B and C are collinear.

ii) (-5, 1), (5, 5), (10, 7)
Solution:
Let A(-5, 1), B(5, 5), C(10, 7)
Slope of \(\frac{1}{2}\), Slope of \(\frac{1}{2}\)
Slope of \(\frac{1}{2}\) = Slope of \(\frac{1}{2}\)
A, B, C are collinear.
Equation of the line is y – 1 = \(\frac{1}{2}\)(x + 5)
⇒ 5y – 5 = 2x + 10
⇒ 2x – 5y + 15 = 0

iii) (a, b + c), (b, c + a), (c, a + b)
Solution:
Let A(a, b + c), B(b, c + a), C(c, a + b)
Slope of \(\overrightarrow{\mathrm{AB}}=\frac{\mathrm{c}+\mathrm{a}-\mathrm{b}-\mathrm{c}}{\mathrm{~b}-\mathrm{a}}=\frac{\mathrm{a}-\mathrm{b}}{-(\mathrm{a}-\mathrm{b})}\) = -1, Slope of \(\stackrel{\rightharpoonup}{\mathrm{AC}}=\frac{\mathrm{a}+\mathrm{b}-\mathrm{b}-\mathrm{c}}{\mathrm{c}-\mathrm{a}}=\frac{-(\mathrm{c}-\mathrm{a})}{\mathrm{c}-\mathrm{a}}\) = -1

Since slope of \(\overrightarrow{\mathrm{AB}}\) = slope of \(\overrightarrow{\mathrm{AC}}\), the point A, B, C are collinear.
Equation of the line is y – (b + c) = -1(x – a)
⇒ x + y = a + b + c

For systematic learning, students can include AP Inter 1st Year English Grammar with Answers and Transformation of Sentences in English Inter 1st Year in their daily study routine.

AP Inter 1st Year English Transformation of Sentences with Answers

The term ‘voice’ in grammar refers to one aspect of the verb. If the verb group in a sentence has ‘be+pp of the verb’, then that sentence is said to be in the passive voice. If any of or both the elements (be+V₃) are missing in the structure of the verb then the sentence is said to be in the active voice.
(వాక్యంలోనిverb group లో ‘be + V₃ ఉంటే, passive voice, be + V₃ లలో ఏ ఒక్కటి లేకున్నా active voice అంటారు.)

If the verb is in the passive form, the subject of that sentence is just the ‘sufferer’ of the action indicated by the verb.
(Verb passive లో ఉంటే ఆ వాక్యంలోని subject, verb సూచించిన పని ఫలితాన్ని అనుభవిస్తుంది. అందుకనే ఈ రూపానికి ‘passive’, ‘సోమరిగా పనిచేయకుండా అని పేరు.)

If the verb group is in the active voice, the subject of that sentence is the doer of the action shown by the verb.
(Verb Active voice ఉన్నపుడు, ఆ వాక్యం యొక్క subject, verb సూచించిన పనిని చేస్తుంది. అందుకే ఈ రూపానికి ‘Active-చురుకుగా పనిచేస్తున్న’ అని పేరు.)

If the doer of an action is either unimportant or unknown, the passive structure is natural.
(పనిచేసిన వారు ప్రదానం కాకున్నా, తెలియకున్నా, passive నిర్మాణాలు సహజంగా ఉంటాయి.)

Transformation of sentences from one voice to the other involves five steps.

1. The object in the Active Voice sentence becomes the subject in the Passive Voice sentence.
Ex :

If there are two objects in the A.V. sentence, either of them can be made the subject in the P.V. sentence.
Ex :

An interesting book was given to her brother by her.

2. A suitable ‘be’ form is to be introduced. This is the most important step. Selection of the right be’ form is based on two factors : (a) the number and person of the subject in the passive voice and (b) the tense of the verb in the Active Voice.

The following table helps one select the right ‘be’ form.

Ex :

3. The main verb in the Active Voice sentence is to be changed into its past participle form. One must know the correct past participle forms of irregular verbs.

Ex:

4. The preposition ‘by’ is used.

5. The subject in the Active voice sentence is made the object of the preposition by’ in the passive voice sentence.
Ex : He broke the glass. (A.V.)

If the subject in the A.V. sentence is either unimportant or a general one, by + object’ may be dropped.
Ex: Someone removed the dead snake. (A.V.)
The dead snake was removed. (PV.)
(by someone’ is dropped)
People call him ‘Babuji.’ (A.V.)
He is called Babuji’. (‘by people’ not necessary) (P V.)

→ Changing Imperative Sentences into Passive

We follow a slightly different method to change imperative sentences from the A.V. to the P.V. (Sentences with dropped subject and with V + object structure, conveying a request or an order are called imperative sentences.)
1. The passive voice sentence begins with Let’.
Ex: Close the door. (A.V.) imperative
Let the door be closed. (P.V.)

2. The object in the A.V. sentence becomes the subject in the P.V. sentence.

3. The ‘be’ form ‘be’ is introduced.
(Whatever be the subject or the tense form, it is always ‘be’ in imperative sentences.)
Ex. Clean the room. (A.V.)
Let the room be cleaned. (P.V.)

4. The M.V. is changed into its Past Participle form.
Ex:

As the subject is not explicit in the imperative sentences, the need to use by’ and its object doesn’t arise, while changing imperative sentences into the passive voice.

Changing Questions into Passive

Carefully study the transformation of the interrogative sentences into the passive voice.
Ex: Who brought this book ? (A.V.)
By whom was this book brought ? (P.V.)

All the steps followed here are exactly like those we follow in changing the declarative sentences. But the word order is different,
By whom + be + Subject + M.V. in PP + etc ?
Now, notice the transformation of Questions with other ‘wh’ words.
Ex : Where did you put my pen? (A.V.)
Where was my pen put ?
‘Wh’ word + be + subject + MV in PP ?

Observe how ‘yes / no’ Questions are rewritten in the passive voice.
Have you solved the problem ? (A.V.)
Has the problem been solved? (P.V.)
Helping Verb + Subject + be + MV in pp + etc ?

More Examples

1. Lalitha does homework. (AV)
Homework is done by Lalitha. (PV)

2. They are eating bananas. (AV)
Bananas are being eaten by them. (PV)

3. Teja has painted the door. (AV)
The door has been painted by Teja. (PV)

4. We saw many animals in the zoo. (AV)
Many animals were seen by us in the zoo. (PV)

5. Suresh was playing cricket. (AV)
Cricket was being played by Suresh. (PV)

6. 1 had bought three books. (AV)
Three books had been bought by me. (PV)

7. I shall watch a movie. (AV)
A movie will be watched by me. (PV)

8. You will have built a house by the end of next year. (AV)
A house will have been built by the end of next year by you. (PV)

9. He is going to post this letter. (AV)
This letter is going to be posted by him. (PV)

10. Close the door. (AV)
Let the door be closed. (PV)

11. Why have you eaten it ? (AV)
Why has it been eaten by you ? (PV)

12. Did you paint the door ? (AV)
Was the door painted by you ? (PV)

13. We can lock the gate. (AV)
The gate can be locked. (PV)

Exercise

Change the voice of the following sentences.

1. Everyone will laugh at us.
Answer:
We will be laughed at by everyone.

2. How did you solve this sum ?
Answer:
How was this sum solved by you?

3. He has lost his wallet and bag.
Answer:
His wallet and bag have been lost by him.

4. Farmers grow crops.
Answer:
Crops are grown by farmers.

5. Tell him to go.
Answer:
Let him be told to go.

6. The jury found the prisoner guilty of murder.
Answer:
The prisoner was found guilty of murder by the jury.

7. I hate people looking at me.
Answer:
I hate being looked at by people.

8. Who taught you English ?
Answer:
By whom were you taught English?

9. I bought her a present.
Answer:
A present was bought for her by me.

10. Who did this?
Answer:
By whom was this done?

11. Consult a good doctor.
Answer:
Let a good doctor be consulted.

12. I will never forget this experience.
Answer:
This experience will never be forgotten by me.

13. The tiger was chasing the deer.
Answer:
The deer was being chased by the tiger.

14. Have you finished the report ?
Answer:
Has the report been finished by you?

15. Somebody stole my pen yesterday.
Answer:
My pen was stolen yesterday by somebody.

16. Nurses look after patients.
Answer:
Patients are looked after by nurses.

17. I saw him opening the parcel.
Answer:
He was seen opening the parcel by me.

18. They are repairing chairs in the hall.
Answer:
Chairs are being repaired in the hall (by them).

19. Do not insult the weak and the oppressed.
Answer:
Let the weak and the oppressed not be insulted.

20. Prashant is solving problems.
Answer:
Problems are being solved by Prashant.

21. She has opened the door.
Answer:
The door has been opened by her.

22. She wrote a poem.
Answer:
A poem was written by her.

(B) Reported Speech or Direct / Indirect Speech

Sometimes it becomes necessary to report a person’s words. It can be done in two different ways. One way is to reproduce the actual words of the speaker. The speaker’s actual words are shown in quotation marks in writing. This kind of reporting is called the ‘Direct Speech’.

The other kind is to express the idea of the speaker in the reporter’s words. This type is referred to as the ‘Indirect Speech’.

(ఒక వ్యక్తి అన్న మాటలు అన్నట్లుగా తిరిగి చెబితే Direct Speech. కేవలం భావం మాత్రమే తిరిగి చెబితే Indirect Speech.)

The sentence in the Direct Speech has two parts. They are (a) the part out side the quotation marks (called the reporting part and (b) the part within the quotation marks (called the reported part).

Ex :

The verb in the reporting part is called “the reporting verb.”

To change a sentence from the direct speech to the indirect speech, the following steps are to be followed.

1. The reporting part is to be written first. This is applicable only when the reporting part is either in the middle or at the end of the sentence.

2. The order of the words in the reporting part is arranged as “Subject + Verb.”
This is applicable only when this order is not followed in the direct speech sentence.
Ex :

3. The reporting verb is changed to words like ordered, requested, enquired, told, asked, advised exclaimed, depending on the meaning of the reported part.
Ex :

4. The quotation marks and the comma between the reporting and reported parts are removed in the indirect speech.
Ex : The boy said,” This plant has grown very tall.” (D.S.)
The boy said that that plant had grown very tall. (I.S.)

5. A suitable connecting word is used to connect the reporting and reported parts. The selection of the right connecting word needs careful observation.

Ex:

6. The pronouns in the reported part are to be changed suitably. This depends on the first speaker, the speaker’s listener, the reporter and the reporter s listener. One has to ensure that the original speaker’s intention is correctly reported.
(Indirect speech లోకి మార్చేటప్పుడు, ఎవరు ఎవరితో అన్న మాటలను తిరిగి ఎవరు ఎవరికీ చెబుతున్నారు అనేదానిని బట్టి pronouns మారతాయి. మొదట మాట్లాడిన వ్యక్తి ఉద్దేశ్యం మారకుండా సరిగ్గా చెప్పడం అవసరం.)

Now notice the following examples carefully :

a) The teacher said to you, “You are late again.” (D.S.)
The teacher told you that you were late again. (I.S.) ‘you’ not changed.

b) The teacher said to me “You have improved your performance.” (D.S.)
The teacher told me that I had improved my performance.
You → I & Your → my

c) The teacher said to Geetha, “You have to submit your assignments tomorrow.” (D.S.)
The teacher told Geetha that she had to submit her assignments the following day. (I.S.)
You → she and your → her

d) The teacher said to Prudhvi, “When will you finish your computer course” ? (D.S.)
The teacher asked Prudhvi when he would finish his computer course. (I.S.)
You → he and your → his

e) The teacher said to the students, “You must consult your parents.” (D.S.)
The teacher told the students that they should consult their parents. (l.S.)
You → they and Your → their

As the examples a, b, c, d and e show you, pronouns change in accordance with the speaker’s reference to a person / persons.

7. In changing sentences into the indirect speech, the tense form of the verbs in the reported part is to be changed. This is the most important part of the transformation. This is done in accordance with the concept popularly known as “the sequence of tenses.”
(Reported part లోని verbs tenses మార్చడం అతి ప్రధాన భాగం. దేనిని ’Sequence of tenses’ అనే నియమానుసారంగా చేయాలి.)

The following table clearly shows when and how to change the tense.

8. The next step in the transformation from the direct to the indirect is to change the adjectives or adverbs showing ‘nearness” to those showing ‘distance’ as explained in the following table :

9. The transformation from the direct to the indirect involves a change in the word order. This principle is applicable to interrogative and exclamatory sentences only. This is exemplified in the table given below.

Note that the wh’ word in exclamatory sentences doesn’t have the meaning of a question. It just emphasises the adjective or adverb. Hence this ‘wh’ word in the direct speech becomes the intensifier ‘very’ in the indirect speech.

10. The last step in this kind of transformation is to change the Question mark’ or the ‘exclamatory mark’ into a full stop in the indirect speech.

You must note that all the Ten steps detailed above are not necessary to follow in the case of all kinds of sentences. You carefully check which steps are necessary and which steps are not necessary to follow in changing a given sentence.

Examples :

A) Statements:
1. Direct : He said to me, “I am happy to be here this evening”.
Indirect : He told me that he was happy to be there that evening.

2. Direct : He said to her, ‘I bought a house’.
Indirect : He told her that he had bought a house.

3. Direct : He said to me, ‘I shall pay your fees tomorrow.
Indirect : He told me that he would pay my fees the next day.

4. Direct : He said, The sun rises in the east’.
Indirect : He said that the sun rises in the east.

B) Questions:

1. Direct : He said, How many brothers have you ?’
Indirect : He asked me how many brothers I had.

2. Direct : “Have you no sense ?’ shouted the man.
Indirect : The man shouted if he had any sense.

3. Direct : The student asked ‘Where can I get an application form ?’
Indirect : The student enquired where he could get an application form.

4. Direct : John said ‘Will you play chess with me ?’
Indirect : John asked me whether I would play chess with him.

5. Direct : The doctor asked the patient, What did you eat today ?’
Indirect : The doctor enquired the patient what he had eaten that day.

6. Direct : Are you the manager ?’ I said.
Indirect : I asked (or enquired) whether he was the manager.

Imperative Sentences:

1. Direct : The teacher said to the boy, Get out of the room’.
Indirect : The teacher asked (or ordered) the boy to get out of the room.

2. Direct : The doctor said to the patient, Do not smoke’.
Indirect : The doctor advised the patient not to smoke.

3. Direct : Please take me to the officer’ said the visitor.
Indirect : The visitor requested them to take him to the officer.

4. Direct : Polonius said to Laertes, ‘Neither a borrower nor a lender be’.
Indirect: Polonius advised Laertes to be neither a borrower nor a lender.

5. Direct : He said, Let us wait for an hour’.
Indirect: He proposed that they should wait for an hour.

6. Direct : He said to her, ‘Don’t go out now’.
Indirect: He asked her not to go out then.

D) Exclamatory Sentences:

1. Direct : He said ‘What a fool I am !’
Indirect : He exclaimed that he was a fool.

2. Direct : ‘How selfish I have been!‘ he said.
Indirect : He exclaimed that he had been very selfish.

3. Direct : He said, ‘Alas, I am ruined!’
Indirect: He exclaimed that he was ruined.

4. Direct : He said, ‘God save the king!’
Indirect : He prayed that God might save the King.

5. Direct : ‘What a lazy girl you are!’ he said to Judy.
Indirect : He exclaimed to Judy that she was a very lazy girl.

6. Direct : ‘May our country be peaceful!’ said they.
Indirect : They expressed their wish that their country might be peaceful.

7. Direct : ‘Hurrah! I have won again’, he said.
Indirect : He exclaimed with joy that he had won again.

8. Direct : He said, ‘Alas! How foolishly have I acted!’
Indirect : He exclaimed with regret that he had acted very foolishly.

9. Direct : The captain said to the goalee, ‘Bravo ! Well done’.
Indirect: The captain applauded the goalee.

10. Direct : The frog said, ‘What a feast it was!’
Indirect: The frog exclaimed that it was a great feast.

More Examples

1. Kiran said, “I am angry.” (Direct)
Kiran said that he was angry. (Indirect)

2. Vijay said, “I’m searching for my wallet.” (Direct)
Vijay said that he was searching for his wallet. (Indirect)

3. He said, “I watched this movie last week.” (Direct)
He said that the had watched that movie the previous week. (Indirect)

4. Latha said, “I have lived here since 1970.” (Direct)
Latha said that she had lived there since 1970. (Indirect)

5. He said, “They had finished the project when I arrived.” (Direct)
He said that they had finished the project when he had arrived.” (Indirect)

6. Shreya said, “I was playing tennis when the accident occurred.” (Direct)
Shreya said that she had been playing tennis when the accident had occurred. (Indirect)

7. She said, “I have been playing the piano for two hours.” (Direct)
She said that she had been playing the piano for two hours. (Indirect)

8. Ajay said, “I had been writing a novel when the light went off.” (Direct)
Ajay said that he had been writing a novel when the light had gone off. (Indirect)

9. Vimal said, “I will shut the window.” (Direct)
Vimal said that he would shut the window. (Indirect)

10. Anji said, “I would buy Rolls Royce if I were rich.” (Direct)
Anji said that he would buy Rolls Royce if he had been rich.” (Indirect)

Exercise

Convert the following sentences into indirect speech.

Question 1.
Parnika will say, “I am not invited.”
Answer:
Parnika will say that she is not invited.

Question 2.
Our science teacher said, “Oil floats on water.”
Answer:
Our science teacher said that oil floats on water.

Question 3.
Eleanor Roosevelt said, “The future belongs to those who believe in the beauty of their dreams.”
Answer:
Eleanor Roosevelt said that the future belongs to those who believe in the beauty of their dreams.

Question 4.
Aarav said to Meghana, “You should not reveal my plan to them.”
Answer:
Aarav told Meghana that she should not reveal his plan to them.

Question 5.
Darshil said to Mayukhi, “I will meet you here tomorrow.”
Answer:
Darshil told Mayukhi that he would meet her there the next day.

Question 6.
Aarushi said, “I was watching TV at 10 pm yesterday.”
Answer:
Aarushi said that she had been watching TV at 10 pm the previous day.

Question 7.
Mahima said, “I have been studying for three hours.”
Answer:
Mahima said that she had been studying for three hours.

Question 8.
We asked the guide, “How far is the museum from here ?”
Answer:
We asked the guide how far the museum was from there.

Question 9.
Angelina said to Jennifer, “When did you submit your project ?”
Answer:
Angelina asked Jennifer when she had submitted her project.

Question 10.
Vedant asked his mother, “Can you make me a sandwich ?”
Answer:
Vedant asked his mother if she could make him a sandwich.

Question 11.
Anokhi said to Sahasra, “Do you like painting ?”
Answer:
Anokhi asked Sahasra if she liked painting.

Question 12.
Aaradhya said to her brother, “Please pass me the book.”
Answer:
Aaradhya requested her brother to pass her the book.

Question 13.
My father said to me, “Don’t waste your time.”
Answer:
My father told me not to waste my time.

Question 14.
My brother said, “Let’s go to a movie”.
Answer:
My brother suggested that we go to a movie. / My brother said that we should go to a movie.

Question 15.
The doctor said to the patient, “Wish you a speedy recovery.”
Answer:
The doctor wished the patient a speedy recovery.

Question 16.
“May God bless our country,” said the priest.
Answer:
The priest prayed that God might bless our country.

Question 17.
Jatin said, “I can solve this problem.”
Answer:
Jatin said that he could solve that problem.

Question 18.
Arunima said, “I may attend the meeting.”
Answer:
Arunima said that she might attend the meeting.

Question 19.
Yashwin said, “I must finish the report today.”
Answer:
Yashwin said that he must finish the report that day.

Question 20.
Yamini said, “I would join the team if I had time.”
Answer:
Yamini said that she would join the team if she had time.

Question 21.
The boy said, ‘‘Mom, please give me some milk.”
Answer:
The boy requested her mother to give him some milk.

Question 22.
Kishore said, “I will go to Delhi tomorrow.”
Answer:
Kishore said that he would go to Delhi the next day.

(C) Degrees of Comparison

Man craves for variety. He knows that variety is the spice of life. Hence he tries to express the same idea in various different ways. The result is the tradition of describing the quality, quantity size etc., of some thing or person in three different ways. Most adjectives and some adverbs have three different forms. They are positive, comparative and superlative. They are known as the three degrees of comparison.

(ఒక వ్యక్తి యొక్క, వస్తువు యొక్క గుణ, పరిమాణాలను మూడు భిన్న రూపాలలో వ్యక్తీకరించడం ఆనవాయితీ. ఈ రూపాలనే Degrees of comparison అని పిలుస్తారు.)

When the comparison is among a minimum of three persons/things, that can be expressed in all the three degrees. If the compared persons are two only, the superlative degree is not possible. When we talk of the quality, quantity of a single item, only positive degree is possible.

Interchange of Degrees of comparison :
Transforming one sentence from the Superlative Degree to Comparative Degree involves the following steps.

1. First, remove ‘the’ that is used before the superlative adjective, while changing it into the comparative.
2. Write the comparative form of the adjective.
3. Then, add than any other’ after the adjective.
Ex : Praveen is the youngest boy in the class. (S.D.)
Praveen is younger than any other boy in the class.(C.D.)

To change that sentence into the positive :

1. Begin the sentence with No other’.
2. Secondly, place the phrase after the superlative adjective after No other’.
3. Use ‘so + adj in positive Degree + as’ after the verb.
4. Place the subject in the S.D after ‘as’ in P.D.
Ex : He is the richest man in our town. (S.D.)

Transforming one sentence from the superlative Degree to comparative or positive Degree involves the following steps.

Sentences with one of phrase in the Superlative degree follow a slightly different wording. This is as explained here under.

Some sentences show the comparison between only two things / persons. In such cases, the superlative is not possible. The interchange is made as shown here.

Examples:

I.
Question 1.
Bengaluru is as cool as Ooty. (Into Comparative)
Answer:
Ooty is not cooler than Bengaluru.

Question 2.
Australia is the biggest island in the world. (Into other degrees)
Answer:
i) Australia is bigger than any other island in the world. (Comparative)
ii) No other island in the world is so big as Australia. (Positive)

Question 3.
The moon is brighter than the stars. (Into Positive)
Answer:
The stars are not so bright as the moon.

Question 4.
There is no vice so bad as drink. (Into Comparative)
Answer:
Drink is worse than any other vice.

Question 5.
Kashmir is the most beautiful place in India. (Into other degrees)
Answer:
i) No other place in India is so beautiful as Kashmir. (Positive)
ii) Kashmir is more beautiful than any other place in India. (Comparative)

Question 6.
Silver is not so precious as gold. (Into Comparative)
Answer:
Gold is more precious than silver.

Question 7.
}Rumble is one of the ablest bowlers. (Into other degrees)
Answer:
i) Very few bowlers are so able as Kumble. (Positive)
ii) Kumble is abler than most other bowlers.(Comparative)

Question 8.
The lion is more ferocious than any other animal. (Into other degrees)
Answer:
i) The lion is the most ferocious of all the animals. (S.D)
ii) No other animal is so ferocious as the lion. (P.D)

Question 9.
The sword is not so mighty as the pen.(Into Comparative)
Answer:
The pen is mightier than the sword. (C.D)

Question 10.
Copper is more useful than most other metals. (Into other degrees)
Answer:
i) Copper is one of the most useful metals. (S.D)
ii) Very few metals are so useful as copper. (P.D)

Question 11.
Mount Everest is higher than any other peak in the world. (Into other degrees)
Answer:
i) Mount Everest is the highest of all the peaks in the world. (S.D)
ii) No other peak in the world is so high as Mount Everest. (P.D)

Question 12.
I like you better than him. (Into Positive)
Answer:
I don’t like him as well as I like you. (P.D)

Question 13.
A deer runs faster than a horse. (Into Positive)
Answer:
A horse does not run so fast as a deer. (P.D)

Question 14.
Sangeetha is the tallest girl in the class. (Into Positive)
Answer:
No other girl in the class is so tall as Sangeetha. (P.D)

Question 15.
This is one of the most powerful earthquakes that occurred. (Into other degrees)
Answer:
i) This is more powerful than most other earthquakes that occurred. (C.D)
ii) Very few earthquakes that occurred are as powerful as the earthquake. (PD)

More Examples

Question 1.
Yashwant is not so tail as Praneeth. (Positive)
Answer:
Praneeth is taller than Yashwanth. (Comparative)
Not possibile. (Superlative)

Question 2.
Few other/very few boys are at least as tall as Praneeth. (Positive)
Answer:
Praneeth is taller than most other boys.
Praneeth is one of the tallest boys.

Question 3.
No other boy is so tall as Praneeth. (Positive)
Answer:
Praneeth is taller than any other boy. (Comparative)
Praneeth is the tallest boy. (Superlative)

Question 4.
Mumbai is bigger than Hyderabad. (Comparative)
Hyderabad is not so/as big as Mumbai. (Positive)

Question 5.
Mumbai is one of the biggest cities in the world. (Superlative)
Mumbai is bigger than most other cities in the world. (Comparative)
Few other cities in the world are as big as Mumbai. (Positive)

Question 6.
Mumbai is the biggest city in India. (Superlative)
Mumbai is bigger than any other city in India.(Comparative) No other city in India is so/as big as Mumbai. (Positive)

Exercise for Practice

Fill in the blanks with the correct form of the adjective in parentheses.

Question 1.
Chinnodu is the ___________ (small) of the three friends.
Answer:
smallest

Question 2.
Peddodu says he is ___________ (brave) than Chinnodu.
Answer:
braver

Question 3.
Bheemudu is the ___________ (tall) and the ___________ (strong) of all.
Answer:
tallest, strongest

Question 4.
Chinnodu finds a rock and says it is ___________ (big) any other rocks he’s ever seen.
Answer:
bigger

Question 5.
The water in the stream feels ___________ (cold) than they expected.
Answer:
colder

Question 6.
Bheemudu says he is the ___________ (brave) because he’s not even cold!
Answer:
bravest

Question 7.
In the meadow, Chinnodu thinks the flowers are the ___________ (pretty) he’s ever seen.
Answer:
prettiest

Question 8.
Peddodu finds a yellow flower and says it is ___________ (pretty) than the others.
Answer:
prettier

Question 9.
Bheemudu spots a purple flower and claims it’s the ___________ (pretty) flower of all.
Answer:
prettiest

Question 10.
At the end of the day, Chinnodu says he is the ___________ (tired) of the three.
Answer:
most tired

Question 11.
Peddodu yawns and says he’s even ___________ (tired) than Chinnodu.
Answer:
more tired

Question 12.
Bheemudu believes he is the ___________ (tired) because he walked the farthest.
Answer:
most tired

Exercise

Rewrite the sentences according the clues given in the brackets.

Question 1.
Rahul is one of the cleverest boys in the class.
(Convert into Comparative and Positive)
Answer:
Comparative: Rahul is cleverer than most other boys in the class.
Positive : Very few boys in the class are as clever as Rahul.

Question 2.
No other English poet is so great as Milton.
(Rewrite using ‘greater’ and ‘greatest’)
Answer:
Comparative: Milton is greater than any other English poet.
Superlative : Milton is the greatest of all English poets.

Question 3.
Keerthana is more intelligent than any other girl in the class.
(Convert into Superlative and Positive)
Answer:
Superlative: Keerthana is the most intelligent girl in the class.
Positive : No other girl in the class is as intelligent as Keerthana.

Question 4.
Very few towns are as much popular as Varanasi.
(Convert into Comparative and Superlative)
Answer:
Comparative : Varanasi is more popular than most other towns.
Superlative : Varanasi is one of the most popular towns.

Question 5.
Chennai is hotter than Mumbai. (Convert into Positive)
Answer:
Positive : Mumbai is not as hot as Chennai.

Question 6.
Hercules was stronger than any other man.
(Rewrite using ‘strongest’ and ‘strong’)
Answer:
Superlative : Hercules was the strongest man.
Positive : No other man was as strong as Hercules.

Question 7.
He is more like his father than his mother. (Convert into Positive)
Answer:
Positive : He is not as like his mother as he is like his father.

Question 8.
No other animal is so faithful as the dog.
(Convert into Comparative and Superlative)
Answer:
Comparative: The dog is more faithful than any other animal.
Superlative : The dog is the most faithful of all animals.

Question 9.
Mount Everest is higher than any other peak in the world.
(Rewrite using ‘highest’ or ‘high’ instead of ‘higher’)
Answer:
Superlative : Mount Everest is the highest peak in the world.
Positive : No other peak in the world is as high as Mount Everest.

Question 10.
Apples are not so cheap as oranges. (Convert into Comparative)
Answer:
Comparative : Oranges are cheaper than apples.

Question 11.
Manasa sings the best.
(Convert into Comparative and Positive)
Answer:
Comparative : Manasa sings better than all others.
Positive : No one sings as well as Manasa.

Question 12.
Vivekananda was more popular than most other Indian saints.
(Convert into Positive and Superlative)
Answer:
Positive : Very few Indian saints were as popular as Vivekananda.
Superlative: Vivekananda was one of the most popular Indian saints.

Question 13.
No one sings so much beautifully as Chitra.
(Rewrite using ‘better’ and ‘best’)
Answer:
Comparative : Chitra sings better than anyone else.
Superlative : Chitra sings the best of all.

Question 14.
Julie is not as pretty as her sister. (Rewrite using ‘prettier’)
Answer:
Comparative : Her sister is prettier than Julie.

Question 15.
Judith painted better than all other participants.
(Convert into Positive and Superlative)
Answer:
Positive : No other participant painted as well as Judith.
Superlative : Judith was the best painter among all the participants.

Question 16.
Sudha sings the best of all. (Write the Positive Degree)
Answer:
No other girl (person) sings as well as she does.

Question 17.
Very few plays are as much as popular as ‘Julius Caesar’.
(Write the Superlative Degree)
Answer:
‘Julius Caesar’ is one of the most popular plays.

(D) Clause Analysis and Synthesis of Sentences

Words are grouped into bigger units in various ways. Depending on their parts and function, the word groups are classified into phrases, clauses and sentences.
(పదసమూహాలలోని భాగాలను, వాటి విధులను అనుసరించి పదసముదాయాలను phrases, clauses and sentences అని వర్గీకరిస్తారు.)

A word group that has no subject, predicate distinction is called the phrase.
Ex: a clever boy (word group – no subject predicate division)
an old house – will have been completed.

A word group that has one subject and one predicate (and only one) with complete or incomplete sense is known as the clause.

Note -1: A predicate is finite verb + something. Therefore, if there is a finite verb in a group of words, that group of words can be identified as a clause.

Note – 2 : To infinites and present participles or past participles without a helping verb before them are not finite verbs. Therefore, they cannot make a group of words a clause.
Ex : I want to go. (To go – not finite – not a clause)
Seeing the police, the crowd dispersed. (Seeing-present participle-No H.V.)
Dejected, she started crying. (Dejected-p.p. No H.V.)

A group of words with one or more subjects and with complete sense is termed as the sentence.
Clauses are of two types. Clauses that have complete sense are main clauses. Clauses that have incomplete sense are subordinate clauses.
Depending on the number of clauses in a sentence and their type, sentences are divided into three kinds.

1. The Simple Sentence: A sentence which contains one main clause and no subordinate clause is called a Simple sentence. It has only one subject and one predicate.
Eg : – The sun rises in east.
1 saw a lion in the zoo.
She completed her work.

2. The Complex Sentence : A sentence which contains one main clause and one or more subordinate clauses is called a Complex sentence.
Eg : – A man, who is dead, tells no lies.
Although she is poor, she is honest.
Where there is a will, there is a way.

3. The Compound Sentence : A sentence which has two or more main clauses is a Compound sentence.
Eg:- Do it or leave it.
She took her dinner and went to bed.
They painted the wall but the paint was washed off.

Conversion from one type of sentence into another requires changing a phrase into a clause or vice versa.

• Simple to compound – a phrase into a main clause.
• Simple to complex – a phrase into a subordinate clause. Compound to simple – a main clause into a phrase.
• Complex to simple – a subordinate clause into a phrase Compound to complex – a main clause into a subordinate clause.
• Complex to compound – a subordinate clause into a main clause.

Conversion of phrases into clauses or vice versa cannot be reduced into a set of simple rules. Hence one has to practise with keen observation the transformation of simple, complex and compound sentences. However, the following lists of words will be of some help to the learners.

The correct use of the following words will help the learner construct a compound sentence.
and, but, then, or, so, yet, still, either or neither… nor; not only … but also; otherwise, else, therefore.

The words listed below are useful in writing complex sentences.
as, though, although, that, so…that, as soon as; as…as, who, what, when, where, which
(Note : ‘So’ is used both in compound and complex sentences. In complex sentences, ‘So’ is followed by “that”.)

The following phrases are helpful in writing simple sentences.
in spite of; besides, owing to; on account of; in the event of, present participles, past participles’ without helping verbs.

Examples:

I. Conversion of Simple Sentences to Compound

1. By using ‘and’:
e.g.: i) Putting off his clothes, he plunged into the river. (Simple)
He put off his clothes and plunged into the river. (Compound)

ii) The farmer went home at sun set. (Simple)
The sun set and the farmer went home.(Compound)

2. By using ‘not only …………… but also’:
e.g.: i) Besides being handsome, she is intelligent. (Simple)
She is not only handsome, but also intelligent. (Compound)

ii) In addition to useful advice, he gave me financial help. (Simple)
He gave me not only useful advice but also financial help. (Compound)

3. By using but, still, yet, etc :
e.g.: i) In spite of hard work, he could not succeed. (Simple)
He worked hard but he could not succeed. (Compound)

ii) In spite of very bad weather, they marched on. (Simple)
The weather was very bad, still they marched on. (Compound)

iii) Not withstanding her poverty, I love her. (Simple)
She is poor, yet I love her. (Compound)

4. By using conjunctions like or, otherwise, else, so, therefore, etc.
e.g.: i) You must pay the fine to get your scooter back. (Simple)
You must pay the fine or you won’t get your scooter back. (Compound)

ii) I would have been robbed but for the timely help of the policeman. (Simple)
The policeman gave me timely help, otherwise I would have been robbed. (Compound)

iii) She failed on account of her laziness. (Simple)
She was lazy, so she failed. (Compound)

iv) I was pleased with him for his honesty. (Simple)
He was honest, therefore 1 was pleased with him. (Compound)

II. Conversion of Compound Sentences into Simple

1, By using a participle :
e.g.: i) His wife died and this added to his misery. (Compound)
His wife died adding to his misery. (Simple)

ii) The players were tired and they decided to stop the play. (Compound)
Having been tired, the players decided to stop the play. (Simple)

2. By using a prepositional phrase :
e.g.: i) He is rich, yet he is not happy. (Compound)
In spite of his being rich, he is not happy.(Simple)

ii) The film is not only interesting but also it conveys a moral. (Compound)
Besides being interesting, the film conveys a moral. (Simple)

III. Conversion of Simple Sentences into Complex Sentences

By expanding a word or a phrase into a clause :
e.g.: i) He pleaded ignorance of the law. (Simple)
He pleaded that he was ignorant of the law. (Complex)

ii) A dead man tells no lies. (Simple)
A man, who is dead, tells no lies. (Complex)

iii) Immediately on reading the telegram, she wept. (Simple)
As soon as she read the telegram, she wept. (Complex)

iv) Being rich, he can buy two cars. (Simple)
As he is rich, he can buy two cars. (Complex)

v) He is working hard to get rank. (Simple)
He is working hard so that he may get rank. (Complex)

vi) The typist worked to the best of his ability. (Simple)
The typist worked as best as he could. (Complex)

vii) In spite of all his riches, he is not happy. (Simple)
Although he is rich, he is not happy. (Complex)

IV. Conversion of Complex Sentences to Simple Sentences

By Converting a clause into a word or a phrase :
e.g.: i) lam responsible for what I do. (Complex)
I am responsible for my actions. (Simple)

ii) The boy proved that he was brilliant. (Complex)
The boy proved his brilliance! (Simple)

iii) As he was ill, he stayed at home. (Complex)
He stayed at home on account of his illness .(Simple)

iv) When the teacher went out, the pupils made a lot of noise. (Complex)
The pupils made a lot of noise in the absence of the teacher. (Simple)

V. Conversion of Compound Sentences to Complex

1. Sentences containing the cumulative conjunction.
e.g.: i) He wanted to win the prize and worked hard .(Compound)
He worked hard so that he might win the prize.(CompIex)

ii) Go to the dentist and he will pull out your painful tooth. (Compound)
If you go to the dentist, he will pull out your painful tooth. (Complex)

2. Sentences containing alternative conjunctions like but, still, nevertheless, however, or otherwise, else, either ………… or, etc.
e.g.: i) Do it or you will suffer. (Compound)
Unless you do it, you will suffer. (Complex)

ii) I have known her for many years but I have never seen her. (Compound)
Although I have known her for many years, I have never seen her. (Complex)

iii) His parents were poor, still they gave him the best education. (Compound)
Though his parents were poor, they gave him the best education. (Complex)

iv) Hurry up, otherwise you will be lat^.(Compound)
Unless you hurry up, you-will be late. (Complex)

VI. Conversion of Complex Sentences to Compound

1. By using cumulative conjunction ‘and’:
e.g.: i) Everybody knows that death is inevitable.(Complex)
Death is inevitable and everybody knows it. (Compound)

ii) I am happy that you have come. (Complex)
You have come and I am happy for it. (Compound)

2. By using alternative conjunction ‘but’:
e.g.: i) Although we lost the match, we fought bravely.(Complex)
We lost the match but we fought bravely/Compound)

ii) I have found my pen that I had lost. (Complex)
I had lost my pen but I have found it. (Compound)

3. By using alternative conjunction ‘or’:
e.g.: i) Unless you control your anger, you won’t prosper. (Complex)
Control your anger or you won’t prosper. (Compound)

4. By using illusive conjunctions like so, therefore etc.,
e.g.: i) As he wants to be a doctor, he has taken Bi.P.C. (Complex)
He wants to be a doctor, therefore, he has taken Bi.P.C. (Compound)

ii) They took rest because they were tired. (Complex)
They were tired, so they took rest. (Compound)

A single sentence can be changed from one form into two other forms as seen above. Similarly, two simple sentences can be synthesised to form one simple, complexor compound sentence.

I. The following examples will illustrate the synthesis.

i) By using a participle
A participle is commonly used when the subjects of the two sentences are the same.
For example,
He was tired, He lay down on the sofa.
Being tired, he lay down on the sofa.

ii) By using a prepositional phrase (The preposition is followed by a noun.)
For example,
The magician made a flower. He used colour papers.
The magician made a flower with colour papers.

iii) By using an adverb or adverb phrase
For example,
He signed the agreement. That was foolish.
He foolishly signed the agreement.

iv) By using an infinitive
For example,
You must hurry. You will catch the train.
You must hurry to catch the train.

II. Two or more simple sentences can be combined into a compound sentence by making use of coordinating or correlative conjunctions.
For example,
His parents could not persuade him to give up smoking. Neither could I.
Neither his parents nor I could persuade him to give up smoking.
He was unwell. He attended office.
He was unwell, however, he attended office.

III. Two or more simple sentences can be combined into a complex sentence making use of subordinating conjunctions or relative pronouns and relative adverbs which act as subordinating conjunctions.
For example,
I said something, Ashwin did not listen to it.
Ashwin did not listen to what I said.
She was rich. She was not popular.
Although she was rich, she was not popular.

Examples

1. Advaith Reddy was late but he worked hard. (Compound Sentence)
Though Advaith Reddy was ill he worked hard. (Complex Sentence)
In spite of being ill, Advaith Reddy worked hard. (Simple Sentence)

2. Ram had high fever, so he was at home. (Compound)
As Ram had high fever he stayed at home. (Complex)
Because of high fever Ram stayed at home. OR (Simple)
Having high fever Ram stayed at home. (Simple)

3. Get up early or you can’t catch the train. (Compound)
If you get up early, you can catch the train. OR (Complex)
Unless you get up early, you can’t catch the train. (Complex)
Get up early to catch the train. OR (Simple)
Without getting up early, you can’t catch the train .(Simple)

4. The tree not only gives fruit, but also gives good shade. (Compound)
The tree gives good shade although it mainly gives fruit. (Complex)
Besides giving fruit, the tree gives good shade. (Simple)

More Examples

Rewrite the sentences combining them with the words given in the brackets.

Question 1.
It was cloudy. I went with an umbrella. (as)
Answer:
I went with an umbrella as it was cloudy.

Question 2.
Jhansi is a queen. She is an archer, (not only… but also)
Answer:
Jhansi is not only a queen, but also an archer.

Question 3.
She was tired. She continued to stitch the dress for me. (although)
Answer:
Although she was tired, she continued to stitch the dress for me.

Question 4.
He came to the hotel. He did not eat anything. (however)
Answer:
He came to the hotel; however, he did not eat anything.

Question 5.
That lady is very old. She cannot walk quickly, (too.. to)
Answer:
That lady is too old to walk quickly.

Question 6.
He saw the principal. He started studying. (Seeing)
Answer:
Seeing the principal, he started studying.

Question 7.
I have appointed an attendant. He will look after my grandfather, (to infinitive)
Answer:
I have appointed an attendant to look after my grandfather.

Question 8.
She is very short. She cannot pluck the fruit, (so… that)
Answer:
She is so short that she cannot pluck the fruit.

Question 9.
We must make it fast. The class will begin. (else)
Answer:
We must make it fast, else the class will begin.

Question 10.
My task is over. I want to go to a movie, (so)
Answer:
My task is over, so I want to go to a movie.

Exercise for Practice

Rewrite the sentences as directed in the parentheses.

Question 1.
In spite of facing many setbacks, he didn’t lose hope. (Change into compound and complex)
Answer:
Compound: He faced many setbacks, but he didn’t lose hope.
Complex : Although he faced many setbacks, he didn’t lose hope.

Question 2.
Although she is an illiterate, she speaks English well. (Change into simple and compound)
Answer:
Simple : Despite being an illiterate, she speaks English well.
OR
In spite of being an illiterate, she speaks English well.
Compound: She is an illiterate, but she speaks English well!

Question 3.
Speak the truth, or I will kill you. (Change into simple and complex)
Answer:
Simple : You must speake the truth to avoid being killed.
Complex : If you do not speak the truth, I will kill you.

Question 4.
If you work hard, you will get first class. (Change into compound and simple)
Answer:
Compound : Work hard, and you will get first class.
Simple : By working hard, you will get first class.

Question 5.
He is strong so he can lift heavy weights. (Change into simple and complex)
Answer:
Simple : Being strong, he can lift heavy weights.
Complex : Because he is strong, he can lift heavy weights.

Question 6.
Owing to his intelligence, he became an IAS officer. (Change into compound and complex)
Answer:
Compound : He was intelligent, and so he became an IAS officer.
Complex : Because he was intelligent, he became an IAS officer.

Question 7.
He finished his meal and went to work. (Change into complex and simple)
Answer:
Complex : After he finished his meal, he went to work.
Simple : Having finished his meal, he went to work.

Question 8.
The chief guest having taken his seat on the dais, the meeting began. (Change into complex and compound)
Answer:
Complex : After the chief guest had taken his seat on the dais, the meeting began.
Compound : The chief guest took his seat on the dais, and the meeting began.

Question 9.
Shrestha is too shy to face strangers. (Change into complex and compound)
Answer:
Complex: Shrestha is so shy that she cannot face strangers.
Compound : Shrestha is very shy, and she cannot face strangers.

Question 10.
Your writing is so illegible that I cannot understand it. (Change into compound and simple)
Answer:
Compound: Your writing is illegible, and I cannot understand it.
Simple : Your writing is too illegible to understand.

Question 11.
Lucy saw a wounded rabbit. (Change into complex)
Answer:
Complex: Lucy saw a rabbit that was wounded

Question 12.
The police recovered the child that was lost. (Change into simple)
Answer:
Simple : The police recovered the lost child.

Question 13.
Take care when crossing the road. (Change into complex)
Answer:
Complex : Take care when you are crossing the road.

Question 14.
We eat that we may live. (Change into simple)
Answer:
Simple : We eat to live.

Question 15.
No sooner did he see us than we disappeared into the night.
(Use ‘Hardly / Scarcely – when ’ or ‘As soon as ’)
Answer:
Hardly had he seen us when we disappeared into the night.
Scarcely had he seen us when we disappeared into the night.
As soon as he saw us, we disappeared into the night.

Question 16.
Scarcely had he entered the room when the clock struck twelve. (Use ‘No sooner – than/As soon as/Hardly – when ’)
Answer:
No sooner had he entered the room than the clock struck twelve.
As soon as he entered the room, the clock struck twelve.
Hardly had he entered the room when the clock struck twelve.

Question 17.
As soon as the thief saw the policeman coming, he took to his heels. (Use No sooner-than, Scarcely/Hardly – when ’)
Answer:
No sooner did the thief see the policeman coming than he took to his heels.
Scarcely had the thief seen the policeman coming when he took to his heels.
Hardly had the thief seen the policeman coming when he took to his heels.

Exercise

Two or more simple sentences can be combined with the help of coordinating conjunctions like, and, therefore, otherwise, as well as, either – or, neither – nor, not only – but also.

Question 1.
They went to the showroom. They bought a car. (Combine with ‘and’)
Answer:
They went to the showroom and bought a car.

Question 2.
The bag was heavy. I couldn’t carry it. (Combine with ‘therefore ’)
Answer:
The bag was heavy; therefore, I couldn’t carry it.

Question 3.
Run fast. You will miss the train. (Combine with ‘otherwise’)
Answer:
Run fast; otherwise, you will miss the train.

Question 4.
Siraj plays hockey. He plays cricket. (combine with ‘as well as ’)
Answer:
Siraj plays hockey as well as cricket.

Question 5.
Ramani is lazy. She is dull. (Combine with ‘either-or’)
Answer:
Ramani is either lazy or dull.

Question 6.
My father doesn’t smoke. He doesn’t drink. (Combine with ‘neither – nor)
Answer:
My father neither smokes nor drinks.

Question 7.
He was a boxer. He lectured at a university. (Combine with ‘not only – but also’)
Answer:
He was not only a boxer but also a university lecturer.

Question 8.
Besides being a singer, she is a dancer. (Combine with ‘not only – but also’)
Answer:
She is not only a singer but also a dancer.

Question 9.
She danced beautifully. I played the piano. (Combine with ‘while’)
Answer:
While she danced beautifully, I played the piano.

Question 10.
She had the required qualifications but she did not get the job. (Change into simple and complex)
Answer:
Simple : In spite of having the required qualifications, she did not get the job.
Complex : Though she had the required qualifications, she did not get the job.

Question 11.
Though she was poor, she would never ask for help. (Change into simple and compound)
Answer:
Simple : In spite of being poor, she would never ask for help.
Compound : She was poor, but she would never ask for help.

Question 12.
Unless he makes the payment in time, his house will be confiscated. (Change into compound and simple)
Answer:
Compound: He must make the payment in time, or his house will be confiscated.
Simple: Failing to make the payment in time, he will have his house confiscated.

Question 13.
Hire a taxi and you will not miss the bus. (Change into complex and simple)
Answer:
Complex : If you hire a taxi, you will not miss the bus.
Simple : By hiring a taxi, you will not miss the bus.

Question 14.
The visitors arrived so we all stood up. (Change into simple and complex)
Answer:
Simple : On the arrival of the visitors, we all stood up.
Complex : When the visitors arrived, we all stood up.

Question 15.
As it was very cold, we stayed indoors. (Change into simple and compound)
Answer:
Simple : Owing to the extreme cold, we stayed indoors.
Compound : It was very cold, so we stayed indoors.

Question 16.
I did not know what to do and called the doctor. (Change into complex and simple)
Answer:
Complex : As I did not know what to do, I called the doctor.
Simple : Not knowing what to do, I called the doctor.

Question 17.
When he heard a cry, he rushed into the room. (Change into compound and simple)
Answer:
Compound : He heard a cry and rushed into the room.
Simple : Hearing a cry, he rushed into the room.

Question 18.
The problem was very difficult and so no one could solve it. (Change into complex and simple)
Answer:
Complex : As the problem was very difficult, no one could solve it.
Simple : Due to the difficulty of the problem, no one could solve it.

Question 19.
The box was so heavy that I couldn’t lift it. (Change into compound and simple)
Answer:
Compound : The box was very heavy, so I couldn’t lift it.
Simple : The box was too heavy for me to lift.

Question 20.
A blind man needs help. (Change into complex)
Answer:
Complex : A man who is blind needs help.

Question 21.
The injured were taken to the nearest hospital. (Change into complex)
Answer:
Complex: Those who were injured were taken to the nearest hospital.

Question 22.
The students hope to pass the exams. (Change into complex)
Answer:
Complex: The students hope that they will pass the exams.

Question 23.
Please wait until told to proceed. (Change into complex)
Answer:
Complex : Please wait until you are told to proceed.

Question 24.
The box is too heavy for me to carry.(Change into complex)
Answer:
Complex : The box is so heavy that I cannot carry it.

(E) Question Tag

Question tags are short questions added to statements.
They seek confirmation or agreement. (ఏకీభావాన్ని లేదా ధృవీకరణను ఆశిస్తూ, ఒక వాక్యాన్ని అనుబంధంగా చేర్చిబడే చిన్న ప్రశ్నలే Question tags.)
Question tags have a fixed structure. It can be represented as:
Helping verb + (n’t) + pronoun?
((n’t) in brackets means it is optional.)
e.g.

1. While constructing a tag, selecting the suitable helping verb is the first and the most important step. If the given sentence has a helping verb, then that helping verb is used in the tag too.
Eg: She has answered all the questions, hasn’t she ?
has → Helping Verb in the sentence used in the tag too.

The sentence doesn’t have a helping verb, if the verb is either in the simple present or in the simple past and in the affirmative, in case of the simple present, the suitable helping verb is “do/does”. If the verb in the sentence is in the simple past, then we must use did’ in the tag.
Eg: a) She hates smokers, doesn’t she ?
hates → Simple present, hence does’ in tag.
b) They grow vegetables, don’t they ?
Simple present
c) He sold his car, did n’t he ?
Sold – Simple past; hence ‘did’ in tag.

2. The second step is to decide whether to use ‘n’t’ or not. The guiding principle is quite simple. If the given sentence doesn’t have ‘no/not, use n’t is the tag. If the given sentence has ‘no/not’, then the tag doesn’t have n’t’.
Eg : a) He was sleeping in the class, was n’t he ?
‘not’ is not used n’t’ in the tag.

b) They don’t grow rice, do they ?
‘n’t’ in sentence ‘n’t’ not used in tag
Remember to use the short form n’t’ in the tag.

3. The last part of the tag is a pronoun. This corresponds to the subject of the given sentence. If the subject of the sentence is a pronoun, the same pronoun is used in the tag. If the sub¬ject is a noun, select a suitable pronoun as shown below.

Eg : a) Gourav is a good cricketer, isn’t he ?
b) Sanya Mirza played well, didn’t she ?
c) Ooty is a beautiful hill station, isn’t it ?
d) Mangoes aren’t good this season, are they ?

4. Note that ‘amn’t is not used. Instead, aren’t is used.
Eg : I am rather slow, aren’t I ?

5. Imperative sentences take will’ as the helping verb in the tags.
Eg : a) Close the door, won’t you? (will + n’t = won’t)
b) Don’t leave the place, will you ?

6. Sentences expressing proposals with the help of “Let us ” have for their tags “Shall we ?”
Eg : Let us walk fast, shall we ?

Examples

1. The boys are playing in the hot sun, aren’t they?
2. The girls have sung well, haven’t they?
3. Hari has not learnt well, has he?
4. Hari is one of the cleverest students, isn’t he?
5. Gouri was not good at Mathematics in her S.S.C, was she?
6. Children were playing at that time, weren’t they?
7. Charan did not meet the principal, did he ?
8. Students do not use their mind power properly, do they?
9. Your mother does not become angry easily, does she?
10. Communication skills will help you a lot, won’t they?
11. He should complete the syllabus in time, shouldn’t he?
12. Reshma cannot solve this problem, can she?
13. He shall see me tomorrow, shan’t he?
14. We have never seen him before, have we?
15. There is no time to spend with them, is there?
16. I am your lecturer, aren’t I?
17. Indu and Bhargavi work hard, don’t they?
18. Ramesh treats reading a physical activity, doesn’t he?
19. Prasad spoke to Murali well, didn’t he?
20. Ravi and Ramesh seldom agree with each other, do they ?
21. He hardly helps others, does he?
22. She scarcely responded to my call, did she?
23. I am not invited, am I?
24. Let’s have a cup of tea, shall we?
25. Let me see your photograph, will you?

Exercise

Add suitable question tags to each of the following sentences given below.

Question 1.
You can’t speak Tamil, _________ ?
Answer:
can you?

Question 2.
Children often play video games on smartphones, _________ ?
Answer:
don’t they?

Question 3.
There aren’t any spiders in the bedroom, _________ ?
Answer:
are there?

Question 4.
He could always borrow the money, _________ ?
Answer:
couldn’t he?

Question 5.
They won’t tell anyone, _________ ?
Answer:
will they?

Question 6.
That was amazing, _________ ?
Answer:
wasn’t it?

Question 7.
I’m quite busy with my schedule today, _________ ?
Answer:
aren’t I?

Question 8.
They hadn’t been to the States before, _________ ?
Answer:
had they?

Question 9.
My father works at the art theatre, _________ ?
Answer:
doesn’t he?

Question 10.
Open the door, _________ ? (polite)
Answer:
will you?

Question 11.
Let’s go to the beach, _________ ?
Answer:
shall we?

Question 12.
Be quiet, _________ ? (irritation)
Answer:
will you?

Question 13.
John and Mary admire each other, _________ ?
Answer:
don’t they?

Question 14.
One shouldn’t say things like that, _________ ?
Answer:
should one?

Question 15.
Owing to the menace of mosquitoes, none of us slept well last night, _________ ?
Answer:
did we?

Question 16.
Don’t spend too much on trifles, _________ ?
Answer:
will you?

Question 17.
Nobody cares how I feel. _________ ?
Answer:
do they?

Question 18.
I think she’s from Chennai, _________ ?
Answer:
isn’t she?

Question 19.
Have another dosa. _________ ?
Answer:
won’t you?

Question 20.
A little progress has been made, _________ ?
Answer:
hasn’t it?

Question 21.
Sivamani plays the drum well. _________ ?
Answer:
doesn’t he?

Question 22.
I can lift this box, _________ ?
Answer:
can’t I ?

Referring to the AP Inter 1st Year Maths Study Material Chapter 9 Straight Lines Exercise 9a Solutions makes it easier to understand complex problems.

AP Inter 1st Year Maths Straight Lines Solutions Exercise 9a

Question 1.
Draw a quadrilateral in the cartesian plane, whose vertices are (-4, 5), (0, 7), (5, -5) and (-4, -2). Also find its area.
Solution:
Let ABCD be the given quadrilateral with vertices A(-4, 5), B(0, 7), C(5, -5) and D(-4, -2). Then by plotting A, B, C and D on the cartesian plane and joining AB, BC, CD and DA.The given quadrilateral can be drawn as

To find the area of quadrilateral ABCD, we draw one diagonal, say AC.
Accordingly area (ABCD)=Area (ΔABC)+Area (ΔACD)

We know that the area of a triangle whose vertices are (x₁, y₁), (x₂, y₂), and (x₃, y₃) is given by
\(\frac{1}{2}\)|x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|
Therefore, area of
ΔABC = \(\frac{1}{2}\)|- 4(7 + 5) + 0(-5 – 5) + 5(5 – 7)| unit²
= \(\frac{1}{2}\)|- 4(12) + 5(-2)| unit²
= \(\frac{1}{2}\) |- 48 – 10 | unit²
= \(\frac{1}{2}\) × 58 unit²
= 29 unit²

Question 2.
The base of an equilateral triangle with side 2a lies along the Y-axis such that the mid point of the base is at the origin. Find the vertices of the triangle.
Solution:
Let ABC be the given equilateral triangle with side 2a.
Accordingly AB = BC = CA = 2 a
Assume that base BC lies along the Y-axis such that the mid-point of BC is at the origin,
i. e., BO = OC = a, where O is the origin.

Now it is clear that the coordinates of point C are (0, a) while the coordinates of point B are (0, -a). It is known that the line joining a vertex at an equilateral triangle with the mid-point of its opposite side is perpendicular. Hence, vertex A lies on the X-axis.

On applying Pythagoras theorem to ΔAOC, we have (AC)² = (OA)² + (OC)²
(2a)² = (OA)² + a²
⇒ 4a² – a² = (OA)²
⇒ (OA)² = 3a²
Coordinates of point A is (±√3a, 0).
Thus the vertices of the given equilateral triangle are
(0, a)(0, -a) and (√3a, 0) (or) (0, a)(0, -a) and (-√3a, 0)

Question 3.
Find the distance between P(x₁, y₁) and Q(x₂, y₂) when
i) PQ is parallel to the Y-axis
ii) PQ is parallel to the X-axis.
Solution:
The given points are P(x₁, y₁) and Q(x₂, y₂)
i) When PQ is parallel to the Y-axis, x₁ = x₂ in this case, distance between P and Q
⇒ \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{\left(y_2-y_1\right)^2}\) = |y₂ – y₁|

ii) When PQ is parallel to the X-axis, y₁ = y₂, in this case, distance between P and Q ⇒ \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{\left(x_2-x_1\right)^2}\) = |x₂ – x₁|

Question 4.
Find a point on the X-axis, which is equidistant from the points (7, 6) and (3, 4).
Solution:
Let (a, 0) be the point on the X-axis that is equidistant from the points (7, 6) and (3, 4) accordingly
⇒ \(\sqrt{(7-a)^2+(6-0)^2}=\sqrt{(3-a)^2+(4-0)^2}\)

⇒ \(\sqrt{49+a^2-14 a+36}=\sqrt{9+a^2-6 a+16}\)

⇒ \(\sqrt{a^2-14 a+85}=\sqrt{a^2-6 a+25}\)

Squaring on both sides, we have
a² = 14a + 85 = a² – 6a + 25
⇒ -14a + 6a = 25 – 85
⇒ -8a = -60
⇒ a = \(\frac{60}{8}=\frac{15}{2}\)
Thus, the required point on the X-axis is (\(\frac{15}{2}\), 0).

Question 5.
Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P(0, -4) and B(8, 0).
Solution:
The coordinates of the mid-point of the line segment joining the points P(0, -4) and B(8, 0) are \(\left(\frac{0+8}{2}, \frac{-4+0}{2}\right)\) = (4, -2)
It is known that the slope (m) of a non-vertical line passing through the points (x₁, y₁) and (x₂, y₂) is m = \(\frac{y_2-y_1}{x_2-x_1}\), x₂ ≠ x₁
Therefore, the slope of the line passing through (0, 0) and (4, -2) is
= \(\frac{-2-0}{4-0}=\frac{-2}{4}=-\frac{1}{2}\)
Hence, the required slope of the line is –\(\frac{1}{2}\).

Question 6.
Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (-1, -1) are the vertices of a right angled triangle.
Solution:
Let the given points be A(4, 4), B(3, 5) and C(-1, -1)
Slope of \(\overline{A B}=\frac{y_2-y_1}{x_2-x_1}=\frac{5-4}{3-4}\) = -1

Slope of \(\overline{\mathrm{AC}}=\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}=\frac{-1-4}{-1-4}\) = 1

Slope of \(\overline{\mathrm{AB}}\) × slope of \(\overline{\mathrm{AB}}\) = -1
∴ AB and AC are perpendicular
∴ ΔABC is a right angled triangle

Question 7.
Find the slope of the line, which makes an angle of 30° with the positive direction of Y- axis measured anti — clockwise.
Solution:
If a line makes an angle of 300 with the positive direction of the Y—axis measured anti-clockwise, then the angle made by the line with the positive direction of the
X-axis measured anti-clockwise is 90° + 30° 120°
Thus the slope of the given line is tan 120° = tan(180° – 60°) = – tan 60° = -√3

Question 8.
Without using distance formula, show that points (-2, -1). (4, 0), (3, 3) and (-3, 2) are the vertices of a parallelogram.
Solution:
Let the points be A (-2, -1), B (4, 0), C (3, 3) and D (-3, 2).
Slope of AB = \(\frac{0+1}{4+2}=\frac{1}{6}\) = and Slope of CD = \(\frac{2-3}{-3-3}=\frac{-1}{-6}=\frac{1}{6}\)
⇒ Slope of AB = Slope of CD ⇒ AB and CD are parallel to each other.
⇒ Slope of BC = Slope of AD ⇒ BC and AD are parallel to each other.
Now, slope of BC = \(\frac{3-0}{3-4}=\frac{3}{-1}\) = -3 and Slope of AD = \(\frac{2+1}{-3+2}=\frac{3}{-1}\) = -3
Therefore, both pairs of opposite sides of a quadrilateral ABCD are parallel. Hence ABCD is a parallelogram.
Thus, points (-2, -1), (4, 0), (3, 3) and (-3, 2) are the vertices of a parallelogram.

Question 9.
Find the angle between the X-axis and the line joining the points (3, -1) and (4, -2).
Solution:
The slope of the line joining the points (3, -1) and (4, -2) is m = \(\frac{-2-(-1)}{4-3}\) = -1
Now, the inclination (θ) of the line joining the points (3, -1) and (4, -2) is given by tan θ = -1
⇒ θ = (180° – 45°) = 135°
Thus, the angle between the X-axis and the line joining the points (3, -1) and (4, -2) is 135°.

Question 10.
The slope of a line is double of the slope of another line. If tangent of the angle between them is 1/3, find the slopes of the lines.
Solution:
Let m₁ and m are the slopes of the two given lines such that m₁ = 2m.
We know that 0 is the angle between the lines l₁ and l₂ with slopes m₁ and m₂ then
tan θ = \(\left|\frac{m_1-m_2}{1+m_1 m_2}\right|\)
It is given that the tangent of the angle between the two lines is 1/3.
\(\frac{1}{3}=\left|\frac{2 m-m}{1+(2 m) m}\right|\)
⇒ \(\frac{1}{3}=\left|\frac{m}{1+2 m^2}\right|\)
⇒ \(\frac{1}{3}=\frac{\mathrm{m}}{1+2 \mathrm{~m}^2}\)
(or)
\(\frac{1}{3}=-\left(\frac{m}{1+2 m^2}\right)=\frac{-m}{1+2 m^2}\)

Case I:
\(\frac{1}{3}=\frac{\mathrm{m}}{1+2 \mathrm{~m}^2}\)
⇒ 2m² + 1 = 3m
⇒ 2m² – 3m + 1 = 0
⇒ 2m² – 2m – m + 1 = 0
⇒ 2m(m – 1) – 1(m – 1) = 0
⇒ (m – 1)(2m – 1) = 0
⇒ m = 1 or m = -½
If m = 1, then m₁ = 2m = 2(1) = 2
If m = -½, then m₁ = 2(½) = 1

Case II:
\(\frac{1}{3}=\frac{-m}{1+2 m^2}\)
⇒ 2m2 + 2m + m + 1 = 0
⇒ 2m(m + 1) + (m + 1) = 0
⇒ (m + l).(2m + 1) = 0
⇒ m = -1 (or) m = -½
If m = -1, then m₁ = 2m = 2 (-1) = – 2
If m = -½, then m₂ = 2m = 2 (-½) = – 1
Hence the slopes of the lines are 1 and 2 or ½ and 1; -1 and -2 or -½ and -1.

Question 11.
A line passes through (x₁, y₁) and (h, k). If slope of the line is m, show that k – y₁ = m (h – x₁).
Solution:
The slope of the line passing through (x₁, y₁) and (h, k) is \(\frac{k-y_1}{h-x_1}\) .
It is given that the slope of the line is m.
∴ \(\frac{k-y_1}{h-x_1}\) = m ⇒ k – y₁ = m(h – x₁).
Hence, k – y₁ = m(h – x₁).