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TS 10th Class Maths Solutions Chapter 6 Progressions InText Questions

March 5, 2024February 29, 2024 by Mahesh

Students can practice TS Class 10 Maths Solutions Chapter 6 Progressions InText Questions to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 6 Progressions InText Questions

Try This

Question 1.
i) Which of these are arithmetic progressions and why ? (P.No. 128)

a) 2, 3, 5, 7, 8, 10, 15, …
Solution:
2, 3, 5, 7, 8, 10, 15, ……. is not an A.P
∵ a₂ – a₁ = 3 – 2 = 1
a₃ – a₂ = 3 – 2 = 1
a₄ – a₃ = 7 – 5 = 1
i.e., The difference between any two successive terms is not same throughout the series.
(or)
Every number is not formed by adding a fixed number to its preceding term.

b) 2, 5, 7, 10, 12, 15, ….
Solution:
The given list does not form an A.R, since each term is not obtained by adding a fixed number to its preceding term.

c) -1, – 3, – 5, -7
-1, – 3, – 5, – 7, … is an A.P.
∵ a₂ – a₁ = -3-(-1) = -3 + 1 = -2
a₃ – a₂ = -5 – (-3) = -5 + 3 = -2
a₄ – a₃ = -7 – (-5) = -7 + 3 = -2
Every number is formed by adding a fixed number to its preceding term.

ii) Write 3 more Arithmetic Progressions.
a) a = -7; d = – 3 and
A.P. is -7, -10, -13, -16, ….
b) a = 15; d = 4 and
A.P. is 15, 19, 23, 27, 31, ….
c) a = 100; d = 50 and
A.P. is 100, 150, 200, 250, ….

Think – Discuss

Question 1.
Think how each of the list given, form an A.P. Discuss with your friends. (Page No. 129)

a) Heights (in cm) of some students of a school standing in a queue in the morning assembly are 147,148,149,…., 157.
Solution:
The given list forms an A.P. since each term starting from the second is obtained by adding a fixed number +1 to its preceding term.

b) Minimum temperatures (in °C) recorded for a week, in the month of January in a city, arranged in ascending order are – 3.1, – 3.0, – 2.9, – 2.8, – 2.7, – 2.6, – 2.5
Solution:
The given list forms an A.R, since every term starting from the second is obtained by adding a fixed number +0.1 to its preceding term.

c) The balance money (in Rs.) after paying 5% of the total loan of Rs. 1000 every month is 950, 900, 850, 800, … , 50.
Solution:
The given list forms an A.P., since each term starting from the second is obtained by adding a fixed number (-50) to its preceding term.

d) Cash prizes (in Rs.) given by a school to the toppers of Classes I to XII are 200, 250, 300, 350, …., 750 respectively.
Solution:
The given list forms an A.P., since each term starting from the second is obtained by adding a fixed number 50 to its preceding term.

e) Total savings (in Rs.) after every month for 10 months when Rs. 50 are saved each month are 50, 100, 150, 200, 250, 300, 350, 400, 450, 500.
Solution:
The given list forms an A.P., since every term starting from the second term is obtained by adding a fixed number 50 to its preceding term.

Question 2.
Find the common difference of each of the above lists. Think when is it positive. (Page No. 129)
Solution:
Common difference d = a₂ – a₁
a) 148 -147 = 1
b) -3.0 – (-3.1) = 0.1
c) 900 – 950 = – 50
d) 250 – 200 = 50
e) 100 – 50 = 50
Common difference is positive when a₂ > a₁

Question 3.
Make a positive Arithmetic progression in which the common difference is a small positive quantity. (Page No. 129)
Solution:
a = 50; d = 0.5 then A.P is 50, 50.5, 51, 51.5, 52, …

Question 4.
Make an A.P. in which the common difference is negative. (Page No. 129)
Solution:
a = 100, d = 1000 then A.P. is 100, 1100, 2100, 3100, 4100, ….

Question 5.
Make an A.P. in which the common difference is negative. (Page No. 129)
Solution:
a = 80, d = -7 then A.P is 80, 73, 66, 59, 52,…….

Do This

Question 1.
Write three examples for finite A.P. and three for infinite A.P. (Page No. 130)
Solution:
Examples for finite A.P.
i) 3, 5, 7, 9, 11 where a = 3; d = 2.

ii) x, x + a, x + 2a, x + 3a, x + 4a, x + 5a
where a = x; d = a.

iii) \(\frac{1}{9}\), \(\frac{2}{9}\), \(\frac{3}{9}\), \(\frac{4}{9}\), \(\frac{5}{9}\), \(\frac{6}{9}\), \(\frac{7}{9}\)
where a = \(\frac{1}{9}\); d = \(\frac{1}{9}\).
Examples for infinite A.P

i) 10, 20, 30, 40,…
where a = 10, d = 10.
ii) 5.5, 6.6, 7,7, 8.8, 9.9,……..
where a = 5.5; d = 1.1.
iii) -100, -95, -90, -85,………
where a = -100, d = 5.

Question 2.
Take any Arithmetic Progression. (Page No. 131)
Solution:
4, 7, 10, 13, 16,……………

Question 3.
Add a fixed number to each and every term of A.P. Write the resulting numbers as a list. Check whether the resulting lists are AP. In each case. (Page No. 131)
Solution:
4, 7, 10, 13, 16, …
Adding 5’ to each term of the above A.P we get
4 + 5, 7 + 5, 10 + 5, 13 + 5, 16 + 5,……..
9, 12, 15, 18, 21,………
In the list obtained the first term
a₁ = 9; a₂ = 12, a₃ = 15,…………
Also a₂ – a₁ = 12 – 9 = 3
a₃ – a₂ = 15 – 12 = 3
a₄ – a₃ = 18 – 15 = 3
……………………
i.e., d = a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = …… = 3
∴ The resulting list forms an AP

Question 4.
Similarly subtract a fixed number from each and every term of A.P. Write the resulting numbers as a list. Check whether the resulting lists are A.P. in each case. (Page No. 131)
Solution:
4, 7, 10, 13, 16,………….
Subtracting ‘2’ from the each term of A.P. in given series, we get

In the list obtained, the first term
a₁ = 2, a₂ = 5, a₃ = 8, a₄ = 11, ……….
Also a₂ – a₁ = 5 – 2 = 3
a₃ – a₂ = 8 – 5 = 3
a₄ – a₃ = 11 – 8 = 3
…………………………….
i.e., d = a₂ – a₁ = a₃ – a₂ = a₄ – a₃ …… = 3
∴ The resulting list forms an A:P

Question 5.
Multiply and divide each term of A.P by a fixed number and write the resulting numbers as a list.
Check whether the resulting lists are A.P in each case. (Page No. 131)
Solution:
4, 7, 10, 13, 16,………….
Multiplying each term by 3. we get

In the list obtained, the first term
a₁ = 12 and a₂ = 21, a₃ = 30………
Also a₂ – a₁ = a₃ – a₂ = ……… = 9
∴ The resulting list also forms an A.P.
Now divide every term by 7, we get
\(\frac{4}{7}\), \(\frac{7}{7}\), \(\frac{10}{7}\), \(\frac{13}{7}\), \(\frac{16}{7}\),……….. is the resulting list.

Question 6.
Check whether the resulting lists are A.P in each case.
Solution:
The first term
a₁ = \(\frac{4}{7}\), a₂ = \(\frac{7}{7}\), a₃ = \(\frac{10}{7}\), a₄ = \(\frac{13}{7}\),………
Also, a₂ – a₁ = \(\frac{7}{7}\) – \(\frac{4}{7}\) = \(\frac{3}{7}\),
a₃ – a₂ = \(\frac{10}{7}\) – \(\frac{7}{7}\) = \(\frac{3}{7}\),
a₄ – a₃ = \(\frac{13}{7}\) – \(\frac{10}{7}\) = \(\frac{3}{7}\),
∴ d = a₂ – a₁= a₃ – a₂ = a₄ – a₃ = … = \(\frac{3}{7}\) and the above list forms an A.P

Question 7.
What is your conclusion? (Page No. 131)
Solution:
If a₁, a₂, a₃, … are in A.P, then
a₁ ± k, a₂ ± k, a₃ ± k, a₄ ± k. ….. are also inA.P
a₁k, a₂k, a₃k,………. are also in A.P
\(\frac{a_1}{\mathrm{k}}\), \(\frac{a_2}{\mathrm{k}}\), \(\frac{a_3}{\mathrm{k}}\) ,…….. are also in A.P.
i.e.. “If each term of an A.P is added/multiplied/divided by a fixed number,the resulting terms also form an A.P” and fixed term is subtraded from each term of an A.P, then the resulting terms also form an A.P

Do This

Question 1.
Find the sum of Indicated number of terms in each of the following A.Ps. (Page No. 143)

i) 16, 11, 6,……… 23 terms.
Solution:
Given: 16, 11, 6,……. S₂₃
t₁ = a = 16; t₂ = 11; t₃ = 6
d = t₂ – t₁ = 11 – 16 = -5
S_n = \(\frac{\mathrm{n}}{2}\)[2a + (n-1)d]
S₂₃ = \(\frac{23}{2}\)[2 × 16 + (23 – 1) × (-5)]
= \(\frac{23}{2}\)[32 + (22) × (-5)]
= \(\frac{23}{2}\)[32 – 110]
= \(\frac{23 \times(-78)}{2}\)
= -23 × 39 = -897

ii) -0.5, -1.0, -1.5,………. 10 terms.
Solution:
Given: -0.5, -1.0, -1.5, …….. S₁₀
a = -0.5
d = t₂ – t₁ = (-1.0) – (-0.5)
= -1.0 + 0.5 = -0.5
S_n = \(\frac{\mathrm{n}}{2}\)[2a + (n – 1)d]
S₁₀ = \(\frac{10}{2}\) [ 2 × (-0.5) + (10 – 1)(-0.5)]
= 5[-1.0 + 9 × (-0.5)]
= 5[-1.0-4.5]
= 5 × (-5.5) = -27.5

iii) -1, \(\frac{1}{4}\), \(\frac{3}{2}\),……… 10 terms
Solution:
Given -1, \(\frac{1}{4}\), \(\frac{3}{2}\),……… S₁₀
a = -1
TS 10th Class Maths Solutions Chapter 6 Progressions InText Questions 3

Do This

Question 1.
Find which of the following are not GP.? (Page No. 149)

Question 1.
6, 12, 24, 48,……..
Solution:
Given : 6, 12, 24, 48,……..
a₁ = a = 6; a₂ = 12; a₃ = 24
\(\frac{\mathrm{a}_2}{\mathrm{a}_1}\) = \(\frac{12}{6}\) = 2 ; \(\frac{a_3}{a_2}\) = \(\frac{24}{12}\) = 2 ;
\(\frac{a_4}{a_3}\) = \(\frac{48}{24}\) = 2
i.e., r = \(\frac{a_2}{a_1}\) = \(\frac{a_3}{a_2}\) = \(\frac{a_4}{a_3}\) = …….. = 2
The given list is of the form
a, ar, ar², ar³,………..
∴ The given numbers are in G.P.

Question 2.
1, 4, 9, 16,………….
Solution:
Given: 1, 4, 9, 16,………
a₁ = a = 1
a₂ = 4; a₃ = 9, a₄ = 16
TS 10th Class Maths Solutions Chapter 6 Progressions InText Questions 4
∴ The given numbers do not form a G.P.

Question 3.
1, -1, 1, -1,……….
Solution:
TS 10th Class Maths Solutions Chapter 6 Progressions InText Questions 5

Question 4.
-4, -20, -100, -500,………
Solution:
Given, -4, -20, -100, -500,……….
a₁ = a = -4, a₂ = -20,
a₃ = -100, a₄ = -500,……….
TS 10th Class Maths Solutions Chapter 6 Progressions InText Questions 6
∴ The given list forms a G.P

Think – Discuss

Question 1.
Explain why each of the lists given is a G.P. (Page No. 149)

i) 1, 4, 16, 64, 256,…………
Here
a = 1 = a₁ ; a₂ = 4; a₃ = 16; a₄ = 64;….
\(\frac{a_2}{a_1}\) = \(\frac{4}{1}\) ; \(\frac{a_3}{a_2}\) = \(\frac{16}{4}\) ; \(\frac{a_4}{a_3}\) = \(\frac{64}{16}\) = 4
i.e., Common ratio r = 4.

ii) 550, 605, 665.5,………..
Solution:
The given series is in G.P Since every term can be obtained by multiplying its preceding term by a fixed number ‘1.1’.
(\(\frac{a_2}{a_1}\) = \(\frac{a_3}{a_2}\) = ……. = 1.1 = r)

iii) 256, 128, 64, 32,………….
Solution:
The given series forms a G.P Since every term, starting from the second can be obtained by multiplying its preceding term by a fixed number \(\frac{1}{2}\).
[∵ r = \(\frac{a_2}{a_1}\) = \(\frac{a_3}{a_2}\) = ………. = \(\frac{1}{2}\)]

iv) 18, 16.2, 14.58, 13.122,……….
Solution:
The given list forms a G.P
Since each term, starting from the second can be obtained by multiplying its preceding term by a fixed number 0.9.
here r = \(\frac{a_2}{a_1}\) = \(\frac{a_3}{a_2}\) = \(\frac{a_4}{a_3}\) = …………. = 0.9

Question 2.
To know about a G.P. what is minimum information that we need ? (Page No. 149)
Solution:
To know whether a number pattern forms a G.P or not, we should check that the ratio between the successive terms is equal or not.

TS 10th Class Maths Solutions Chapter 6 Progressions Optional Exercise

March 5, 2024February 29, 2024 by Mahesh

Students can practice TS Class 10 Maths Solutions Chapter 6 Progressions Optional Exercise to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 6 Progressions Optional Exercise

Question 1.
Which term of the AP:
121, 117, 113,…, is the first negative term?
[Hint: Find n for a_n < 0]
Solution:
Given A.P. : 121, 117, 113,……….
a = 121 and d = a₂ – a₁
= 117 – 121 = -4
Let the n^th term be the first negative term of the given G.P
Then, a_n = 0
⇒ a_n + (n – 1)d < 0
⇒ 121 + (n – 1)(-4) < 0
⇒ 121 – 4n + 4 < 0
⇒ 121 – 4n < 0
⇒ -4n < -125 ⇒ 4n > 125
⇒ n > \(\frac{125}{4}\)
n > 31.25
∴ When n = 32, the term becomes negative, (or) 32^nd term is the first negative term of the given A.P.

Question 2.
The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of theAP. ?
Solution:
Given in an A.P : t₃ + t₇ = 6
t₃ + t₇ = 8
a_n = t_n = a + (n – 1) d
Let the terms of the A.R be
(a – 4d), (a – 3d), (a – 2d), (a – d), a(a + d), (a + 2d), (a + 3d),….
Then a₃ = a – 2d; a₇ = a + 2d
So a₃ + a₇ = a – 2d + a + 2d = 6
⇒ 2a = 6 (or) a = 3
Also, a₃ . a₇ = (a – 2d) (a + 2d)
⇒ a² – 4d² = 8
⇒ 3² – 4d² = 8
⇒ 9 – 4d² = 8
⇒ 4d² = 9 – 8
⇒ d² = \(\frac{1}{4}\)
∴ d = ± \(\frac{1}{2}\)
Taking d = \(\frac{1}{2}\)
TS 10th Class Maths Solutions Chapter 6 Progressions Optional Exercise 6

Question 3.
A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2\(\frac{1}{2}\) m apart, what is the length of the wood required for the rungs ?
[Hint : Number of rungs = \(\frac{250}{25}\) + 1]
TS 10th Class Maths Solutions Chapter 6 Progressions Optional Exercise 8
Solution:
Given : A ladder with rungs separated by a distance between the rungs = 2\(\frac{1}{2}\)m = 250 cm
∴ Number of rungs = \(\frac{250}{25}\) + 1
= 10 + 1 = 11
Length of the bottom and top rungs = 45 cm and 25 cm
where a = 45; last term l = a₁₁ = 25 and d = 2 cm
∴ S_n = (a + l) = \(\frac{11}{2}\)(45 + 25)
= \(\frac{11}{2}\) × 70
= 35 × 11 = 385 cm
∴ Length of wood required for total rungs = 385 cm.

Question 4.
The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the
numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. And find this value of x.
[Hint : S_x-1 = S₄₉ – S_x]
Solution:
Given: Houses with numbers from 1 to 49.
x is a number x such that,
S_x-1 = S₄₉ – S_x
We know that, S_x = \(\frac{x}{2}\)[2a + (x – 1)d]
Where a = 1; d = a₂ – a₁ = 2 – 1 = 1
TS 10th Class Maths Solutions Chapter 6 Progressions Optional Exercise 9
⇒ x(x – 1) + x(x + 1) = 2 × 1225
⇒ x + x² + x = 2450
⇒ 2x² = 2450
x² = \(\frac{2450}{2}\) = 1225
⇒ x = \(\sqrt{1225}\) = ±35
⇒ x = 35
(∵ x is a counting number)

Question 5.
A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete.
Each step has a rise of \(\frac{1}{4}\)m and a tread of \(\frac{1}{2}\) m. (See Fig.). Calculate the total volume of concrete required to build the terrace.
[Hint : Volume of concrete required to build the first step = \(\frac{1}{4}\) × \(\frac{1}{2}\) × 50 m³]
TS 10th Class Maths Solutions Chapter 6 Progressions Optional Exercise 10
Solution:
Given No. of steps = 15
i.e., n = 15
Volume of concrete required to build the second step = (\(\frac{1}{4}\) + \(\frac{1}{4}\)) × \(\frac{1}{2}\) × 50
= \(\frac{1}{2}\) × \(\frac{1}{2}\) × 50
= \(\frac{25}{2}\) m³
TS 10th Class Maths Solutions Chapter 6 Progressions Optional Exercise 11
∴ Total volume of the concrete required
= 750 m³

Question 6.
150 workers were engaged to finish a piece of work in a certain number of days. Four workers dropped from the work in the second day. Four workers dropped in third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.
[Let the no.of days to finish the work is ‘x’ then
150x = \(\frac{x+8}{2}\)[2 × 150 + (x + 8 – 1)(-4)]
Solution:
Let the number of days to finish the work is x.
No. of workers engaged = 150
Given four workers dropped from the work in the second day.
Four workers dropped in the third day and so on.
∴ Common difference d = -4
8 more days to finish the work
n = x + 8
a = 150
S_n = 150x
\(\frac{\mathrm{n}}{2}\)[2a + (n – 1)d] = 150x
\(\left(\frac{x+8}{2}\right)\)[2 × 150 + (x + 8 – 1)(-4)] = 150x
(x + 8)[150 + (x + 7)(-2)] = 150x
(x + 8) [150 – 2x – 14] = 150x
(x + 8)(136 – 2x) = 150x
136x – 2x² + 1088 – 16x = 150x
-2x² + 120x + 1088 – 150x = 0
-2x² – 30x + 1088 = 0
2x² + 30x – 1088 = 0
x² + 15x – 544 = 0
x² – 17x + 32x – 544 = 0
x(x – 17) + 32(x – 17) = 0
(x – 17) (x + 32) = 0
Either x – 17 = 0 or x + 32 = 0
x = 17 (or) x = -32
Number of days can’t be negative.
∴ x = 17
∴ Number of days to complete the work is
17 + 8 = 25

Question 7.
A machine costs ₹ 5,00,000. If the value depreciates 15% in the first year, 13\(\frac{1}{2}\) % in the second year, 12% in the third year and so on. What will be its value at the end of 10 years, when all the percent ages will be applied to the original cost?
Solution:
Given : Cost price of a machine = ₹ 5,00,000
Depreciation during the years
TS 10th Class Maths Solutions Chapter 6 Progressions Optional Exercise 12
Sum of the depreciations =
15 + 13\(\frac{1}{2}\) +12+……..+10 terms
a = 15; d = a₂ – a₁
= 13\(\frac{1}{2}\) – 15 = -1\(\frac{1}{2}\) = –\(\frac{3}{2}\)

∴ Cost after 10 years = (100 – 82.5)% of 5,00,000
= 17.5% of 5,00,000
= \(\frac{17.5 \times 500000}{100}\)
= Rs. 87,500
∴ The value at the end of 10 years will be Rs. 87,500.

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5

February 29, 2024February 28, 2024 by Mahesh

Students can practice TS Class 10 Maths Solutions Chapter 6 Progressions Ex 6.5 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 6 Progressions Exercise 6.5

Question 1.
For each geometric progression find the common ratio ‘r’, and then find a_n.

i) 3, \(\frac{3}{2}\), \(\frac{3}{4}\), \(\frac{3}{8}\),………..
Solution:
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5 1

ii) 2, -6, 18, -54,…………
Solution:
Given
G.P. 2, -6, 18, -54,…………
a = 2, r = \(\frac{a_2}{a_1}\) = \(\frac{-6}{2}\) = -3
a_n = a. r^n-1 = 2 × (-3)^n-1
∴ r = -3; a_n = 2(-3)^n-1

iii) -1, -3, -9, -27,………..
Solution:
Given G.P. = – 1, – 3, – 9, – 27, ….
a = -1, r = \(\frac{a_2}{a_1}\) = \(\frac{-3}{-1}\) = 3
a_n = a . r^n-1 = (-1) × 3^n-1
∴ r = 3, a_n = (-1) × 3^n-1

iv) 5, 2, \(\frac{4}{5}\), \(\frac{8}{25}\),……….
Solution:
Given G.P: 5, 2, \(\frac{4}{5}\), \(\frac{8}{25}\),…………
a = 5, r = \(\frac{\mathrm{a}_2}{\mathrm{a}_1}\) = \(=\frac{2}{5}\)
a_n = a . r^n-1 = 5 × \(\left(\frac{2}{5}\right)^{\mathrm{n}-1}\)
∴ r = \(\frac{2}{5}\) ; a_n = 5\(\left(\frac{2}{5}\right)^{\mathrm{n}-1}\)

Question 2.
Find the 10^th and n^th term of G.P.:
5, 25, 125,……..
Solution:
Given G.P. = 5, 25, 125,……….
a = 5, r = \(\frac{a_2}{a_1}\) = \(\frac{25}{5}\) = 5
a_n = a. r^n-1 = 5 × 5^n-1 = 5^1+n-1 = 5ⁿ
∴ a₁₀ = a . r⁹ = 5 × 5⁹ = 5¹⁰
∴ a₁₀ = 5¹⁰; a_n = 5ⁿ

Question 3.
Find the indicated term of each geometric progression.

i) a₁ = 9 ; r = \(\frac{1}{3}\) ; find a₇.
Solution:
a_n = a . r^n-1
a₇ = 9 × \(\left(\frac{1}{3}\right)^{7-1}\) = 3² \(\left(\frac{1}{3}\right)^6\) = \(\frac{3^2}{3^6}\) = \(\frac{1}{3^4}\)
∴ a₇ = \(\frac{1}{3^4}\)

ii) a₁ = -12 ; r = \(\frac{1}{3}\) ; find a₆.
Solution:
a_n = a. r^n-1
a₆ = (-12) × \(\left(\frac{1}{3}\right)^{6-1}\)
= (-12) × \(\frac{1}{3^5}\) = \(\frac{-4 \times 3}{3^5}\) = \(\frac{-4}{3^4}\)
∴ a₆ = \(\frac{-4}{3^4}\)

Question 4.
Which term of the G.P.

i) 2, 8, 32,………. is 512?
Solution:
Given GP. :2, 8, 32,………. is 512
a = 2; r = \(\frac{a_2}{a_1}\) = \(\frac{8}{2}\) = 4
Let the n^th term of G.P. be 512
a_n = a . r^n-1
512 = 2 × (4)^n-1
2⁹ = 2 × (2²)^n-1
2⁹ = 2¹ × 2^2(n-1)
2⁹ = 2^2n-2+1
2⁹ = 2^2n-1
[∵ bases are equal, exponents are also equal]
∴ 2n = 9 + 1
n = \(\frac{10}{2}\) = 5
∴ 512 is the 5^th term of the given G.P

ii) \(\sqrt{3}\), 3, 3\(\sqrt{3}\),……. is 729?
Solution:
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5 2
[∵ bases are equal, exponents are also equal]
\(\frac{1}{2}\)(n-1) = 6 – 1
⇒ n – 1 = 5 × 2
⇒ n = 10 + 1 = 11
∴ 729 is the 11^th term of the given G.P

iii) \(\frac{1}{3}\), \(\frac{1}{9}\), \(\frac{1}{27}\),……….. is \(\frac{1}{2187}\) ?
Solution:
Given
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5 3
[∵ bases are equal, exponents are also equal]
∴ 7^th term of GP is \(\frac{1}{2187}\)

Question 5.
Find the 12^th term of a G.P whose 8^th term is 192 and the common ratio is 2.
Solution:
Given a G.P such that a₈ = 192 and
r = 2
a_n = a . r^n-1
∴ a₈ = a(2)^8-1 = 192
a.2⁷ = 192
⇒ a = \(\frac{192}{2^7}\) = \(\frac{192}{128}\) = \(\frac{12}{8}\) = \(\frac{3}{2}\)
∴ a₁₂ = a. r¹¹ = \(\frac{3}{2}\) × (2)¹¹
= 3 × 2¹⁰ = 3 × 1024 = 3072.

Question 6.
The 4^th term of a geometric progressions is \(\frac{2}{3}\) and the seventh term is \(\frac{16}{81}\). Find the geometric series.
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5 4
Now substituting r = \(\frac{2}{3}\) in equation (1) we get,
a\(\left(\frac{2}{3}\right)^3\) = \(\frac{2}{3}\) ⇒ a = \(\frac{2}{3}\) × \(\frac{3}{2}\) × \(\frac{3}{2}\) × \(\frac{3}{2}\) = \(\frac{9}{4}\) ×
∴ The G.P is a, ar, ar², ar³,…………..
\(\frac{9}{4}\), \(\frac{9}{4}\) × \(\frac{2}{3}\) ; \(\frac{9}{4}\) × \(\left(\frac{2}{3}\right)^2\) …….. = \(\frac{9}{4}\), \(\frac{3}{2}\), 1,……..

Question 7.
If the geometric progressions 162, 54, 18 …. and \(\frac{2}{81}\), \(\frac{2}{27}\), \(\frac{2}{9}\),……….. have their n^th term equal, find the value of n.?
Solution:
Given G.P. :
162, 54, 18,…….. and \(\frac{2}{81}\), \(\frac{2}{27}\), \(\frac{2}{9}\),………
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5 5
Given that n^th terms are equal
a_n = a. r^n-1
⇒ 162 × \(\left(\frac{1}{3}\right)^{\mathrm{n}-1}\) = \(\frac{2}{81}\) × (3)^n-1
⇒ 3^n-1 × 3^n-1 = 162 × \(\frac{81}{2}\)
⇒ 3^n-1+n-1 = 81 × 81
⇒ 3^2n-2 = 3⁸ [a^m . aⁿ = a^m+n]
⇒ 2n – 2 = 8
[∵ bases are equal, exponents are also equal]
2n = 8 + 2
∴ n = \(\frac{10}{2}\) = 5
The 5^th terms of the two G.Ps are equal.

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.4

February 29, 2024February 28, 2024 by Mahesh

Students can practice TS Class 10 Maths Solutions Chapter 6 Progressions Ex 6.4 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 6 Progressions Exercise 6.4

Question 1.
In which of the following situations, does the list of numbers involved in form a G.P. ?

i) Salary of Sharmila, when her salary is Rs. 5,00,000 for the first year and expected to receive yearly increase of 10%.
Solution:
Given : Sharmila’s yearly salary = Rs. 5,00,000
Rate of annual increment = 10%
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.4 1
Here, a = a₁ = 5,00,000

Every term starting from the second can be obtained by multiplying its preceding term by a fixed number \(\frac{11}{10}\).
∴ r = common ratio = \(\frac{11}{10}\)
Hence the situation forms a G.R

ii) Number of bricks needed to make each step, if the stair case has total 30 steps. Bottom step needs 100 bricks and each successive step needs 2 bricks less than the previous step.
Solution:
Given : Bricks needed for the bottom step = 100.
Each successive step needs 2 bricks less than the previous step.
∴ Second step from the bottom needs = 100 – 2 = 98 bricks.
Third step from the bottom needs = 98 – 2 = 96 bricks
Fourth step from the bottom needs = 96 – 2 = 94 bricks.
Here the numbers are
100, 98, 96, 94, ………….
Clearly this is an A.P. but not G.R

iii) Perimeter of the each triangle, when the mid-points of sides of an equilateral triangle whose side is 24 cm are joined to form another triangle, whose mid-points in turn are joined to form still another triangle and the process continues indefinitely.
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.4 3
Solution:
Perimeter of the 1^st equilateral triangle = 3 × 24 = 72 cm
Perimeter of the 2^nd equilateral triangle = 3 × 12 = 36 cm
Perimeter of the 3^rd equilateral triangle = 3 × 6 = 18 cm
Successive terms are obtained by dividing with 2 the preceding term except first term.
∴ The above situation is a G.P.

Question 2.
Write three terms of the G.P. when the first term ‘a’ and the common ratio ‘r’ are given.

i) a = 4; r = 3.
Solution:
The terms are a, ar, ar², ar³.
∴ 4, 4 × 3, 4 × 3², 4 × 3³
∴ 4, 12, 36, 108,…………

ii) a = \(\sqrt{5}\); r = \(\frac{1}{5}\)
Solution:
The terms of a G.P are:
a, ar, ar², ar³,……….
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.4 4

iii) a = 81; r = \(-\frac{1}{3}\).
Solution:
The terms of a G.P are:
a, ar, ar²,…………
⇒ 81, 81 × \(\left(\frac{-1}{3}\right)\), 81 × \(\left(\frac{-1}{3}\right)^2\)
⇒ 81, -27, 9,…………

iv) a = \(\frac{1}{64}\), r = 2.
Solution:
Given: a = \(\frac{1}{64}\) ; r = 2
a₁ = a = \(\frac{1}{64}\)
a₂ = ar = \(\frac{1}{64}\) × 2 = \(\frac{1}{32}\)
a₃ = ar² = \(\frac{1}{64}\) × 2² = \(\frac{1}{64}\) × 4 = \(\frac{1}{16}\)
∴ The G.P. is \(\frac{1}{16}\), \(\frac{1}{32}\), \(\frac{1}{16}\),………….

Question 3.
Which of the following are G.P.? If they are G.P, write three more terms.

i) 4, 8, 16,………….
Solution:
Given : 4, 8, 16,………….
where, a₁ = 4; a₂ = 8; a₃ = 16,………….
\(\frac{a_2}{a_1}\) = \(\frac{8}{4}\) = 2
\(\frac{\mathrm{a}_3}{\mathrm{a}_2}\) = \(\frac{16}{8}\) = 2
∴ r = \(\frac{a_2}{a_1}\) = \(\frac{a_3}{a_2}\) = 2
Hence 4,8,16,……… is a G.P
where a = 4 and r = 2
a⁴ = a. r³ = 4 × 2³ = 4 × 8 = 32
a⁵ = a.r⁴ = 4 × 2⁴ = 4 × 16 = 64
a⁶ = a. r⁵ = 4 × 2⁵ = 4 × 32 = 128

ii) \(\frac{1}{3}\), \(\frac{-1}{6}\), \(\frac{1}{12}\)………, (x ≠ 0)
Solution:
Given: t₁= \(\frac{1}{3}\); t₂ = \(\frac{-1}{6}\): t₃ = \(\frac{1}{12}\),………….
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.4 5
Hence the ratio is common between any two successive terms.
∴ \(\frac{1}{3}\), \(\frac{-1}{6}\), \(\frac{1}{12}\),………. is a G.P.

iii) 5, 55, 555,………….
Solution:
Given: t₁ = 5, t₂ = 55, t₃ = 555
\(\frac{\mathrm{t}_2}{\mathrm{t}_1}\) = \(\frac{55}{5}\) = 11
\(\frac{\mathrm{t}_3}{\mathrm{t}_2}\) = \(\frac{555}{55}\) = \(\frac{111}{11}\)
∴ \(\frac{t_2}{t_1}\) ≠ \(\frac{t_3}{t_2}\)

iv) -2, -6, -18,…………..
Solution:
Given : t₁ = -2; t₂ = -6; t₃ = -18
\(\frac{\mathrm{t}_2}{\mathrm{t}_1}\) = \(\frac{-6}{-2}\) = 3; \(\frac{t_3}{t_2}\) = \(\frac{-18}{-6}\) = 3
∴ r = \(\frac{t_2}{t_1}\) = \(\frac{t_3}{t_2}\) = …….. = 3
∴ -2, -6, -18,……….. is a G.P.
where a = -2 and r = 3
a_n =a.r^n-1
a₄ = ar³ = (-2) × 3³ = -2 × 27 = -54
a₅ = ar⁴ = (-2) × 3⁴ = – 2 × 81 = -162
a₆ = a.r⁵ = (-2) × 3⁵ = -2 × 243 = -486

v) \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{6}\), …………..
Solution:
Given: t₁ = \(\frac{1}{2}\), t₂ = \(\frac{1}{4}\), t₃ = \(\frac{1}{6}\)
\(\frac{t_2}{t_1}\) = \(\frac{\frac{1}{4}}{\frac{1}{2}}\) = \(\frac{1}{4}\) × \(\frac{2}{1}\) = \(\frac{1}{2}\)
\(\frac{t_3}{t_2}\) = \(\frac{\frac{1}{6}}{\frac{1}{4}}\) = \(\frac{1}{6}\) × \(\frac{4}{1}\) = \(\frac{2}{3}\)
∴ \(\frac{\mathrm{t}_2}{\mathrm{t}_1}\) ≠ \(\frac{t_3}{t_2}\)
i.e., \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{6}\),………… is not a G.P.

vi) 3, -3², 3³,…………
Solution:
Given : t₁ = 3; t₂ = -3², t₃ = 3³,….
\(\frac{t_2}{t_1}\) = \(\frac{-3^2}{3}\) = -3
\(\frac{\mathrm{t}_3}{\mathrm{t}_2}\) = \(\frac{3^3}{-3^2}\) = -3
∴ r = \(\frac{t_2}{t_1}\) = \(\frac{t_3}{t_2}\) = …….. = -3
i.e., every term is obtained by multiplying its preceding term by a fixed number -3.
3, -3², 3³, ……… forms a G.P
where a = 3; r = -3
aⁿ = a.r^n-1
∴ a₄ = 3 × (-3)^4-1 = 3 × (-3)³ = -81
a₅ = 3 × (-3)⁴ = 3 × 81 = 243
a₆ = 3 × (-3)⁵ = 3 × (-243) = -729

vii) x, 1, \(\frac{1}{x}\),……… (x ≠ 0)
Solution:
Given:
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.4 7

viii) \(\frac{1}{\sqrt{2}}\), -2, \(\frac{8}{\sqrt{2}}\),………..
Solution:
Given :
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.4 9
Given terms are not in G.P.

ix) 0.4, 0.04, 0.004,………
Solution:
Given : t₁ = 0.4; t₂ = 0.04; t₃ = 0.004,…………..
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.4 10

Question 4.
Find x so that x, x + 2, x + 6 are consecutive terms of a geometric progression.
Solution:
Given x, x + 2 and x + 6 are in G.P but read it as x, x + 2 and x + 6.
∴ r = \(\frac{t_2}{t_1}\) = \(\frac{t_3}{t_2}\)
⇒ \(\frac{x+2}{x}\) = \(\frac{x+3}{x+2}\)
⇒ (x + 2)² = x(x + 6)
⇒ x² + 4x + 4 = x² + 6x
⇒ 4x – 6x = -4 ⇒ -2x = -4
∴ x = 2

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

February 29, 2024February 28, 2024 by Mahesh

Students must practice these Maths 1B Important Questions TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type to help strengthen their preparations for exams.

TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type

Question 1.
If Q(h, k) is foot of the perpendicular from P(x₁, y₁) on the straight line ax + by + c = 0 then show that \(\frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{-\left(a x_1+b y_1+c\right)}{a^2+b^2}\). [May ’14, ’07; Mar. ’03]
Solution:
Let A(x₁, y₁), P(h, k)
‘P’ lies in ax + by + c = 0 then
ah + bk + c = 0
ah + bk = -c
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q1

Question 2.
Find the foot of the perpendicular from (-1, 3) on the straight line 5x – y – 18 = 0. [May ’13 (old), ’07: Mar. ’03]
Solution:
Given the equation of the straight line is 5x – y – 18 – 0
Comparing with ax + by + c = 0, we get
a = 5, b = -1, c = -18
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q2
Let the given point A(x₁, y₁) = (-1, 3)
Let (h, k) is the foot perpendicular from the point A(-1, 3) on line 5x – y – 18 = 0

∴ Foot of the perpendicular P(h, k) = (4, 2)

Question 3.
If Q(h, k) is the image of the point p(x1, y1) with respect to the straight line ax + by + c = 0 then, show that \(\frac{h-\mathbf{x}_1}{\mathbf{a}}=\frac{\mathbf{k}-\mathbf{y}_1}{\mathbf{b}}=\frac{-2\left(a \mathbf{x}_1+\mathbf{b y}_1+\mathbf{c}\right)}{a^2+b^2}\). [Mar. ’19 (AP); Mar. ’13, ’04, ’93; May ’06, ’01]
Solution:
Let A(x₁, y₁), B(h, k)
‘C’ is the midpoint of \(\overline{\mathrm{AB}}\) then
C = \(\left(\frac{\mathrm{x}_1+\mathrm{h}}{2}, \frac{\mathrm{y}_1+\mathrm{k}}{2}\right)\)
‘B’ is the image of A, then midpoint ‘C’ lies on ax + by + c = 0
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q3

Question 4.
Find the image of (1, -2) with respect to the straight line 2x – 3y + 5 = 0. [Mar. ’13]
Solution:
Given the equation of the straight line is 2x – 3y + 5 = 0
Comparing with ax + by + c = 0, we get
a = 2, b = -3, c = 5
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q4
Let the given point A(x₁, y₁) = (1, -2)
Now, B(h, k) be the image of A(1, -2) with respect to the straight line 2x – 3y + 5 = 0
\(\frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{-2\left(a x_1+b y_1+c\right)}{a^2+b^2}\)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q4.1
∴ Image of B(h, k) = (-3, 4)

Question 5.
Find the equation of the straight lines passing through the point (-3, 2) and making an angle of 45° with the straight line 3x – y + 4 = 0. [Mar. ’09]
Solution:
Given, equation of the straight line is 3x – y + 4 = 0 ……..(1)
slope of the line (1) is m = \(\frac{-3}{-1}\) = 3
Let the given point A = (-3, 2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q5
Let the slope of the required line is m₂ = m
∴ The equation of the required line is passed through (-3, 2), and having slope m is y – y₁ = m(x – x₁)
y – 2 = m(x + 3) …….(2)
y – 2 = mx + 3m
mx – y + (3m + 2) = 0
Given that the angle between the lines is 45°,
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q5.1
Squaring on both sides
⇒ 10(m² + 1) = 2(9m² + 6m + 1)
⇒ 10m² + 10 = 18m² + 12m + 2
⇒ 8m² + 12m – 8 = 0
⇒ 2m² + 3m – 2 = 0
⇒ 2m² + 4m – m – 2 = 0
⇒ (m + 2) (2m – 1) = 0
⇒ m + 2 = 0 (or) 2m – 1 = 0
⇒ m = -2 (or) \(\frac{1}{2}\)
If m = -2,
(2) ⇒ y – 2 = m(x + 3)
y – 2 = -2x – 6
2x + y + 4 = 0
If m = \(\frac{1}{2}\),
(2) ⇒ y – 2 = \(\frac{1}{2}\)(x + 3)
2y – 4 = x + 3
x – 2y + 7 = 0
∴ The equation of the straight line is 2x + y + 4 = 0
∴ The equation of straight line is x – 2y + 7 = 0
∴ The required equations of the straight line are, 2x + y + 4 = 0, x – 2y + 7 = 0

Question 6.
Find the equations of the straight lines passing through the point of intersection of the lines 3x + 2y + 4 = 0, 2x + 5y = 1 and whose distance from (2, -1) is 2. [May ’09; Mar. ’09]
Solution:
Given equations of the straight lines are
3x + 2y + 4 = 0 ……….(1)
2x + 5y – 1 = 0 ………(2)
Let the given point A = (2, -1)
∴ The equation of the straight line passing through the point of intersection of lines is (1) & (2) is L₁ + λL₂ = 0
(3x + 2y + 4) + λ(2x + 5y – 1) = 0 ……(3)
3x + 2y + 4 + 2λx + 5λy – λ = 0
(3 + 2λ)x + (2 + 5λ)y + (4 – λ) = 0
Given that the perpendicular distance from point A(2, -1) to the line (3) is 2.
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q6
squaring on both sides
(-λ + 4)² = 29λ² + 32λ + 13
λ² + 16 – 8λ = 29λ² + 32λ + 13
28λ² + 40λ – 3 = 0
28λ² + 42λ – 2λ – 3 = 0
14λ(2λ + 3) – 1(2λ + 3) = 0
(2λ + 3) (14λ – 1) = 0
(2λ + 3) = 0, 14λ – 1 = 0
λ = \(\frac{-3}{2}\); λ = \(\frac{1}{14}\)
Case I: If λ = \(\frac{-3}{2}\), then
The equation of the straight line is from (3)
(3x + 2y + 4) + (\(\frac{-3}{2}\)) (2x + 5y – 1) = 0
\(\frac{6 x+4 y+8-6 x-15 y+3}{2}=0\)
-11y + 11 = 0
y – 1 = 0
If λ = \(\frac{1}{14}\), then
The equation of the straight line is from (3)
(3x + 2y + 4) + (\(\frac{1}{14}\)) (2x + 5y – 1) = 0
\(\frac{42 x+28 y+56+2 x+5 y-1}{14}=0\)
44x + 33y + 55 = 0
4x + 3y + 5 = 0
∴ The required equations of the straight lines are y – 1 = 0 & 4x + 3y + 5 = 0

Question 7.
Find the orthocentre of the triangle whose vertices are (-5, -7), (13, 2), and (-5, 6). [Mar. ’16 (AP) ’12, ’03; May ’98]
Solution:
Let A(-5, -7), B(13, 2), C(-5, 6) are the given points.
The equation of the altitude \(\overline{\mathbf{A D}}\).
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q7

Question 8.
Find the circumcentre of the triangle whose vertices are (-2, 3), (2, -1) and (4, 0). [Mar. ’17 (AP), ’13 (Old), ’11; May ’15 (TS), ’02, ’92]
Solution:
Let A(-2, 3), B(2, -1), C(4, 0) are the given points
Let S(α, β) be the circumcentre of the triangle ABC
then, SA = SB = SC
Now, SA = SB
⇒ SA² = SB²
⇒ (α + 2)² + (β – 3)² = (α – 2)² + (β + 1)²
⇒ α² + 4 + 4α + β² + 9 – 6β = α² + 4 – 4α + β² + 1 + 2β
⇒ 4α + 9 – 6β + 4α – 1 – 2β = 0
⇒ 8α – 8β + 8 = 0
⇒ α – β + 1 = 0 …….(1)
Also SB = SC
⇒ SB² = SC²
⇒ (α – 2)² + (β + 1)² = (α – 4)² + (β – 0)²
⇒ α² + 4 – 4α + β² + 2β + 1 = α² + 16 – 8α + β²
⇒ 4 – 4α + 2β + 1 – 16 + 8α = 0
⇒ 4α + 2β – 11 = 0 ……(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q8

Question 9.
Find the orthocentre of the triangle whose sides are given by x + y + 10 = 0, x – y – 2 = 0, and 2x + y – 7 = 0. [May ’13]
Solution:
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q9

Question 10.
Find the circumcentre of the triangle whose sides are 3x – y – 5 = 0, x + 2y – 4 = 0 and 5x + 3y + 1 = 0. [May ’11, ’05; Mar. ’06]
Solution:
Given, the equations of the straight lines are
3x – y – 5 = 0 ……..(1)
x + 2y – 4 = 0 ……..(2)
5x + 3y + 1 = 0 ……..(3)
Vertex A: Solving (1) & (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q10

Let S(α, β) be the circumcentre then SA = SB = SC
Now, SA = SB
⇒ SA² = SB²
⇒ (α – 1)² + (β + 2)² = (α – 2)² + (β – 1)²
⇒ α² – 2α + 1 + β² + 4 + 4β = α² + 4 – 4α + β² + 1 – 2β
⇒ -2α + 4β + 4α + 2β = 0
⇒ 2α + 6β = 0
⇒ α + 3β = 0 ……..(4)
Also SB = SC
⇒ SB² = SC²
⇒ (α – 2)² + (β – 1)² = (α + 2)² + (β – 3)²
⇒ α² – 2α + 1 + β² – 2β + 1 = α² + 4α + 4 + β² – 6β + 9
⇒ 4α + 9 – 6β + 4α – 1 + 2β = 0
⇒ 8α – 4β – 8 = 0
⇒ 2α – β + 2 = 0 ……..(5)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q10.2

Question 11.
Find the equations of the straight lines passing through the point (1, 2) and make an angle of 60° with the line √3x + y + 2 = 0. [May ’03; B.P.]
Solution:
Given, equation of the straight line is √3x + y + 2 = 0 ……….(1)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q11
Let the given point A(x₁, y₁) = (1, 2)
Given that θ = 60°
Let the slope of the required straight line is = m
∴ The equation of the straight line is passed through A(1, 2) and having slope ‘m’ is y – y₁ = m(x – x₁)
y – 2 = m(x – 1) …….(2)
y – 2 = mx – m
mx – y – m + 2 = 0
Let ‘θ’ be the angle between the lines (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q11.1
Squaring on both sides
⇒ m² + 1 = (√3m – 1)²
⇒ m² + 1 = 3m² + 1 – 2√3m
⇒ 2m² – 2√3m = 0
⇒ m² – √3m = 0
⇒ m(m – √3) = 0
⇒ m = 0 (or) m – √3 = 0
⇒ m = 0 (or) m = √3
Case 1: If m = 0 then, the equation of a required straight line is,
from (2), y – 2 = 0(x – 1)
y – 2 = 0
y = 2
Case 2: If m = √3 then, the equation of a required straight line is
from (2), y – 2 = √3(x – 1)
y – 2 = √3x – √3
√3x – y + 2 – √3 = 0
∴ The equations of the required straight lines are y = 2 and √3x – y + 2 – √3 = 0

Question 12.
If p and q are the lengths of the perpendiculars from the origin to the straight lines x sec α + y cosec α = a and x cos α – y sin α = a cos 2α prove that 4p² + q² = a². [Mar. ’13 (Old), ’08; May ’13]
Solution:
Given, the equations of the straight lines are
x sec α + y cosec α – a = 0 …….(1)
x cos α – y sin α = a cos 2α ……….(2)
Let the point p(x₁, y₁) = (0, 0)
Now, p = the perpendicular distance from the point p(0, 0) to the straight line (1)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q12
q = the perpendicular distance from the point p(0, 0) to the straight line (2)

L.H.S = 4p² + q²
= 4(a sin α cos α)² + (a cos 2α)²
= 4a² sin²α cos²α + a² cos² 2α
= a²(2 sin α cos α)² + a² cos² 2α
= a² sin² 2α + a² cos² 2α
= a²(sin² 2α + cos² 2α)
= a²
= R.H.S
∴ 4p² + q² = a²

Some More Maths 1B Straight Lines Important Questions

Question 13.
Find the equation of the straight line which makes an angle 135° with the positive X-axis measured counterclockwise and passing through the point (-2, 3).
Solution:
Given that, the inclination of a line is θ = 135°
The slope of a straight line is, m = tan θ
= tan 135°
= tan (180° – 45°)
= -tan 45°
= -1
Let the given point A(x1, y1) = (-2, 3)
∴ The equation of the straight line passing through the point (-2, 3) and having slope ‘-1’ is,
y – y₁ = m(x – x₁)
y – 3 = -1(x + 2)
y – 3 = -x – 2
x + y – 1 = 0

Question 14.
Find the equation of the straight line passing through the point (2, 3) and making non-zero intercepts on the axes of coordinates whose sum is zero. [Mar. ’12; May ’09]
Solution:
The equation of a straight line in the intercept form is
\(\frac{x}{a}+\frac{y}{b}\) = 1 ………(1)
Given that, the sum of the intercepts = 0
a + b = 0
b = -a
From (1), \(\frac{x}{a}+\frac{y}{-a}=1\)
x – y = a ……(2)
Since equation (2) passes through the point (2, 3) then,
2 – 3= a
∴ a = -1
Substitute the value of ‘a’ in equation (2)
x – y = -1
x – y + 1 = 0
∴ The equation is x – y + 1 = 0.

Question 15.
Find the equation of the straight line passing through points (4, -3) and perpendicular to the line passing through points (1, 1) and (2, 3).
Solution:
Let A(1, 1), B(2, 3), C(4, -3) are the given points
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q3
Slope of the perpendicular line = \(\frac{-1}{m}=\frac{-1}{2}\)
∴ The equation of the straight line passing through C(4, -3) and having slope \(\frac{-1}{2}\) is y – y₁ = \(\frac{1}{m}\) (x – x₁)
y + 3 = \(\frac{-1}{2}\) (x – 4)
2y + 6 = -x + 4
x + 2y+ 6 – 4 = 0
x + 2y + 2 = 0

Question 16.
Transform the equation 3x + 4y + 12 = 0 into (i) slope-intercept form (ii) intercept form (iii) normal form.
Solution:
(i) Slope-intercept form:
Given, the equation of the straight line is 3x + 4y + 12= 0
4y = -3x – 12
y = \(\frac{-3 x}{4}-\frac{12}{4}=\left(\frac{-3}{4}\right) x+(-3)\)
which is in the form of y = mx + c
Slope, m = \(\frac{-3}{4}\), y-intercept, c = -3
(ii) Intercept form:
Given, the equation of the straight line is 3x + 4y + 12 =0
3x + 4y = -12 × 1
\(\frac{3 x}{-12}+\frac{4 y}{-12}=1\)
\(\frac{x}{-4}+\frac{y}{-3}=1\)
which is in the form of \(\frac{x}{a}+\frac{y}{b}=1\)
∴ x-intercept, a = -4, y-intercept, b = -3
(iii) Normal form:
Given the equation of the straight line is, 3x + 4y + 12 = 0
3x + 4y= -12
-3x – 4y = 12
On dividing both sides with \(\sqrt{a^2+b^2}\)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q4
which is in the form x cos α + y sin α = p
∴ cos α = \(\frac{-3}{5}\), sin α = \(\frac{-4}{5}\), p = \(\frac{12}{5}\)

Question 17.
Find the distance between the parallel straight lines 5x – 3y – 4 = 0, 10x – 6y – 9 = 0. [Mar. ’13(old), ’09, ’02; May ’12]
Solution:
Given, the equations of the straight lines are
5x – 3y – 4 = 0
10x – 6y – 9 = 0 …….(2)
2(5x – 3y – 4) = 0
10x – 6y – 8 = 0 ………(1)
Comparing (1) with ax + by + c₁ = 0, we get
a = 10, b = -6, c₁ = -8
Comparing (2) with ax + by + c₂ = 0, we get
a = 10, b = -6, c₂ = -9
Distance between the parallel lines =
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q5

Question 18.
Find the value of p, if the straight lines 3x + 7y – 1 = 0 and 7x – py + 3 = 0 are mutually perpendicular.
Solution:
Given, the equations of the straight lines are
3x + 7y – 1 = 0 ……..(1)
7x – py + 3 = 0 ……..(2)
Slope of the line (1) is m₁ = \(\frac{-3}{7}\)
Slope of the line (2) is m₂ = \(\frac{-7}{-p}=\frac{7}{p}\)
Since the given lines are perpendicular then,
m₁ × m₂ = -1
\(\frac{-3}{7} \times \frac{7}{p}\) = -1
p = 3

Question 19.
Find the value of p, if the lines 3x + 4y = 5, 2x + 3y = 4, px + 4y = 6 are concurrent. [May ’06]
Solution:
Given, the equations of the straight lines are,
3x + 4y – 5 = 0 ……..(1)
2x + 3y – 4 = 0 ……..(2)
px + 4y – 6 = 0 …….(3)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q7
∴ The point of intersection of the straight lines (1) & (2) is (-1, 2)
Now, given lines are concurrent then, the point of intersection lies on the line
p(-1) + 4(2) – 6 = 0
-p + 8 – 6 = 0
-p + 2 = 0
p = 2

Question 20.
Find the value of p, if the lines 4x – 3y – 7 = 0, 2x + py + 2 = 0, 6x + 5y – 1 = 0 are concurrent.
Solution:
Given, the equations of the straight lines are,
4x – 3y – 7 = 0 …….(1)
2x + py + 2 = 0 ……..(2)
6x + 5y – 1 = 0 ……….(3)
Solving (1) & (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q8
∴ The point of intersection of the straight lines (1) & (3) is, (1, -1).
Since the given lines are concurrent, then the point of intersection (1, -1) lies on line (2).
2(1) + p(-1) + 2 = 0
2 – p + 2 = 0
4 – p = 0
p = 4

Question 21.
Show that the lines 2x + y – 3 = 0, 3x + 2y – 2 = 0 and 2x – 3y – 23 = 0 are concurrent and find the point of concurrent.
Solution:
Given, the equations of the straight lines are,
2x + y – 3 = 0 …….(1)
3x + 2y – 2 = 0 ……….(2)
2x – 3y – 23 = 0 ………(3)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q9
∴ The point of intersection of the straight lines (1) & (2) is (4, -5).
Now substitute the point (4, -5) in equation (3)
2(4) – 3(-5) – 23 = 0
8 + 15 – 23 = 0
23 – 23 = 0
0 = 0
∴ Given lines are concurrent.
∴ Point of concurrence = (4, -5).

Question 22.
A straight line meets the coordinate axes in A and B. Find the equation of the straight line, when \(\overline{\mathbf{A B}}\) is divided in the ratio 2 : 3 at (-5, 2).
Solution:
The equation of the straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) ……..(1)
The straight line (1) meets the X-axis at A(a, 0), Y-axis at B(0, b).
Now, P(-5, 2) divides \(\overline{\mathbf{A B}}\) in the ratio 2 : 3
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q10

Question 23.
A straight line meets the coordinate axes in A and B. Find the equation of the straight line when \(\overline{\mathbf{A B}}\) is divided in the ratio 1 : 2 at (-5, 4).
Solution:
The equation of the straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) ……..(1)
The straight line (1) meets the X-axis at A(a, 0), Y-axis at B(0, b).
Now, P(-5, 4) divides \(\overline{\mathbf{A B}}\) in the ratio 1 : 2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q11

Question 24.
If non-zero numbers a, b, c are in harmonic progression, then show that the equation \(\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0\) represents a family of concurrent lines and find the point of concurrency.
Solution:
Given that, a, b, c are in harmonic progression.
then, \(\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\)
\(\frac{2}{b}-\frac{1}{a}=\frac{1}{c}\)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q12
bx + ay + 2a – b = 0
b(x – 1) + a(y + 2) = 0
(x – 1) + \(\frac{a}{b}\) (y + 2) = 0
This is of the form L₁ + λL₂ = 0
∴ \(\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0\) represents a set of lines passing through the point of intersection of
L₁ = x – 1 = 0 ……..(1)
L₂ = y + 2 = 0 ………(2)
Solve (1) & (2)
from (1), x – 1 = 0 ⇒ x = 1
from (2) y + 2 = 0 ⇒ y = -2
∴ The point of concurrence = (1, -2).
∴ \(\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0\) represents a set of concurrent lines.

Question 25.
If a, b, c are in arithmetic progression, then show that the equation ax + by + c = 0 represents a family of concurrent lines and find the point of concurrency.
Solution:
Given that, a, b, c are in arithmetic progression, then
2b = a + c
c = 2b – a
Now, ax + by + c = 0
ax + by + (2b – a) = 0
ax + by + 2b – a = 0
a(x – 1) + b(y + 2) = 0
x – 1 + \(\frac{b}{a}\) (y + 2) = 0
This is of the form L₁ + λL₂ = 0
∴ ax + by + c = 0 represents a set of lines passing through the point of intersection of
L₁ = x – 1 = 0 …….(1)
L₂ = y + 2 = 0 ……..(2)
Solve (1) & (2)
from (1), x – 1 = 0 ⇒ x = 1
from (2), y + 2 = 0 ⇒ y = -2
∴ The point of concurrence = (1, -2).
∴ ax + by + c = 0 represents a set of concurrent lines.

Question 26.
A straight line parallels the line y = √3x passes through Q(2, 3), and cuts the line 2x + 4y – 27 = 0 at p then, finds the length of PQ.
Solution:
Given the equation of the straight line is y = √3x
This is of the form y = mx
⇒ m = √3
⇒ tan θ = √3
⇒ θ = 60°
Since \(\overline{\mathrm{PQ}}\) is parallel to the straight line y = √3x
∴ Slope of \(\overline{\mathrm{PQ}}\) = √3
Inclination of a straight line \(\overline{\mathrm{PQ}}\), 0 = 60°
Given point Q(x₁, y₁) = (2, 3)
∴ The coordinates of P = (x₁ + r cos θ, y₁ + r sin θ)
= (2 + r cos 60°, 3 + r sin 60°)
= (2 + r(\(\frac{1}{2}\)), 3 + \(\frac{\sqrt{3} r}{2}\)) where |r| = PQ
Since the point, ‘p’ lies on the straight line 2x + 4y – 27 = 0
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q14

Question 27.
A straight line with slope 1 passes through Q(-3, 5) and meets the line x + y – 6 = 0 at P. Find the distance PQ.
Solution:
Given, equation of the straight line is x + y – 6 = 0 ……….(1)
The slope of the straight line m = 1
Given point Q(x₁, y₁) = (-3, 5)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q15
∴ The equation of the straight line \(\overline{\mathrm{PQ}}\) having slope ‘1’ and passes through the point Q(-3, 5) is (y – y₁) = m(x – x₁)
y – 5 = 1(x + 3)
y – 5 = x + 3
x – y + 8 = 0 …….(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q15.1

Question 28.
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:
Given, equation of the straight line is x + y – 7 = 0 ……..(1)
Given point Q(x₁, y₁) = (2, 3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q16
Given that the straight line makes an angle \(\frac{3 \pi}{4}\) with the ‘-ve’ direction of the X-axis, then the straight line makes an angle π – \(\frac{3 \pi}{4}\) = \(\frac{\pi}{4}\) = 45° with the positive direction of X-axis.
∴ The slope of the straight line is m = tan θ
= tan 45°
= 1
The equation of the straight line \(\overline{\mathrm{PQ}}\) having slope ‘1’ and passing through the point Q(2, 3) is,
y – y₁ = m(x – x₁)
y – 3 = 1(x – 2)
y – 3 = x – 2
x – y + 1 = 0 …….(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q16.1

Question 29.
Find the foot of the perpendicular drawn from (4, 1) upon the straight line 3x – 4y + 12 = 0. [May ’08; June ’02]
Solution:
Given the equation of the straight line is 3x – 4y + 12 = 0.
Comparing with ax + by + c = 0, we get
a = 3, b = -4, c = 12
Let the given point A(x₁, y₁) = (4, 1)
Let (h, k) is the foot perpendicular from the point A(4, 1) on the line 3x – 4y + 12 = 0
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q17

Question 30.
Find the foot of the perpendicular drawn from (3, 0) upon the straight line 5x + 12y – 41 = 0.
Solution:
Given the equation of the straight line is 5x + 12y – 41 = 0
Comparing with ax + by + c = 0, we get
a = 5, b = 12, c = -41
Let the given point A(x₁, y₁) = (3, 0)
Let (h, k) is the foot of the perpendicular from the point A(3, 0) on the line 5x + 12y – 41 = 0
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q18

Question 31.
Find the image of the point (1, 2) in the straight line 3x + 4y – 1 = 0. [Mar. ’07, ’02, ’97; May ’04, ’97]
Solution:
Given the equation of the straight line is 3x + 4y – 1 = 0
Comparing with ax + by + c = 0, we get
a = 3, b = 4, c = -1
Let the given point A(x₁, y₁) = (1, 2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q19

Question 32.
x – 3y – 5 = 0 is the perpendicular bisector of the line segment joining points A, B. If A = (-1, -3) find the coordinates of B. [Mar. ’13 (Old)]
Solution:
Given, equation of the straight line is x – 3y – 5 = 0
Comparing with ax + by + c = 0, we get
a = 1, b = -3, c = -5
Let the given point A(x₁, y₁) = (-1, -3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q20
x – 3y – 5 = 0 is the perpendicular bisector of the line segment joining the points A, B then, B(h, k) is the image of A(-1, -3) with respect to the straight line x – 3y – 5 = 0.

Question 33.
If 2x – 3y – 5 = 0 is the perpendicular bisector of the line segment joining (3, -4) and (α, β), find α + β.
Solution:
Given, the equation of the straight line is 2x – 3y – 5 = 0
Comparing with ax + by + c = 0, we get
a = 2, b = -3, c = -5
Let the given point A(x₁, y₁) = (3, -4)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q21
2x – 3y – 5 = 0 is the perpendicular bisector of the line segment joining the points A, B then, B(α, β) is the image of A(3, -4) with respect to the straight line 2x – 3y – 5 = 0.

∴ Image of A(3, 4) is (α, β) = (-1, 2)
∴ α + β = -1 + 2 = 1

Question 34.
Find the orthocentre of the triangle whose vertices are (-2, -1), (6, -1), and (2, 5). [May ’12; Mar. ’07, ’04]
Solution:
Let the given vertices are A(-2, -1), B(6, -1) & C(2, 5)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q22

Question 35.
Find the orthocentre of the triangle whose vertices are (5, -2), (-1, 2), and (1, 4).
Solution:
Let the given points are A(5, -2), B(-1, 2) & C(1, 4)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q23

Question 36.
Find the circumcentre of the triangle whose vertices are (1, 3), (0, -2), and (-3, 1). [Mar. ’18 (AP); May ’06]
Solution:
Let A(1, 3), B(0, -2), C(-3, 1) are the given points.
Let S(α, β) be the circumcentre of the triangle ABC then, SA = SB = SC
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q24
Now, SA = SB
⇒ SA² = SB²
⇒ (α – 1)² + (β – 3)² = (α – 0)² + (β + 2)²
⇒ α² – 2α + 1 + β² + 9 – 6β = α² + β² + 4 + 4β
⇒ 4 + 4β + 2α – 9 + 6β – 1 = 0
⇒ 2α + 10β – 6 = 0
⇒ α + 5β – 3 = 0 ……(1)
Also SB = SC
⇒ SB² = SC²
⇒ (α – 0)² + (β + 2)² = (α + 3)² + (β – 1)²
⇒ α² + β² + 4 + 4β = α² + 9 + 6α + β² + 1 – 2β
⇒ 9 + 6α – 2β + 1 – 4 – 4β = 0
⇒ 6α – 6β + 6 = 0
⇒ α – β + 1 = 0 ………(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q24.1

Question 37.
Find the circumcentre of the triangle whose vertices are (1, 3), (-3, 5), and (5, -1). [Mar. ’18 (TS)]
Solution:
Let A(1, 3), B(-3, 5), C(5, -1) are the given points.
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q25
Let S(α, β) be the circumcentre of the triangle ABC then, SA = SB = SC
Now, SA = SB
⇒ SA² = SB²
⇒ (α – 1)² + (β – 3)² = (α + 3)² + (β – 5)²
⇒ α² – 2α + 1 + β² + 9 – 6β = α² + 9 + 6α + β² + 25 – 10β
⇒ 6α + 25 – 10β – 1 + 2α + 6β = 0
⇒ 8α – 4β + 24 = 0
⇒ 2α – β + 6 = 0 ……(1)
Also SB = SC
⇒ SB² = SC²
⇒ (α + 3)² + (β – 5)² = (α – 5)² + (β + 1)²
⇒ α² + 9 + 6α + 25 + β² – 10β = α² + 25 – 10α + β² + 1 + 2β
⇒ 9 + 6α – 10β + 10α – 1 – 2β = 0
⇒ 16α – 12β + 8 = 0
⇒ 4α – 3β + 2 = 0 ………(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q25.1

Circumcentre S = (-8, -10)

Question 38.
Find the circumcentre of the triangle whose vertices are (1, 0), (-1, 2), and (3, 2). [May ’13 (Old)]
Solution:
Let A(1, 0), B(-1, 2), C(3, 2) are the given points.
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q26
Let S(α, β) be the circumcentre of the triangle ABC then, SA = SB = SC
Now, SA = SB
⇒ SA² = SB²
⇒ (α – 1)² + (β – 0)² = (α + 1)² + (β – 2)²
⇒ α² – 2α + 1 + β² = α² + 2α + 1 + β² + 4 – 4β
⇒ 2α – 4β + 4 + 1 – 1 + 2α = 0
⇒ 4α – 4β + 4 = 0
⇒ α – β + 1 = 0 ……….(1)
Also SB = SC
⇒ SB² = SC²
⇒ (α + 1)² + (β – 2)² = (α – 3)² + (β – 2)²
⇒ α² + 2α + 1 + β² + 4 – 4β = α² + 9 – 6α + β² + 4 – 4β
⇒ 1 + 2α – 9 + 6α = 0
⇒ 8α – 8 = 0
⇒ α – 1 = 0
⇒ α = 1
Substitute α = 1 in (1), we get
1 – β + 1 = 0
⇒ -β = -2
⇒ β = 2
∴ Circumcentre S = (1, 2)

Question 39.
If the equations of the sides of a triangle are 7x + y – 10 = 0, x – 2y + 5 = 0, and x + y + 2 = 0, find the orthocenter of the triangle. [May ’09, ’00, ’97]
Solution:
Given, the equations of the straight lines are
7x + y – 10 = 0 …….(1)
x – 2y + 5 = 0 …….(2)
x + y + 2 = 0 ……….(3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q27

Question 40.
Find the orthocentre of the triangle whose sides are given by 4x – 7y + 10 = 0, x + y = 5, and 7x + 4y = 15.
Solution:
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q28

Question 41.
Find the orthocentre of the triangle formed by the lines x + 2y = 0, 4x + 3y – 5 = 0 and 3x + y = 0. [Mar. ’10]
Solution:
Given, the equations of the straight lines are,
x + 2y = 0 ……..(1)
4x + 3y – 5z = 0 ……….(2)
3x + y = 0 ………(3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q29

Question 42.
Find the circumcentre of the triangle formed by the straight lines x + y = 0, 2x + y + 5 = 0, and x – y = 2.
Solution:
Given, the equations of the straight lines are
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q30

∴ Vertex C = (-1, -3)
Let S(α, β) be the circumcentre of the triangle then, SA = SB = SC

Now, SA = SB
⇒ SA² = SB²
⇒ (α – 1)² + (β + 1)² = (α + 5)² + (β – 5)²
⇒ α² – 2α + 1 + β² + 1 + 2β = α² + 25 + 10α + β² + 25 – 10β
⇒ 50 + 10α – 10β – 1 + 2α – 2β – 1 = 0
⇒ 12α – 12β + 48 = 0
⇒ α – β + 4 = 0 ……..(4)
Also SB = SC
⇒ SB² = SC²
⇒ (α + 5)² + (β – 5)² = (α + 1)² + (β + 3)²
⇒ α² + 25 + 10α + β² + 25 – 10β = α² + 2α + 1 + β² + 6β + 9
⇒ 25 + 25 + 10α – 10β – 1 = -2α – 9 – 6β = 0
⇒ 8α – 16β + 40 = 0
⇒ α – 2β + 5 = 0 ……..(5)
Solving (4) & (5)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q30.3
α = -3; β = 1
∴ Circumcentre S = (-3, 1)

Question 43.
Find the circumcentre of the triangle formed by the straight lines x + y + 2 = 0, 5x – y – 2 = 0, and x – 2y + 5 = 0. [May ’08]
Solution:
Given, the equations of the straight lines are
x + y + 2 = 0 ………(1)
5x – y – 2 = 0 ………(2)
x – 2y + 5 = 0 ………(3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q31

Let S(α, β) be the circumcentre of the triangle then, SA = SB = SC
Now, SA = SB
⇒ SA² = SB²
⇒ (α + 3)² + (β – 1)² = (α – 0)² + (β + 2)²
⇒ α² + 9 + 6α + β² + 1 – 2β = α² + β² + 4 + 4β
⇒ 6α – 2β + 9 + 1 – 4 – 4β = 0
⇒ 6α – 6β + 6 = 0
⇒ α – β + 1 = 0 ……..(4)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q31.2
Also SB = SC
⇒ SB² = SC²
⇒ (α + 0)² + (β + 2)² = (α – 1)² + (β – 3)²
⇒ α² + β² + 4 + 4β = α² – 2a + 1 + β² – 6β + 9
⇒ 4 + 4β – 1 + 2α – 9 + 6β = 0
⇒ 2α + 10β – 6 = 0
⇒ α + 5β – 3 = 0 ……..(5)
Solving (4) & (5)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q31.3

Question 44.
Find the equations of the straight lines passing through the point (-10, 4) and make an angle θ with the line x – 2y = 10 such that tan θ = 2.
Solution:
Given, equation of the straight line is x – 2y – 10 = 0 ……..(1)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q32
Let the given point P(x₁, y₁) = (-10, 4)
Given that, tan θ = 2
cos θ = \(\frac{1}{\sqrt{5}}\)

Let the slope of the required straight line = m.
∴ The equation of the straight line is passed through the point P(-10, 4) and having slope ‘m’ is y – y₁ = m(x – x₁)
y – 4 = m(x + 10)
y – 4 = mx + 10m
mx – y + 10m + 4 = 0
If ‘θ’ be the angle between the lines (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q32.2
Squaring on both sides
5(m² + 1) = 5(m + 2)²
5m² + 5 = 5m² + 20 + 20m (or) m = \(\frac{1}{0}\)
20m + 15 = 0
20m = -15
m = \(\frac{-3}{4}\) or m = \(\frac{1}{0}\)
Case 1: If m = \(\frac{-3}{4}\) then, the equation of the straight line is,
from (2), y – 4 = \(\frac{-3}{4}\) (x + 10)
4y – 16 = -3x – 30
3x + 4y – 16 + 30 = 0
3x + 4y + 14 = 0
Case 2: If m = \(\frac{1}{0}\) then, the equation of straight line is
from (2), y – 4 = \(\frac{1}{0}\) (x + 10)
0 = x + 10
x + 10 = 0
∴ Required equations of the straight lines are 3x + 4y + 10 = 0, x + 10 = 0

Question 45.
The base of an equilateral triangle is x + y – 2 = 0 and the opposite vertex is (2, -1). Find the equations of the remaining sides. [Mar. ’02]
Solution:
Given that base of an equilateral triangle is, x + y – 2 = 0 ……(1)
Let the opposite vertex is, A = (2, -1)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q33
Since, Triangle ABC is an equilateral triangle B then,
A = B = C = 60°
Let the slope of the side \(\overline{\mathrm{AB}}\) = m
∴ The equation of the side \(\overline{\mathrm{AB}}\) passing through A(2, -1) and having slope ‘m’ is y – y₁ = m(x – x₁)
y + 1 = m(x – 2) ………(2)
y + 1 = mx – 2m
mx – y – 2m – 1 = 0
Let ‘θ’ be the angle between the lines (1) & (2) then,
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q33.1
∴ The equations of the required straight lines are
from (2), y + 1 = 2 ± √3(x – 2).

Question 46.
Prove that the points (1, 11), (2, 15), and (-3, -5) are collinear and find the equation of the straight line containing them.
Solution:
Let A(1, 11), B(2, 15), C(-3, -5) are the given points.
The equation of the straight line \(\overline{\mathrm{AB}}\) is,
(y – y₁) (x₂ – x₁) = (x – x₁)(y₂ – y₁)
⇒ (y – 11) (2 – 1) = (x – 1) (15 – 11)
⇒ (y – 11) (1) = (x – 1) (4)
⇒ y – 11 = 4x – 4
⇒ 4x – y + 7 = 0 ……….(1)
Now, substituting the point C(-3, -5) in equation (1)
4(-3) – (-5) + 7 = 0
⇒ -12 + 5 + 7 = 0
⇒ -7 + 7 = 0
⇒ 0 = 0
∴ The point C(-3, -5) lie on the straight line 4x – y + 7 = 0.
∴ Given points are collinear.
∴ The equation of the straight line is 4x – y + 7 = 0.

Question 47.
Find the angle which the straight line y = √3x – 4 makes with the Y-axis. [Mar. ’19 (AP & TS)]
Solution:
Given the equation of the straight line is y = √3x – 4
⇒ √3x – y – 4 = 0
Comparing the given equation with ax + by + c = 0, then
a = √3, b = -1, c = -4
The slope of a straight line √3x – y – 4 = 0 is
m = \(\frac{-a}{b}=\frac{-\sqrt{3}}{-1}\) = √3
tan θ = √3
∴ θ = 60° = \(\frac{\pi}{3}\)
The angle which the straight line y = √3x – 4 makes with X-axis = \(\frac{\pi}{3}\)
The angle which the straight line y = √3x – 4 makes with the Y-axis is = 180° – (90° + 60°)
= 180° – 150°
= 30°
= \(\frac{\pi}{6}\)

Question 48.
Write the equations of the straight lines parallel to the X-axis and (i) at a distance of 3 units above the X-axis and (ii) at a distance of 4 units below the X-axis.
Solution:
(i) The equation of the straight lines parallel to the X-axis, at a distance of 3 units above the X-axis is y = 3
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q3
(ii) The equation of the straight line parallel to the X-axis at a distance of 4 units below the X-axis is y = -4

Question 49.
Write the equations of the straight lines parallel to the Y-axis and (i) at a distance of 2 units from the Y-axis to the right of it (ii) at a distance of 5 units from the Y-axis to the left of it.
Solution:
(i) The equation of the straight line parallel to the Y-axis and at a distance of 2 units from the Y-axis to the right of it is x = 2.
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q4
(ii) The equation of the straight line parallel to the Y-axis and at a distance of 5 units from the Y-axis to the left of it is x = -5.

Question 50.
Find the equation of the straight line which makes an angle \(\frac{\pi}{4}\) with the positive X-axis in the positive direction and which passes through the point (0, 0).
Solution:
Given that, inclination of a straight line is θ = \(\frac{\pi}{4}\)
The slope of a line is m = tan θ
= tan \(\frac{\pi}{4}\)
= tan 45°
= 1
Let the given point A(x₁, y₁) = (0, 0)
∴ The equation of the straight line passing through A(0, 0) and having slope ‘1’ is y – y₁ = m(x – x₁)
⇒ y – o = 1(x – 0)
⇒ y = x – o
⇒ y = x
⇒ x – y = 0

Question 51.
Find the equation of the straight line which makes an angle of 135° with the positive X – axis in the positive direction and which pass through the point (3, -2).
Solution:
Given that, the inclination of a straight line is θ = 135°
The slope of a line is m = tan θ
= tan 135°
= tan (180° – 45°)
= tan 45°
= -1
Let the given point A(x₁, y₁) = (3, -2)
∴ The equation of the straight line passing through A(3, -2) and having slope ‘-1’ is y – y₁ = m(x – x₁)
⇒ y + 2 = -1(x – 3)
⇒ y + 2 = -x + 3
⇒ x + y + 2 – 3 = 0
⇒ x + y – 1 = 0

Question 52.
Find the equation of the straight line which makes an angle of 150° with the positive X-axis in the positive direction and the Y-intercept is 2.
Solution:
Given the inclination of a straight line is θ = 150°
The slope of a line is m = tan θ
= tan 150°
= -cot 60°
= \(\frac{-1}{\sqrt{3}}\)
y-intercept, c = 2.
The equation of the straight line having slope \(\frac{-1}{\sqrt{3}}\) and y-intercept ‘2’ is y = mx + c
⇒ y = \(\frac{-1}{\sqrt{3}}\)x + 2
⇒ y = \(\frac{-x+2 \sqrt{3}}{\sqrt{3}}\)
⇒ x + √3y – 2√3 = 0

Question 53.
Find the equation of the straight line which makes an angle \(\tan ^{-1}\left(\frac{2}{3}\right)\) with the positive X-axis in the positive direction and the y-intercept is 3.
Solution:
Given, inclination of a straight line is θ = \(\tan ^{-1}\left(\frac{2}{3}\right)\)
tan θ = \(\frac{2}{3}\)
Slope of a line is m = tan θ = \(\frac{2}{3}\)
y-intercept, c = 3
∴ The equation of a straight line having slope \(\frac{2}{3}\) and y-intercept ‘3’ is y = mx + c
⇒ y = \(\frac{2}{3}\) (x) + 3
⇒ y = \(\frac{2 x+9}{3}\)
⇒ 3y = 2x + 9
⇒ 2x – 3y + 9 = 0

Question 54.
Prove that the points (a, b + c), (b, c + a), (c, a + b) are collinear and find the equation of the straight line containing them.
Solution:
Let A(a, b + c), B(b , c + a), C(c, a + b) are the given points
The equation of the straight line \(\overline{\mathrm{AB}}\) is (y – y₁) (x₂ – x₁) = (x – x₁)(y₂ – y₁)
⇒ (y – b – c) (b – a) = (x – a) (c + a – b – c)
⇒ (y – b – c) (b – a) = (x – a) (a – b)
⇒ (y – b – c) (b – a) = -(x – a) (b – a)
⇒ y – b – c = -x + a
⇒ x + y – a – b – c = 0 ……..(1)
Now, Substituting the point C(c, a + b) in equation (1)
x + y – a – b – c = 0
⇒ c + a + b – a – b – c = o
⇒ 0 = 0
∴ Point C(c, a + b) lies on the straight line x + y – a – b – c = 0.
∴ Given points are collinear.
∴ The straight line equation is x + y – a – b – c = 0.

Question 55.
A(10, 4), B(-4, 9) and C(-2, -1) are the vertices of a triangle. Find the equations of (i) \(\overline{\mathrm{AB}}\) (ii) the median through A (iii) the altitude through B (iv) the perpendicular bisector of the side \(\overline{\mathrm{AB}}\).
Solution:
(i) Equation of the straight line \(\overline{\mathrm{AB}}\) is (y – y₁) (x₂ – x₁)= (x – x₁)(y₂ – y₁)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q10
⇒ (y – 4)(-4 – 10) = (x – 10) (9 – 4)
⇒ (y – 4) (-14) = (x – 10) (5)
⇒ -14y + 56 = 5x – 50
⇒ 5x + 14y – 106 = 0
(ii) The equation of the median through ‘A’:
Since ‘D’ is the midpoint of BC, then
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q10.1
The equation of the median through A is the equation of the straight line \(\overline{\mathrm{AD}}\),
(y – y₁)(x₂ – x₁) = (x – x₁)(y₂ – y₁)
⇒ (y – 4) (-3 – 10) = (x – 10)(4 – 4)
⇒ (y – 4) (-13) = (x – 10)(0)
⇒ (y – 4)(-13) = 0
⇒ y – 4 = 0
(iii) The equation of the altitude through ‘B’:
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q10.2
= \(\frac{5}{12}\)
Since \(\overline{\mathrm{BE}} \perp \overline{\mathrm{AC}}\), then
slope of \(\overline{\mathrm{BE}}\) = \(\frac{-1}{m}=\frac{-1}{\frac{5}{12}}=\frac{-12}{5}\)
The equation of the altitude through B is, the equation of the straight line passing through B(-4, 9) and having slope \(\frac{-12}{5}\) is y – y₁ = m(x – x₁)
⇒ y – 9 = \(\frac{-12}{5}\) (x + 4)
⇒ 5y – 45 = -12x – 48
⇒ 12x + 5y + 3 = 0
(iv) The equation of the perpendicular bisector of side \(\overline{\mathrm{AB}}\):
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q10.3
∴ The equation of the perpendicular bisector of \(\overline{\mathrm{AB}}\) is, the equation of the straight line passing through F(3, \(\frac{13}{2}\)) and having slope \(\frac{14}{5}\) is y – y₁ = m(x – x₁)
⇒ y – \(\frac{13}{2}\) = \(\frac{14}{5}\)(x – 3)
⇒ \(\frac{2 y-13}{2}=\frac{14}{5}(x-3)\)
⇒ 10y – 65 = 28x – 84
⇒ 28x – 10y – 19 = 0

Question 56.
A straight line passing through A(1, -2) makes an angle \(\tan ^{-1}\left(\frac{4}{3}\right)\) with the positive direction of the X-axis in the anti-clockwise sense. Find the points on the straight line whose distance from A is 5.
Solution:
Given point (x1, y1) = A(1, -2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q11
The inclination of the straight line is θ = \(\tan ^{-1}\left(\frac{4}{3}\right)\)
tan θ = \(\frac{4}{3}\)
sin θ = \(\frac{4}{5}\)
cos θ = \(\frac{3}{5}\)
Distance, |r| = 5 units
Required point = (x₁ + |r| cos θ, y₁ + |r| sin θ)
= (x₁ ± r cos θ, y₁ ± r sin θ)
= (1 ± 5 . \(\frac{3}{5}\), -2 ± 5 . \(\frac{4}{5}\))
= (1 ± 3, -2 ± 4)
= (1 + 3, (-2) + 4), (1 – 3, -2 – 4)
= (4, 2), (-2, -6)

Question 57.
Find the sum of the squares of the intercepts of the line 4x – 3y = 12 on the axes of coordinates. [Mar. ’18 (AP)]
Solution:
Given, the equation of the straight line is 4x – 3y = 12
\(\frac{x}{3}+\frac{y}{-4}=1\) which is of the form \(\frac{x}{a}+\frac{y}{b}=1\)
∴ x-intercept, a = 3, y-intercept, b = -4
Now, the Sum of the squares of the intercepts = a² + b²
= 3² + (-4)²
= 9 + 16
= 25

Question 58.
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:
The intercepts of a straight line on the coordinate axes are a, b.
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q13

This is of the form x cos α + y sin α = p
∴ p = the length of the perpendicular drawn from the origin to the line = \(\frac{|a b|}{\sqrt{a^2+b^2}}\)

Question 59.
If the product of the intercepts made by the straight line x tan α + y sec α = 1 (0 ≤ α ≤ \(\frac{\pi}{2}\)) on the coordinate axes is equal to sin α, find α.
Solution:
Given, the equation of the straight line is x tan α + y sec α = 1
\(\frac{x}{\frac{1}{\tan \alpha}}+\frac{y}{\frac{1}{\sec \alpha}}=1\)
\(\frac{x}{\cot \alpha}+\frac{y}{\cos \alpha}=1\)
This is of the form \(\frac{x}{a}+\frac{y}{b}=1\)
∴ x-intercept (a) = cot α
y-intercept (b) = cos α
Given that, a product of the intercepts is equal to sin α.
cot α × cos α = sin α
⇒ \(\frac{\cos \alpha}{\sin \alpha}\) × cos α = sin α
⇒ \(\frac{\cos ^2 \alpha}{\sin ^2 \alpha}\) = 1
⇒ cot²α = 1
⇒ tan²α = 1
⇒ tan α = 1
⇒ α = \(\frac{\pi}{2}\) (∵ 0 ≤ α ≤ \(\frac{\pi}{2}\))

Question 60.
A straight line passing through A(-2, 1) makes an angle of 30° with \(\overline{\mathrm{OX}}\) in the positive direction. Find the points on the straight line whose distance from A is 4 units.
Solution:
Given point A(x₁, y₁) = (-2, 1)
Inclination of straight line = 30°
distance |r| = 4
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q15
∴ Required points = (x₁ + |r| cos θ, y₁ + |r| sin θ)
= (x₁ ± r cos θ, y₁ ± r sin θ)
= (-2 ± 4 cos 30°, 1 ± 4 sin 30°)
= (-2 ± 4\(\left(\frac{\sqrt{3}}{2}\right)\), 1 ± 4(\(\frac{1}{2}\)))
= (-2 ± 2√3 ,1 ± 2)
= (-2 + 2√3, 1 + 2), (-2 – 2√3, +1 – 2)
= (-2 + 2√3, 3), (-2 + 2√3, -1)

Question 61.
A straight line whose inclination with the positive direction of the X-axis measured in the anti-clockwise sense is \(\frac{\pi}{3}\) makes a positive intercept on the Y-axis. If the straight line is at a distance of 4 from the origin; find its equation.
Solution:
Given that, inclination of a straight line is θ = \(\frac{\pi}{3}\)
Let ‘l’ is the required straight line.
Now, ‘N’ is the foot of the perpendicular from the origin to the straight line ‘L’
∴ ∠XON = 150°
∴ a = 150°
Given that, p = 4
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q16
The equation of the straight line ‘L’ in the normal form is x cos α + y sin α = p
⇒ x cos (150°) + y sin (150°) = 4
⇒ x cos(180 – 30) + y sin (180 – 30) = 4
⇒ -x cos 30° + y sin 30° = 4
⇒ \(-x\left(\frac{\sqrt{3}}{2}\right)+y\left(\frac{1}{2}\right)=4\)
⇒ -√3x + y = 8
⇒ √3x – y + 8 = 0

Question 62.
Find the ratio in which the straight line 2x + 3y – 10 = 0 divides the join of the points (2, 3) and (2, 10).
Solution:
Given equation of the straight line is L = 2x + 3y – 10 = 0
Comparing the equation with ax + by + c = 0, we get
a = 2, b = 3, c = -10
Let the given points are A(x₁, y₁) = (2, 3) and B(x₂, y₂) = (2, 10)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q17

Question 63.
State whether (3, 2) and (-4, -3) are on the same side or opposite sides of the straight line 2x – 3y + 4 = 0.
Solution:
Given equation of the straight line is L = 2x – 3y + 4 = 0
Let the given points are A(x₁, y₁) = (3, 2) and B(x₂, y₂) = (-4, -3)
Now, l₁₁ = L(3, 2)
= 2(3) – 3(2) + 4
= 6 – 6 + 4
= 4 > 0
l₂₂ = L(-4, -3)
= 2(-4) – 3(-3) + 4
= -8 + 9 + 4
= 5 > 0
Since l₁₁ 0 and l₂₂ > 0, the given points are on the same side of the straight line, L = 0.

Question 64.
Find the ratio’s in which (i) the X-axis and (ii) the Y-axis divide the line segment \(\overline{\mathrm{AB}}\) joining A(2, -3) and B(3, -6).
Solution:
Given points are A(x₁, y₁) = (2, -3) and B(x₂, y₂) = (3, -6)
(i) The X-axis divides the line segment AB in the ratio is,
-y₁ : y₂ = -(-3) : -6
= 3 : -6
= 1 : -2
= -1 : 2
(ii) Y-axis divides \(\overline{\mathrm{AB}}\) in the ratio is,
-x₁ : x₂ = -2 : 3

Question 65.
Find the equation of the straight line passing through the point of intersection of the lines x + y + 1 = 0 and 2x – y + 5 = 0 and containing the point (5, -2).
Solution:
Given, the equations of the straight lines are
x + y + 1 = 0 ……….(1)
2x – y + 5 = 0 ………..(2)
Let the given point be (5, -2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q20
∴ The point of intersection of lines (1) & (2) is A = (-2, 1)
Now, the equation of the straight line \(\overline{\mathrm{AB}}\) is (y – y1)(x2 – x1) = (x – x1)(y2 – y1)
⇒ (y – 1) (5 + 2) = (x + 2) (-2 – 1)
⇒ (y – 1)(7) = (x + 2)(-3)
⇒ 7y – 7 = -3x – 6
⇒ 3x + 7y – 1 = 0

Question 66.
Find the ratio in which the straight line 3x + 4y = 6 divides the join of the points (2, -1) and (1, 1).
Solution:
Given equation of the straight line is L = 3x + 4y – 6 = 0
Comparing the equation with ax + by + c = 0, we get
a = 3, b = 4, c = -6
Let the given points are A(x₁, y₁) = (2, -1) and B(x₂, y₂) = (1, 1)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q21

Question 67.
Transform the equation (2 + 5k)x – 3(1 + 2k)y + (2 – k) = 0 into the form L₁ + λL₂ = 0 and find the point of concurrency of the family of straight lines represented by the equation.
Solution:
Given, equation is (2 + 5k)x – 3(1 + 2k)y + (2 – k) = 0
⇒ 2x + 5kx – 3y – 6ky + 2 – k = 0
⇒ (2x – 3y + 2) + k(5x – 6y – 1) = 0
This is of the form l₁ + λl₂ = 0
where l₁ = 2x – 3y + 2 = 0 ……..(1)
l₂ = 5x – 6y – 1 = 0 ……..(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q22
∴ Point of concurrence = (5, 4)

Question 68.
Transform the equation (k + 1)x + (k + 2)y + 5 = 0, into the form L₁ + λL₂ = 0 and find the point of concurrency of the family of straight lines represented by the equation.
Solution:
Given, equation is (k + 1)x + (k + 2)y + 5 = 0
⇒ kx + x + ky + 2y + 5 = 0
⇒ (x + 2y + 5) + k(x + y) = 0
This is of the form l₁ + λl₂ = 0
where l₁ = x + 2y + 5 = 0 ………(1)
l₂ = x + y = 0 ……..(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q23
∴ Point of concurrence = (5, -5)

Question 69.
Find the area of the triangle formed by the straight line x – 4y + 2 = 0 and the coordinate axes.
Solution:
Given, equation of the straight line is x – 4y + 2 = 0 ………(1)
Comparing with ax + by + c = 0, we get
a = 1, b = -4, c = 2
∴ The area of the triangle formed by the straight line (1) & the co-ordinate axis = \(\frac{c^2}{2|a b|}=\frac{2^2}{2|1(-4)|}\) = \(\frac{1}{2}\)

Question 70.
Find the area of the triangle formed by the straight line 3x – 4y + 12 = 0 and the coordinate axes.
Solution:
Given, equation of the straight line is 3x – 4y + 12 = 0 ………(1)
Comparing with ax + by + c = 0, we get
a = 3, b = -4, c = 12
∴ The area of the triangle formed by the straight line and the co-ordinate axis = \(\frac{c^2}{2|a b|}=\frac{12^2}{2|3(-4)|}=\frac{12^2}{2|-12|}=\frac{144}{2(12)}\)
= 6 sq. units.

Question 71.
Find the equation of the straight line passing through the points (-1, 2) and (5, -1) and the area of the triangle formed by it with the axes of coordinates. [May ’15 (TS)]
Solution:
Let the given points are A(x₁, y₁) = (-1, 2) & B(x₂, y₂) = (5, -1)
The equation of the straight line passing through the points A(-1, 2) & B(5, -1) is (y – y₁)(x₂ – x₁) = (x – x₁)(y₂ – y₁)
⇒ (y – 2) (5 + 1) = (x + 1) (-1 – 2)
⇒ (y – 2)(6) = (x + 1)(-3)
⇒ 2y – 4 = -x – 1
⇒ x + 2y – 3 = 0 ……..(1)
Comparing equation (1) with ax + by + c = 0, we get
a = 1, b = 2, c = -3
∴ The area of the triangle formed by the straight line (1) and the co-ordinate axis = \(\frac{c^2}{2|a b|}=\frac{9}{2|1(2)|}=\frac{9}{2(2)}=\frac{9}{4}\) sq. units

Question 72.
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:
Given, equation of the straight line is 3x – 5y + a = 0 ……..(1)
Comparing (1) with ax + by + c = 0, we get
a = 3, b = -5, c = a
Let the given points are A(x₁, y₁) = (1, 2) & B(x₂, y₂) = (3, 4)
Given that, the points (1, 2) & (3, 4) lie on the same direction 3x – 5y + a = 0, then -l₁₁ : l₂₂ < 0
⇒ \(\frac{-\mathrm{L}_{11}}{\mathrm{~L}_{22}}\) < 0
⇒ \(\frac{-\left(a x_1+b y_1+c\right)}{\left(a x_2+b y_2+c\right)}\) < 0
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q27

Question 73.
Find the angle between the lines 2x + y + 4 = 0 and y – 3x = 7.
Solution:
Given, the equation of the straight lines is
2x + y + 4 = 0 ……..(1)
y – 3x – 7 = 0
⇒ 3x – y + 7 = 0 …….(2)
Comparing (1) with a₁x + b₁y + c₁ = 0, we get
a₁ = 2, b₁ = 1, c₁ = 4
Comparing (2) with a₂x + b₂y + c₂ = 0, we get
a₂ = 3, b₂ = -1, c₂ = 7
If ‘θ’ is the angle between the lines (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q28

Question 74.
Find the angle between the lines √3x + y + 1 = 0 and x + 1 = 0.
Solution:
Given, the equation of the straight lines are
√3x + y + 1 = 0 …….(1)
x + 1 = 0 ………(2)
Comparing (1) with a₁x + b₁y + c₁ = 0, we get
a₁ = √3, b₁ = 1, c₁ = 1
Comparing (2) with a₂x + b₂y + c₂ = 0, we get
a₂ = 1, b₂ = 0, c₂ = 1
If ‘θ’ is the angle between the lines (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q29

Question 75.
Find the angle between the lines ax + by = a + b, a(x – y) + b(x + y) = 2b.
Solution:
Given, the equation of the straight lines are
ax + by = a + b
ax + by – a – b = 0 …….(1)
a(x – y) + b(x + y) = 2b
ax – ay + bx + by = 2b
(a + b)x + (-a + b)y – 2b = 0 ………(2)
Comparing (1) with a₁x + b₁y + c₁ = 0, we get
a₁ = a, b₁ = b, c₁ = -a – b
Comparing (2) with a₂x + b₂y + c₂ = 0, we get
a₂ = a + b, b₂ = b – a, c₂ = -2b
If ‘θ’ is the angle between the lines (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q30

Question 76.
Find the equation of a straight line perpendicular to the line 5x – 3y + 1 = 0 and pass through the point (4, -3). [Mar. ’15 (TS)]
Solution:
Given, the equation of the straight line is 5x – 3y + 1 = 0
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q31
Slope of the given line is m = \(\frac{-5}{-3}=\frac{5}{3}\)
Slope of the perpendicular line = \(\frac{-1}{m}=\frac{-1}{\frac{5}{3}}=\frac{-3}{5}\)
Let the given point A(x₁, y₁) = (4, -3)
Equation of the straight line passing through (4, -3) and having slope \(\frac{-3}{5}\) is,
y – y₁ = \(\frac{-1}{m}\)(x – x₁)
⇒ y + 3 = \(\frac{-3}{5}\)(x – 4)
⇒ 5y + 15 = -3x + 12
⇒ 3x + 5y + 3 = 0

Question 77.
Find the value of k, if the straight line 6x – 10y + 3 = 0 and kx – 5y + 8 = 0 are parallel.
Solution:
Given, the equation of the straight line is
6x – 10y + 3 = 0 …….(1)
kx – 5y + 8 = 0 ……(2)
Slope of the line (1) is m1 = \(\frac{-a}{b}=\frac{-6}{-10}=\frac{3}{5}\)
Slope of the line (2) is m2 = \(\frac{-k}{-5}=\frac{k}{5}\)
Since the given lines are parallel then m₁ = m₂
⇒ \(\frac{3}{5}\) = \(\frac{k}{5}\)
⇒ k = 3

Question 78.
(-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:
Let ABCD be a square.
Let the given point be A(-4, 5).
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q33
Equation of the diagonal \(\overline{\mathrm{AC}}\) is, 7x – y + 8 = 0 is given
Slope of the diagonal, \(\overline{\mathrm{AC}}\) = 7
Since in a square diagonals \(\overline{\mathrm{AC}}\) & \(\overline{\mathrm{BD}}\) are perpendicular
then slope of the diagonal \(\overline{\mathrm{BD}}\) = \(\frac{-1}{m}=\frac{-1}{7}\)
∴ The equation of the diagonal \(\overline{\mathrm{BD}}\) is y – y1 = \(\frac{-1}{m}\)(x – x1)
⇒ y – 5 = \(\frac{-1}{7}\)(x + 4)
⇒ 7y – 35 = -x – 4
⇒ x + 7y – 31 = 0

Question 79.
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.
Given that, opposite vertices of a square are A(-1, 1) and B(5, 3).
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q34
The equation of straight line passing through E & slope -3 is y – y₁ = \(\frac{-1}{m}\)(x – x₁)
⇒ y – 2 = -3(x – 2)
⇒ y – 2 = -3x + 6
⇒ 3x + y – 8 = 0

Question 80.
Show that lines x – 7y – 22 = 0, 3x + 4y + 9 = 0 and 7x + y – 54 = 0 form a right angled isosceles triangle.
Solution:
Given, the equations of the straight lines are
x – 7y – 22 = 0 …..(1)
3x + 4y + 9 = 0 …….(2)
7x + y – 54 = 0 ………(3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q35
Let ‘A’ be the angle between the lines (1) & (3) then,

∴ A = 90°, B = 45°, C = 45°
∴ Given lines form a right angle isosceles triangle.

Question 81.
If ab > 0, find the area of the rhombus enclosed by the four straight lines ax ± by ± c = 0.
Solution:
Given, the equations of the straight lines are
ax + by + c = 0 ……..(1)
ax + by – c = 0 ………(2)
ax – by + c = 0 ……..(3)
ax – by – c = 0 ………(4)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q36

Question 82.
Find the area of the parallelogram whose sides are 3x + 4y + 5 = 0, 3x + 4y – 2 = 0, 2x + 3y + 1 = 0 and 2x + 3y – 7 = 0.
Solution:
Given, the equations of the straight lines are
3x + 4y + 5 = 0 …….(1)
3x + 4y – 2 = 0 ………(2)
2x + 3y + 1 = 0 ……….(3)
2x + 3y – 7 = 0 ………..(4)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q37

Question 83.
Find the incentre of the triangle whose vertices are (1, √3), (2, 0), and (0, 0).
Solution:
Given, A (x₁, y₁) = (1, √3), B(x₂, y₂) = (2, 0), C(x₃, y₃) = (0, 0)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q38

Question 84.
Find the incentre of the triangle whose sides are x = 1, y = 1, and x + y = 1.
Solution:
Given, the equation of the straight lines are
x = 1 …….(1)
y = 1 ………(2)
x + y = 1 ……..(3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q39
Vertex A:
Solving (1) & (3)
from (1), x = 1
from (3), x + y = 1
⇒ 1 + y = 1
⇒ y = 0
∴ Vertex A = (1, 0)

Vertex B:
Solving (1) & (2)
from (1), x = 1
from (2), y = 1
∴ Vertex B = (1, 1)
Vertex C:
Solving (2) & (3)
from (2), y = 1
from (3), x + y = 1
⇒ x + 1 = 1
⇒ x = 0
∴ Vertex C = (0, 1)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q39.2

Question 85.
Find the values of k, if the angle between the straight lines kx + y + 9 = 0 and 3x – y + 4 = 0 is \(\frac{\pi}{4}\).
Solution:
Given equations are
kx + y + 9 = 0 ……..(1)
3x – y + 4 = 0 ………(2)
Comparing (1) with a₁x + b₁y + c₁ = 0, we get
a₁ = k, b₁ = 1, c₁ = 9
Comparing (2) with a₂x + b₂y + c₂ = 0, we get
a₂ = 3, b₂ = -1, c₂ = 4
Given that, θ = \(\frac{\pi}{4}\)
Let ‘θ’ is the angle between the lines (1) & (2) then
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q40

Squaring on both sides
⇒ (k² + 1)(10) = 2(3k – 1)²
⇒ 10k² + 10 = 18k² + 2 – 12k
⇒ 8k² – 12k – 8 = 0
⇒ 2k² – 3k – 2 = 0
⇒ 2k² – 4k + k – 2 = 0
⇒ (k – 2)(2k + 1) = 0
⇒ k = 2 or k = \(\frac{-1}{2}\)

Question 86.
Find the equation of the straight line passing through the origin and also through the point of intersection of the lines 2x – y + 5 = 0 and x + y + 1 = 0.
Solution:
Given equations of the straight lines are
2x – y + 5 = 0 ………(1)
x + y + 1 = 0 ………(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q41
The point of intersection of lines (1) & (2) is P(-2, 1).
The equation of the straight line passing through the point O(0, 0) & P(-2, 1) is (y – y₁)(x₂ – x₁) = (x – x₁)(y₂ – y₁)
⇒ (y – 0) (-2 – 0) = (x – 0) (1 – 0)
⇒ -2y = x
⇒ x + 2y = 0

Question 87.
Find the equation of the straight line perpendicular to the line 2x + 3y = 0 and pass through the point of intersection of the lines x + 3y – 1 = 0 and x – 2y + 4 = 0.
Solution:
Given equations of the straight lines are
x + 3y – 1 = 0 ……..(1)
x – 2y + 4 = 0 …….(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q42
∴ The point of intersection of lines (1) & (2) is P(-2, 1).
Given equation of the straight line 2x + 3y = 0 ……..(3)
Slope of line is m = \(\frac{-2}{3}\)
Since the required line is perpendicular to the line (3),
Slope of required line is \(\frac{-1}{m}=\frac{-1}{\frac{-2}{3}}=\frac{3}{2}\)
The equation of the straight line passing through P(-2, 1) & slope \(\frac{3}{2}\) is y – y₁ = \(\frac{-1}{m}\) (x – x₁)
⇒ y – 1 = \(\frac{3}{2}\) (x + 2)
⇒ 2y – 2 = 3x + 6
⇒ 3x – 2y + 8 = 0

Question 88.
Find the value of ‘a’ if the distances of the points (2, 3) and (-4, a) from the straight line 3x + 4y – 8 = 0 are equal.
Solution:
Given the equation of the straight line is 3x + 4y – 8 = 0
Let the given points be P = (2, 3), Q (-4, a)
Now, the perpendicular distance from the point P(2, 3) to the straight line 3x + 4y – 8 = 0 is
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q43
The perpendicular distance from the point Q(-4, a) to the straight line 3x + 4y – 8 = 0 is,

Given that two perpendicular distances are equal then,
2 = \(\frac{|4 a-20|}{5}\)
⇒ |4a – 20| = 10
⇒ 2|2a – 10| = 10
⇒ 2a – 10 = ±5
⇒ 2a – 10 = 5; 2a – 10 = -5
⇒ 2a = 15; 2a = 5
⇒ a = \(\frac{15}{2}\); a = \(\frac{5}{2}\)
∴ a = \(\frac{15}{2}\) (or) \(\frac{5}{2}\)

Question 89.
Two adjacent sides of a parallelogram are given by 4x + 5y = 0 and 7x + 2y = 0 and one diagonal is 11x + 7y = 9. Find the equations of the remaining sides and the other diagonal.
Solution:
Given that, two adjacent sides of a parallelogram are given by
4x + 5y = 0 …….(1)
7x + 2y = 0 ………(2)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q44

⇒ -6y – 8 = 21x – 35
⇒ 21x + 6y – 27 = 0
⇒ 7x + 2y – 9 = 0
The equation of the diagonal \(\overline{\mathrm{BD}}\) is,
(y – y₁)(x₂ – x₁) = (x – x₁)(y₂ – y₁)
⇒ (y – 0)(1 – 0) = (x – 0)(1 – 0)
⇒ y(1) = x(1)
⇒ y = x
⇒ x – y = 0
∴ Two adjacent sides of a parallelogram are 4x + 5y – 9 = 0, 7x + 2y – 9 = 0.
The equation of one of the diagonals is x – y = 0.

Question 90.
Find the incentre of the triangle formed by the straight lines x + 1 = 0, 3x – 4y = 5, and 5x + 12y = 27.
Solution:
Given, the equation of the straight lines are
x + 1 = 0 ………(1)
3x – 4y = 5 ………..(2)
5x + 12y = 27 ………..(3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q45
Vertex A:
Solving (1) & (3)
from (1), x + 1 = 0
⇒ x = -1
from (3), 5(-1) + 12y = 27
⇒ -5 + 12y = 27
⇒ 12y = 32
⇒ y = \(\frac{32}{12}\) = \(\frac{8}{3}\)
∴ Vertex A = (-1, \(\frac{8}{3}\))
Vertex B:
Solving (1) & (2)
from (1), x + 1 = 0
⇒ x = -1
from (2), 3(-1) – 4y = 5
⇒ -3 – 4y = 5
⇒ -4y = 8
⇒ y = -2
∴ Vertex B = (-1, -2)
Vertex C:
Solving (2) & (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q45.1

Question 91.
Find the equation of the straight line parallel to the line 3x + 4y = 7 and passing through the point of intersecting of the lines x – 2y – 3 = 0, x + 3y – 6 = 0. [Mar. ’16 (TS)]
Solution:
Given, the equation of the straight lines are
x – 2y – 3 = 0 …….(1)
x – 3y – 6 = 0 …….(2)
Solving (1) & (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q46
∴ The point of intersection of lines (1) & (2) is \(\mathrm{P}\left(\frac{21}{5}, \frac{3}{5}\right)\)
Given equation of the straight line is 3x + 4y = 7 …….(3)
Slope of line is m = \(\frac{-3}{4}\)
Since the required line is parallel to line 3.
Slope of required line = m = \(\frac{-3}{4}\)
∴ The equation of the straight line passing through \(\mathrm{P}\left(\frac{21}{5}, \frac{3}{5}\right)\) & Slope = \(\frac{-3}{4}\) is y – y₁ = m(x – x₁)
⇒ \(y-\frac{-3}{5}=\frac{-3}{4}\left(x-\frac{21}{5}\right)\)
⇒ \(\frac{5 y-3}{5}=\frac{-3}{4}\left(\frac{5 x-21}{5}\right)\)
⇒ 5y – 3 = \(\frac{-3}{4}\)(5x – 21)
⇒ 20y – 12 = -15x + 63
⇒ 15x + 20y – 75 = 0
⇒ 3x + 4y – 15 = 0

Question 92.
Find the value of p, if the straight lines 3x + py – 1 = 0, 7x – 3y + 3 = 0, are mutually perpendicular. [Mar. ’19 (TS)]
Solution:
Given lines are
3x + py – 1 = 0 ……..(1)
7x – 3y + 3 = 0 ………(2)
Slope of (1) is m₁ = \(\frac{-3}{p}\)
Slope of (2) is m₂ = \(\frac{-7}{-3}\) = \(\frac{7}{3}\)
Since (1) & (2) are perpendicular then
m₁ × m₂ = -1
⇒ \(\frac{-3}{p} \times \frac{7}{3}=-1\)
⇒ \(\frac{-7}{p}\) = -1
⇒ p = 7

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3

February 29, 2024February 28, 2024 by Mahesh

Students can practice TS Class 10 Maths Solutions Chapter 6 Progressions Ex 6.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 6 Progressions Exercise 6.3

Question 1.
Find the sum of the following APs :

i) 2, 7, 12, ………., to 10 terms.
Solution:
Given A.P. : 2, 7, 12, ……… to 10 term
a = 2; d = a₂ – a₁ = 7 – 2 = 5; n = 10
s_n = \(\frac{\mathrm{n}}{2}\)[2a + (n – 1)d]
s₁₀ = – \(\frac{10}{2}\)[2 × 2 +(10 – 1)5]
= 5[4 + 9 × 5]
= 5 [4 + 45]
= 5 × 49 = 245

ii) – 37, – 33, – 29, …… to 12 terms.
Solution:
Given A.P :
– 37, – 33, – 29, …., to 12 terms
a = – 37; d = a₂ – a₁ = (- 33) – (- 37)
= – 33 + 37 = 4 and n = 12
S_n = \(\frac{n}{2}\)[2a+ (n – 1)d]
s₁₂ = \(\frac{12}{2}\)[2 × (-37) + (12 – 1)4]
= 6[-74 + 11 × 4]
= 6[-74 + 44] = 6 × (- 30) = -180

iii) 0.6, 1.7, 2.8, …, to 100 terms.
Solution:
Given A.P : 0.6, 1.7, 2.8, …., S₁₀₀
a = 0.6; d = a₂ – a₁ = 1.7 – 0.6 = 1.1 and n = 100
s = \(\frac{\mathrm{n}}{2}\)[2a + (n – 1)d]
s₁₀₀ = \(\frac{100}{2}\)[2 × 0.6 + (100 – 1) × 1.1]
= 50[1.2 + 99 × 1.1]
= 50[1.2 + 108.9]
= 50 × 110.1 = 5505.

iv) \(\frac{1}{15}\), \(\frac{1}{12}\), \(\frac{1}{10}\) ….. to 11 terms.
Solution:
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 1

Question 2.
Find the sums given below :

i) 7 + 10\(\frac{1}{2}\) + 14 + ….. + 84
Solution:
Given A.P. : 7 + 10\(\frac{1}{2}\) + 14 + …… + 84
∴ a = 7, d = a₂ – a₁ = 10\(\frac{1}{2}\) – 7 = 3\(\frac{1}{2}\)
and the last term l = a_n
But, a_n = a +(n – 1) d
∴ 84 = 7 + (n – 1) × 3\(\frac{1}{2}\)
⇒ 84 – 7 = (n – 1) × \(\frac{7}{2}\)
⇒ n – 1 = 77 × \(\frac{2}{7}\) = 22
⇒ n = 22 + 1 = 23
Now, S_n = \(\frac{\mathrm{n}}{2}\)(a + l)
Where a = 7; l = 84
S₂₃ = \(\frac{23}{2}\)(7 + 84)
= \(\frac{23}{2}\) × 91 = \(\frac{2093}{2}\) = 1046\(\frac{1}{2}\)

ii) 34 + 32 + 30 + ……. + 10
Solution:
Given A.P : 34 + 32 + 30 + …….. + 10
Here a = 34; d = a₂ – a₁ = 32 – 34 = -2
and the last term ‘l‘ = a_n = 10
But, a_n = a + (n – 1) d
⇒ 10 = 34 + (n – 1)(-2)
⇒ 10 – 34 = -2n + 2
⇒ -2n = -24 – 2
⇒ n = \(\frac{-26}{-2}\) ∴ n = 13
Also, S_n = \(\frac{\mathrm{n}}{2}\)(a + l)
Where a = 34, l = 10
S₁₃ = \(\frac{13}{2}\)(34 + 10)
\(\frac{13}{2}\) × 44 = 13 × 22 = 286

iii) -5 + (-8) + (-11) + …. + (-230).
Solution:
Given AP:
-5 + (-8) + (-11) + …….. + (-230)
Here first term, a = -5;
Last term, l = a_n = -230:
d = a₂ – a₁ = (-8) – (-5)
= -8 + 5 = -3
But, a_n = a + (n – 1) d
∴ (-230) = (-5) + (n – 1) × (-3)
⇒ -230 + 5 = -3n + 3
⇒ -3n + 3 = -225
⇒ -3n = -225 – 3 ⇒ 3n = 228
∴ n = \(\frac{228}{3}\) = 76
Now S_n = \(\frac{\mathrm{n}}{2}\)(a + l)
S₇₆ = \(\frac{76}{2}\)[(-5) + (-230)]
= 38 × (-235) = -8930

Question 3.
In an AP:

i) Given a = 5, d = 3, a_n = 50, find n and S_n.
Solution:
Given:
a = 5; d = 3 ; a_n = a + (n – 1)d = 50
⇒ 50 = 5 + (n – 1)3
⇒ 50 – 5 = 3n – 3
⇒ 3n = 45 + 3
⇒ n = \(\frac{48}{3}\) = 16
S_n = \(\frac{\mathrm{n}}{2}\)(a + l)
S₁₆ = \(\frac{16}{2}\)[5 + 50] = 8 × 55 = 440.

ii) Given a = 7, a₁₃ = 35, find d and S₁₃.
Solution:
Given : a = 7
a₁₃ = a + 12d = 35
⇒ 7 + 12d = 35
⇒ 12d = 35 – 7
⇒ d = \(\frac{28}{12}\) = \(\frac{7}{3}\)
Now, S_n = \(\frac{\mathrm{n}}{2}\)(a + l)
S₁₃ = \(\frac{13}{2}\)[7 + 35]
= \(\frac{13}{2}\) × 42 = 13 × 21 = 273

iii) Given a₁₂ = 37, d = 3, find a and S₁₂.
Solution:
Given : a₁₂ = a + 11d = 37
d = 3
So, a₁₂ = a + 11 × 3 = 37
a + 33 = 37
a = 37 – 33 = 4
Now, S_n = \(\frac{\mathrm{n}}{2}\)(a + l)
S₁₂ = \(\frac{12}{2}\)[4 + 37] = 6 × 41 = 246.

iv) Given a₃ = 15, S₁₀ = 125, find d and a₁₀.
Solution:
Given : a₃ = a + 2d = 15
⇒ a = 15 – 2d —- (1)
S₁₀ = 125 but take S₁₀ as 175
i.e., S₁₀ = 175
We know that,
S_n = \(\frac{\mathrm{n}}{2}\)(a + l)
⇒ S₁₀ = \(\frac{10}{2}\)[a + l]
⇒ 175 = \(\frac{10}{2}\)[2a + 9d]
⇒ 175 = 5[2a + 9d]
⇒ 35 = 2(15 – 2d) + 9d [∵ a = 15 – 2d]
⇒ 35 = 30 – 4d + 9d
⇒ 35 – 30 = 5d ⇒ d = \(\frac{5}{5}\) = 1
Substituting d = 1 in equation (1) we get
a = 15 – 2 × 1 = 15 – 2 = 13
Now, a_n = a + (n – 1) d
a₁₀ = a + 9d = 13 + 9 × 1
= 13 + 9 = 22
∴ a₁₀ = 22 ; d = 1.

v) Given a = 2, d = 8, S_n = 0, find n and a_n.
Solution:
Given : a = 2; d = 8 and S_n = 90
But S_n = \(\frac{\mathrm{n}}{2}\)[2a + (n – 1) d]
⇒ 90 = \(\frac{\mathrm{n}}{2}\)[2 × 2 + (n – 1) × 8]
⇒ 90 = \(\frac{\mathrm{n}}{2}\)[4 + 8n – 8]
⇒ 90 = \(\frac{\mathrm{n}}{2}\)[8n – 4]
⇒ 90 = \(\frac{4 n}{2}\)[2n – 1]
⇒ 90 = 2n[2n – 1]
⇒ 4n² – 2n – 90 = 0
⇒ 2(2n² – n – 45) = 0
⇒ 2n² – n – 45 = 0
⇒ 2n² – 10n + 9n – 45 = 0
⇒ 2n(n – 5) + 9(n – 5) = 0
⇒ (n – 5) (2n + 9) = 0
⇒ n – 5 = 0 (or) 2n + 9 = 0
⇒ n = 5 (or) n = \(\frac{-9}{2}\) (discarded)
∴ n = 5
Now a_n = a₅ = a + 4d = 2 + 4 × 8
= 2 + 32 = 34.

vi) Given a_n = 4, d = 2, S_n = -14, find n and a.
Solution:
Given a_n = a + (n – 1) d = 4 —– (1)
d = 2; S_n = -14
From (1); a + (n – 1) 2 = 4
a = 4 – 2n + 2 ⇒ a = 6 – 2n
S_n = \(\frac{\mathrm{n}}{2}\) [a + a_n]
-14 = \(\frac{\mathrm{n}}{2}\) [(6 – 2n) + 4] [∵ a = 6 – 2n]
-14 × 2 = n(10 – 2n)
⇒ 10n – 2n² = -28
⇒ 2n² – 10n – 28 = 0
⇒ n² – 5n – 14 = 0
⇒ n² – 7n + 2n – 14 = 0
⇒ n(n – 7) + 2(n – 7) = 0
⇒ (n – 7) (n + 2) = 0
⇒ n = 7 (or) n = – 2
∴ n = 7
Now a = 6 – 2n = 6 – 2 × 7
= 6 – 14 = -8
∴ a = – 8; n = 7.

vii) Given l = 28, S = 144 and there are total 9 terms. Find a.
Solution:
Given :
l = a₉ = a + 8d = 28 and S₉ = 144
But, S_n = \(\frac{9}{2}\)(a + l)
144 = \(\frac{9}{2}\)[a + 28]
⇒ 144 × \(\frac{2}{9}\) = a + 28
⇒ a + 28 = 32
⇒ a = 32 – 28 = 4.

Question 4.
The first and the last terms of an A.P are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum ?
Solution:
Given A.P in which a = 17
Last term = l = 350
Common difference, d = 9
We know that, a_n = a + (n – 1) d
350 = 17 + (n – 1)9
⇒ 350 = 17 + 9n – 9
⇒ 9n = 350 – 8
⇒ n = \(\frac{342}{9}\) = 38
Now, S_n = \(\frac{n}{2}\)(a + l)
S₃₈ = \(\frac{38}{2}\)[17 + 350]
= 19 × 367 = 6973
∴ n = 38
S_n = 6973.

Question 5.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
Given A.P in which
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 2
Substituting d = 4 in equation (1), we get a + 4 = 14
⇒ a = 14 – 4 = 10
Now, S_n = \(\frac{n}{2}\)[a + l] (or) \(\frac{n}{2}\) [2a (n – 1) d]
∴ S₅₁ = \(\frac{51}{2}\)[2 × 10 + (51 – 1) × 4]
= \(\frac{51}{2}\)[20 + 50 × 4]
= \(\frac{51}{2}\) × (20 + 200)
= \(\frac{51}{2}\) × 220
= 51 × 110 = 5610.

Question 6.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms. (T.S. & A.P. Mar. 15, 16)
Solution:
Given :
A.P such that S₇ = 49
S₁₇ = 289
We know that,
S_n = \(\frac{n}{2}\)[2a + (n – 1)d]
S₇ = 49 = \(\frac{7}{2}\)[2a + (7 – 1) d]
⇒ \(\frac{49}{7}\) = \(\frac{1}{2}\)[2a + 6d]
⇒ 7 = a + 3d —– (1)
Also, S₁₇ = 289 = \(\frac{17}{2}\) [2a + (17 – 1)d]
\(\frac{289}{17}\) = \(\frac{1}{2}\)(2a + 16d)
⇒ 17 = a + 8d —— (2)
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 3
⇒ d = \(\frac{10}{5}\)
Substituting d = 2 in equation (1), we get
a + 3 × 2 = 7
⇒ a = 7 – 6 = 1
∴ a = 1; d = 2
Now, S_n = \(\frac{\mathrm{n}}{2}\)[2a + (n – 1)d]
S_n = \(\frac{\mathrm{n}}{2}\) [2 × 1 + (n -1)2]
= \(\frac{\mathrm{n}}{2}\) [2 + 2n – 2)d] = \(\frac{\mathrm{n} .2 \mathrm{n}}{2}\)
∴ Sum of first n terms S_n = n².
Shortcut : S₇ = 49 = 7²
S₁₇ = 289 = 17²
∴ S_n = n²

Question 7.
Show that a₁, a₂, ……. a_n, …. form an AP where an is defined as below :
i) a_n = 3 + 4n
ii) a_n = 9 – 5n. Also find the sum of the first 15 terms in each case.
Solution:
i) Given a_n = 3 + 4n
Then a₁ = 3 + 4 × 1 = 3 + 4 = 7
a₂ = 3 + 4 × 2 = 3 + 8 = 11
a₃ = 3 + 4 × 3 = 3 + 12 = 15
a₄ = 3 + 4 × 4 = 3 + 16 = 19
Now the pattern is 7, 11, 15, ……..
where a = a₁ = 7; a₂ = 11; a₃ = 15, …. and a₂ – a₁ = 11 – 7 = 4;
a₃ – a₂ = 15 -11 = 4; Here d = 4. Hence a₁, a₂, …., a_n … forms an A.P.
s₁₅ = \(\frac{15}{2}\) [2 × 7 + (15 – 1)4]
= \(\frac{15}{2}\) [14 + 14.4] = \(\frac{15}{2}\) [70]
= 15 × 35 = 525.

ii) a_n = 9 – 5n
Solution:
Given : a_n = 9 – 5n
a₁ = 9 – 5 × 1 = 9 – 5 = 4
a₂ = 9 – 5 × 2 = 9 – 10 = -1
a₃ = 9 – 5 × 3 = 9 – 15 = -6
a₄ = 9 – 5 × 4 = 9 – 20 = -11
………………………………..
Also a₂ – a₁ = -1 – 4 = – 5
a₃ – a₂ = – 6 – (-1) = – 6 + 1 = -5
a₄ – a₃ = – 11 – (- 6)
= -11 + 6
= -5
∴ d = a₂ – a₁ = a₃ – a₂
= a₄ – a₃ = ……. = -5
Thus the difference between any two succes-sive terms is constant.
Hence {a_n} forms A.P.
S₁₅ = \(\frac{15}{2}\)[2 × 4 + (15 – 1) × (-5)]
= \(\frac{15}{2}\) [8 + 14 (-5)] = \(\frac{15}{2}\) × -62
= 15 × – 31 = – 465.

Question 8.
If the sum of the first n terms of an AP is 4n – n², what is the first term (remember the first term is S₁) ? What is the sum of first two terms ? What is the second term ? Similarly, find the 3^rd, the 10^th and the n^th terms.
Solution:
Given an A.P in which S<sub?n = 4n – n²
Taking n = 1 we get
S₁ = 4 × 1 – 1² = 4 – 1 = 3
n = 2; S₂ = a₁ + a₂ = 4 × 2 – 2²
= 8 – 4 = 4
n = 3; S₃ = a₁ + a₂ + a₃
= 4 × 3 – 3² = 12 – 9 = 3
n = 4; S₄ = a₁ + a₂ + a₃ + a₄
= 4 × 4 – 4² = 16 – 16 = 0
Hence, S₁ = a₁ = 3
∴ a₂ = S₂ – S₁ = 4 – 3 = 1
a₃ = S₃ – S₂ = 3 – 4 = -1
a₄ = S₄ – S₃ = 0 – 3 = -3
So, d = a₂ – a₁ = 1 – 3 = -2
Now, a₁₀ = a + 9d [∴ a_n = a + (n – 1) d]
= 3 + 9 × (- 2)
= 3 – 18 = -15
a_n = 3 + (n – 1) × (-2)
= 3 – 2n + 2 = 5 – 2n.

Question 9.
Find the sum of the first 40 positive integers divisible by 6.
Solution:
The given numbers are the first 40 positive multiples of 6
⇒ 6 × 1, 6 × 2, 6 × 3, ……., 6 × 40
⇒ 6, 12, 18, …… 240
S_n = \(\frac{\mathrm{n}}{2}\) [a + l]
= \(\frac{40}{2}\) [6 + 240]
= 20 × 246 = 4920
∴ S₄₀ = 4920.

Question 10.
A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.
Solution:
Given :
Total l Sum of all cash prizes = ₹ 700
Each prize differs by ₹ 20
Let the prizes (in ascending order) be
x, x + 20, x + 40, x + 60, x + 80, x + 100, x + 120
∴ Sum of the prizes = S₇ = \(\frac{\mathrm{n}}{2}\) [a + l]
⇒ 700 = \(\frac{7}{2}\) [x + x + 120]
⇒ 700 × \(\frac{2}{7}\) = 2x + 120
⇒ 100 = x + 60
⇒ x = 100 – 60 = 40
∴ The prizes are 160, 140, 120, 100, 80, 60, 40.

Question 11.
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g : a section of Class I will plant 1 tree, a section of Class 11 will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students ?
Solution:
Given : Classes : From I to XII
Section : 3 in each class.
∴ Trees planted by each class = 3 × class number
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 4
∴ Total trees planted = 3 + 6 + 9 + 12 + ………+ 36; n = 12
∴ S_n = \(\frac{\mathrm{n}}{2}\) [a + l]
S₁₂ = \(\frac{12}{2}\) [3 + 36]
= 6 × 39 = 234
∴ Total plants = 234

Question 12.
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, … as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles ? (Take π = \(\frac{22}{7}\))
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 6
[Hint: Length of successive semi-circles is l₁, l₂, l₃, l₄,……… with centres at A, B, A, B,…., respectively.]
Solution:

Given : l₁; l₂, l₃, l₄, …….l₁₃ are the semi-circles
with centres alternately at A and B; with radii
r₁ = 0.5 cm [1 × 0.5]
r₂ = 1.0 cm [2 × 0.5]
r₃ = 1.5 cm [3 × 0.5]
r₄ = 2.0 cm [4 × 0.5]
………………………..
r₁₃ = 13 × 0.5 = 6.5
[∴ Radii are in A.P. as a₁ = 0.5 and d = 0.5]
Now, the total length of the spiral = l₁ + l₂ +….. + l₁₃ [… 13 given]
But circumference of a semi-circle is πr.
∴ Total length of the spiral = π × 0.5 + π × 1.0 + ……… + π × 6.5
= π × \(\frac{1}{2}\)[1 + 2 + 3 + ……. + 13]
Sum of the first n – natural numbers is \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)]

Question 13.
200 logs are stacked in the following manner : 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row ?
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 7
Solution:
Given : Total logs = 200
Number of logs stacked in the first row = 20
Number of logs stacked in the second row = 19
Number of logs stacked in the third row = 18
………………………………
The number series is 20, 19, 18, …. is an A.P.
where a = 20 and d = a₂ – a₁ = 19 – 20 = -1
Also, S_n = 200
∴ S_n = \(\frac{n}{2}\) [2a + (n – 1)d]
200 = \(\frac{n}{2}\) [2 × 20 + (n – 1) × (-1)]
200 = \(\frac{n}{2}\) [40 – n + 1]
200 = \(\frac{n}{2}\) [41 – n]
400 = 41n – n²
⇒ n² – 41n + 400 = 0
⇒ n² – 25n – 16n + 400 = 0
⇒ n(n – 25) – 16(n – 25) = 0
⇒ (n – 25) (n – 16) = 0
⇒ n = 25 (or) 16
There can’t be 25 rows as we are starting with 20 logs in the first row.
∴ Number of rows must be 16.
∴ n = 16 and t₁₆ = a + (n – 1) d = 20 + (16 – 1) (-1) = 20 – 15 = 5

Question 14.
In a bucket and ball race, a bucket is placed at the starting point, which is 5 m from the first ball, and the other balls are placed 3 m apart in a straight line. There are ten balls in the line.
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 8
A competitor starts from the bucket, picks up the nearest ball, runs back with it, drops it in the bucket, runs back to pick up the next ball, runs to the bucket to drop it in, and she continues in the same way until all the balls are in the bucket. What is the total distance the competitor has to run ?
[Hint : To pick up the first ball and the second ball, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)]
Solution:
Given : Balls are placed at an equal distance of 3 m from one another.
Distance of first ball from the bucket = 5 m
Distance of second ball from the bucket = 5 + 3 = 8m (5 + 1 × 3)
Distance of third ball from the bucket = 8 + 3 = 11 m (5 + 2 × 3)
Distance of fourth ball from the bucket = 11 + 3 = 14m (5 + 3 × 3)
…………………………………………………..
∴ Distance of the tenth ball from the bucket = 5 + 9 × 3 = 5 + 27 = 32m.
1^st ball: Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 5 = 10 m.
2^nd ball : Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 8 = 16 m.
3^rd ball : Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 11 = 22 m.
……………………………………………………
10^th ball: Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 32 = 64 m
Total distance = 10 m + 16 m + 22 m +………….+ 64 m.
Clearly, this is an A.P in which a = 10; d = a₂ – a₁ = 16 – 10 = 6 and n = 10.
S_n = \(\frac{\mathrm{n}}{2}\) [2a + (n – 1)d]
= \(\frac{10}{2}\) [2 × 10 + (10 – 1) × 6]
= 5[20 + 54] = 5 × 74 = 370 m
∴ Total distance = 370 m.

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

February 29, 2024February 28, 2024 by Mahesh

Students must practice these Maths 1B Important Questions TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type to help strengthen their preparations for exams.

TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type

Question 1.
Find the condition for the points (a, 0), (h, k), and (0, b), where ab ≠ 0, to be collinear. [Mar. ’10]
Solution:
Let A(a, 0), B(h, k), C (0, b) be the given points.
Slope of \(\overline{\mathrm{AB}}=\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}=\frac{\mathrm{k}-0}{\mathrm{~h}-\mathrm{a}}=\frac{\mathrm{k}}{\mathrm{h}-\mathrm{a}}\)
Slope of \(\overline{\mathrm{BC}}=\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}=\frac{\mathrm{b}-\mathrm{k}}{0-\mathrm{h}}=\frac{\mathrm{b}-\mathrm{k}}{-\mathrm{h}}\)
Since points A, B, C are collinear, then
Slope of \(\overline{\mathrm{AB}}\) = Slope of \(\overline{\mathrm{BC}}\)
\(\frac{k}{h-a}=\frac{b-k}{-h}\)
⇒ -hk = (h – a) (b – k)
⇒ -hk = bh – hk – ab + ak
⇒ bh + ak = ab
⇒ \(\frac{b h}{a b}+\frac{a k}{a b}=1\)
\(\frac{h}{a}+\frac{k}{b}=1\), which is the required condition.

Question 2.
Find the value of x, if the slope of the line passing through (2, 5) and (x, 3) is 2. [Mar. ’18 (AP & TS); May ’12; B.P.]
Solution:
Let A = (2, 5), B = (x, 3) are the given points.
Given, Slope of \(\overline{\mathrm{AB}}\) = 2
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q2

Question 3.
Find the value of y, if the line joining the points (3, y) and (2, 7) is parallel to the line joining the points (-1, 4) and (0, 6). [Mar. ’17 (TS), ’14, ’08]
Solution:
Let A = (3, y), B = (2, 7), C = (-1, 4), and D = (0, 6) are the given points.
Given,
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q3
Since, the lines \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) are parallel then the
Slope of \(\overline{\mathrm{AB}}\) = Slope of \(\overline{\mathrm{CD}}\)
⇒ y – 7 = 2
⇒ y = 9

Question 4.
Find the equation of the straight line which make 150° with the X-axis in the positive direction and which pass through the point (-2, -1). [May ’04]
Solution:
Given that inclination of a straight line is θ = 150°
The slope of a line is, m = tan θ
= tan (150°)
= tan (90° + 60°)
= -cot 60°
= \(\frac{-1}{\sqrt{3}}\)
Let the given point A(x₁, y₁) is (-2, -1).
∴ The equation of the straight line passing through A(-2, -1) and having slope \(\frac{-1}{\sqrt{3}}\) is y – y₁ = m(x – x₁)
⇒ y + 1 = latex]\frac{-1}{\sqrt{3}}[/latex] (x + 2)
⇒ √3(y + 1) = -1(x + 2)
⇒ √3y + √3 = -x – 2
⇒ x + √3y + √3 = -2
⇒ x + √3y + √3 + 2 = 0

Question 5.
Find the equations of the straight lines passing through the origin and making equal angles with the coordinate axes. [May ’05]
Solution:
Let l₁, l₂ are the equations of the straight lines passing through the origin and making equal angles with the co-ordinate axes
Case I: Inclination of a straight line l₁ is θ = 45°
Slope of a line l₁ is, m = tan θ = tan 45° = 1
Let the given point O = (0, 0)
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q5
∴ The equation of a straight line l₁ passing through O(0, 0) and having slope ‘1’ is y – y₁ = m(x – x₁)
y – 0 = 1(x – 0)
⇒ y = x
⇒ x – y = 0
Case II: Inclination of a straight line l₂ is θ = 135°
The slope of a line l₂ is, m = tan θ
= tan 135°
= tan (90° + 45°)
= -tan 45°
= -1
Let the given point O = (0, 0)
∴ The equation of a straight line l₂ passing through O(0, 0) and having slope ‘-1’ is y – y₁ = m(x – x₁)
⇒ y – 0 = -1(x – 0)
⇒ y = -x
⇒ x + y = 0
∴ Required equations of the straight lines are x – y = 0; x + y = 0

Question 6.
Find the equation of the straight line passing through (-4, 5) and cutting off equal and non-zero intercepts on the coordinate axes. [Mar. ’15 (TS) ’13 (Old), ’07, ’00; May ’10, ’08]
Solution:
The equation of a straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) …….(1)
Given that, the straight line making equal intercepts on the co-ordinate axis, then a = b
From (1),
\(\frac{x}{a}+\frac{y}{a}=1\)
x + y = a ……….(2)
Since equation (2) passes through the point (-4, 5) then,
-4 + 5 = a
∴ a = 1
Substitute the value of ’a’ in equation (2)
∴ x + y = 1

Question 7.
Find the equation of the straight line passing through (-2, 4) and making non-zero intercepts whose sum is zero. [Mar. ’15 (AP), ’13; May ’15 (TS), ’02]
Solution:
The equation of a straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) ……..(1)
Given that, the straight line-making intercepts whose sum is ‘0’
i.e., a + b = 0
b = -a
From (1)
\(\frac{x}{a}+\frac{y}{-a}=1\)
x – y = -a …….(2)
Since equation (2) passes through the point (-2, 4) then,
-2 – 4 = a
∴ a = -6
Substitute the value of ‘a’ in equation (2)
x – y = -6
x – y + 6 = 0

Question 8.
Find the equation of the straight line passing through the point (3, -4) and making X and Y-intercepts which are in the ratio 2 : 3. [Mar. ’08]
Solution:
The equation of a straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) ……..(1)
Given that, the ratio of intercepts = 2 : 3
X-intercept = 2a
Y-intercept = 3a
From (1),
\(\frac{x}{2 a}+\frac{y}{3 a}=1\)
\(\frac{3 x+2 y}{6 a}=1\)
3x + 2y = 6a ……..(2)
Since equation (2) passes through point (3, -4) then,
3(3) + 2(-4)= 6a
⇒ 9 – 8 = 6a
⇒ 6a = 1
⇒ a = \(\frac{1}{6}\)
Substitute the value of ‘a’ in equation (2)
3x + 2y = 6(\(\frac{1}{6}\))
∴ 3x + 2y = 1

Question 9.
Find the equation of the straight line passing through the points \(\left(a t_1^2, 2 at_1\right)\) and \(\left(a t_2^2, 2 at_2\right)\). [Mar. ’14. ’04; May ’15 (AP), ’00]
Solution:
Let A\(\left(a t_1^2, 2 at_1\right)\) and B\(\left(a t_2^2, 2 at_2\right)\) are the given points
The equation of the straight line passing through the points A\(\left(a t_1^2, 2 at_1\right)\) and B\(\left(a t_2^2, 2 at_2\right)\) is
(y – y₁) (x₂ – x₁) = (x – x₁) (y₂ – y₁)
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q9

Question 10.
Find the equation of the straight line passing through A(-1, 3) and (i) parallel (ii) perpendicular to the straight line passing through B(2, -5) and C(4, 6). [May ’12; Mar. ’11]
Solution:
A(-1, 3), B(2, -5), C(4, 6) are the given points.
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q10
(i) Slope of the parallel line = m = \(\frac{11}{2}\)
∴ The equation of the straight line passing through A(-1, 3) and having slope \(\frac{11}{2}\) is y – y₁ = m(x – x₁)
y – 3 = \(\frac{11}{2}\)(x + 1)
⇒ 2y – 6 = 11x + 11
⇒ 11x – 2y + 17 = 0
(ii) Slope of the perpendicular line = \(\frac{-1}{m}=\frac{-1}{11 / 2}=\frac{-2}{11}\)
∴ The equation of the straight line passing through A(-1, 3) and having slope \(\frac{-2}{11}\) is y – y₁ = \(\frac{-1}{m}\) (x – x₁)
y – 3 = \(\frac{-2}{11}\) (x + 1)
⇒ 11y – 33 = -2x – 2
⇒ 2x + 11y – 31 = 0

Question 11.
A(10, 4), B(-4, 9) and C(-2, -1) are the vertices of a triangle. Find the equation of the altitude through B. [May ’13 (Old)]
Solution:
Slope of \(\overline{\mathrm{AC}}\) is,
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q11
The equation of the altitude through B is, the equation of the straight line passing through B(-4, 9) and having slope \(\frac{-12}{5}\) is
y – y₁ = m(x – x₁)
y – 9 = \(\frac{-12}{5}\) (x + 4)
5y – 45 = -12x – 48
12x + 5y + 3 = 0

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. [May ’90]
Solution:
Let a, b be the intercepts of a line.
∴ The line cuts the X-axis at A(a, 0), Y-axis at B(0, b)
Midpoint of \(\overline{\mathrm{AB}}\) is,
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q12

Question 13.
Find the angle made by the straight line y = -√3x + 3 with the positive direction of the X-axis measured in the counterclockwise direction. [May ’94]
Solution:
Given, the equation of the straight line is y = -√3x + 3
Comparing this equation with y = mx + c, We get
m = -√3 (∵ m = tan θ)
⇒ tan θ = -√3
⇒ tan θ = tan \(\frac{2 \pi}{3}\)
⇒ θ = \(\frac{2 \pi}{3}\)
∴ The angle made by the straight line is θ = \(\frac{2 \pi}{3}\)

Question 14.
Transform the equation √3x + y = 4 into (i) slope-intercept form (ii) intercept form (iii) normal form. [May ’16 (TS)]
Solution:
Given, the equation of the straight line is √3x + y = 4
(a) Slope-intercept form:
√3x + y = 4
y = -√3x + 4 which is of the form y = mx + c
where, slope (m) = -√3, y-intercept (c) = 4
(b) Intercept form:
Given, the equation of the straight line is,
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q14
which is in the form \(\frac{x}{a}+\frac{y}{b}=1\)
∴ x-intercept (a) = \(\frac{4}{\sqrt{3}}\), y-intercept (b) = 4
(c) Normal form:
Given the equation of the straight line is √3x + y = 4
On dividing both sides with
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q14.1

which is in the form x cos α + y sin α = p
∴ α = \(\frac{\pi}{6}\), p = 2

Question 15.
Transform the equation x + y + 1 = 0 into normal form. [Mar. ’17 (AP), ’08; May ’10; B.P.; Mar. ’18 (TS)]
Solution:
Given, equation of the straight line is x + y + 1 = 0
x + y = -1
– x – y = 1
On dividing both sides with
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q15

Question 16.
Transform the equation 4x – 3y + 12 = 0 into (i) slope-intercept form (ii) intercept form (iii) normal form.
Solution:
(a) Slope-intercept form:
Given equation of the line is 4x – 3y + 12 = 0
3y = 4x + 12
y = \(\frac{4 x+12}{3}=\left(\frac{4}{3}\right) x+4\)
which is in the form of y = mx + c
∴ Slope = \(\frac{4}{3}\), y-intercept = 4
(b) Intercept form:
Given equation is 4x – 3y + 12 = 0
\(\frac{x}{-3}+\frac{y}{4}=1\)
which is in the form of \(\frac{x}{a}+\frac{y}{b}=1\)
∴ x-intercept = -3, y-intercept = 4
(c) Normal form:
Given the equation of the straight line is 4x – 3y = -12
-4x + 3y = 12
On dividing both sides by \(\sqrt{a^2+b^2}\)
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q16

Question 17.
Transform the equation x + y – 2 = 0 into (i) slope-intercept form (ii) intercept form (iii) normal form. [Mar. ’12]
Solution:
(a) Slope-intercept form:
Given equation of the straight line is, x + y – 2 = 0
y = -x + 2
which is in the form of y = mx + c
∴ Slope, m = -1, y-intercept, c = 2
(b) Intercept form:
Given equation of the straight line is, x + y – 2 = 0
x + y = 2
\(\frac{x+y}{2}\) = 1
\(\frac{x}{2}+\frac{y}{2}\) = 1
which is of the form \(\frac{x}{a}+\frac{y}{b}=1\)
∴ x-intercept (a) = 2, y-intercept (b) = 2
(c) Normal form:
Given equation of the straight line is, x + y – 2 = 0
x + y = 2
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q17

Question 18.
Transform the equation √3x + y + 10 = 0 into (i) slope-intercept form (ii) intercept form (iii) normal form. [May ’04]
Solution:
(a) Slope-intercept form:
Given the equation of the straight line is, √3x + y + 10 = 0
y = -√3x – 10 = -√3x + (-10)
which is in the form of y = mx + c
∴ Slope, m = -√3, y-intercept, c = -10
(b) Intercept form:
Given the equation of the straight line is √3x + y + 10 = 0
√3x + y = -10 × 1
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q18
(c) Normal form:
Given equation of the straight line is, √3x + y + 10 = 0
√3x + y = -10
-√3x – y = 10

Question 19.
If the area of the triangle is formed by the straight lines, x = 0, y = 0, and 3x + 4y = a [a > 0] is ‘6’. Find the value of ‘a’. [May ’11, Mar. ’09, ’07]
Solution:
Given equations of the straight lines are (a > 0) 3x + 4y = a, x = 0 and y = 0
Comparing with ax + by + c = 0, we get
a = 3, b = 4, c = -a
The area of the triangle formed by this line and the co-ordinate axis is equal to \(\frac{c^2}{2|a b|}\)
Given that area of the triangle = 6
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q19

Question 20.
Find the value of p, if the straight lines x + p = 0, y + 2 = 0 and 3x + 2y + 5 = 0 are concurrent. [Mar. ’17 (TS), ’13; May ’15 (TS)]
Solution:
Given, the equation of the straight lines
x + p = 0 ……..(1)
y + 2 = 0 ………(2)
3x + 2y + 5 = 0 ………(3)
Solving (2) & (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q20
∴ Point of intersection of the lines (2) & (3) is (\(\frac{-1}{3}\), -2)
since given lines are concurrent, then, the point of intersection (\(\frac{-1}{3}\), -2) lies on (1)
x + p = 0
⇒ \(\frac{-1}{3}\) + p = 0
⇒ p = \(\frac{1}{3}\)

Question 21.
Find the ratio in which the straight line 2x + 3y = 5 divides the line joining the points (0, 0) and (-2, 1). [Mar. ’14]
Solution:
Given the equation of the straight line is L = 2x + 3y – 5 = 0
Comparing the equation with ax + by + c = 0, we get
a = 2, b = 3, c = 5
Let the given points are A(x₁, y₁) = (0, 0) and B(x₂, y₂) = (-2, 1)
Required ratio = \(\frac{-\left(a x_1+b {y}_1+c\right)}{a x_2+b y_2+c}\)
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q21

Question 22.
Find the distance between the parallel straight lines 3x + 4y – 3 = 0 and 6x + 8y – 1 = 0. [Mar. ’19 (AP); May ’13]
Solution:
Given, equations of the straight lines are 3x + 4y – 3 = 0, 6x + 8y – 1 = 0
6x + 8y – 6 = 0 …..(1)
6x + 8y – 1 = 0 …..(2)
Comparing (1) with ax + by + c₁ = 0, we get
a = 6, b = 8, c₁ = -6
Comparing (2) with ax + by + c₂ = 0, we get
a = 6, b = 8, c₂ = -1
Distance between the parallel lines =
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q22

Question 23.
Find the equation of ‘k’, if the angle between the straight lines 4x – y + 7 = 0 and kx – 5y – 9 = 0 is 45°. [Mar. ’12, ’08, ’82; May ’11, ’02]
Solution:
Given, the equations of the straight lines are
4x – y + 7 = 0 ……..(1)
kx – 5y – 9 = 0 ……..(2)
Comparing (1) with a₁x + b₁y + c₁ = 0, we get
a₁ = 4, b₁ = -1, c₁ = 7
Comparing (2) with a₂x + b₂y + c₂ = 0, we get
a₂ = k, b₂ = -5, c₂ = -9
Given that, θ = 45°
If ‘θ’ is the angle between the given lines then,
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q23

Squaring on both sides
⇒ 17(k² + 25) = 2(4k + 5)²
⇒ 17k² + 425 = 2(16k² + 25 + 40k)
⇒ 17k² + 425 = 32k² + 50 + 80k
⇒ 15k² + 80k – 375 = 0
⇒ 3k² + 16k – 375 = 0
⇒ 3k² + 25k – 9k – 75 = 0
⇒ k(3k + 25) – 3(3k + 25) = 0
⇒ (3k + 25) (k – 3) = 0
⇒ 3k + 25 = 0; k – 3 = 0
⇒ 3k = -25; k = 3
⇒ k = 3 or \(\frac{-25}{3}\)

Question 24.
Find the equation of the straight line parallel to the line 2x + 3y + 7 = 0 and pass through the point (5, 4). [Mar. ’13, ’03]
Solution:
Given, the equation of the straight line is 2x + 3y + 7 = 0
Given points (5, 4)
The equation of the straight line parallel to 2x + 3y + 7 = 0 is 2x + 3y + k = 0
Since equation (1) passes through the point (5, 4) then,
2(5) + 3(4) + k = 0
⇒ 10 + 12 + k = 0
⇒ 22 + k = 0
⇒ k = -22
∴ The required equation of the straight line is 2x + 3y – 22 = 0

Question 25.
Find the value of k, if the straight lines y – 3kx + 4 = 0 and (2k – 1)x – (8k – 1)y – 6 = 0 are perpendicular. [Mar. ’10]
Solution:
Given, the equations of the straight lines are
y – 3kx + 4 = 0 ………(1)
(2k – 1)x – (8k – 1)y – 6 = 0 ……….(2)
Slope of the line (1) is m₁ = \(\frac{-(-3 k)}{1}\) = 3k
Slope of the line (2) is m₂ = \(\frac{-(2 k-1)}{-(8 k-1)}=\frac{(2 k-1)}{(8 k-1)}\)
Since the given lines are perpendicular then m₁ × m₂ = -1
\(3 \mathrm{k}\left(\frac{2 \mathrm{k}-1}{8 \mathrm{k}-1}\right)\) = -1
⇒ 3k(2k – 1) = -1(8k – 1)
⇒ 6k² – 3k = -8k + 1
⇒ 6k² + 5k – 1 = 0
⇒ 6k² + 6k – k – 1 = 0
⇒ 6k(k + 1) – 1(k + 1) = 0
⇒ (k + 1)(6k – 1) = 0
⇒ k + 1 = 0 (or) 6k – 1 = 0
⇒ k = -1 (or) k = \(\frac{1}{6}\)

Question 26.
Find the perpendicular distance from the point (3, 4) to the straight line 3x – 4y + 10 = 0. [Mar. ’16 (AP); May. ’15 (AP)]
Solution:
Given the equation of the straight line is 3x – 4y + 10 = 0
Comparing with ax + by + c = 0, we get
a = 3; b = -4, c = 40
Let the given point p(x₁, y₁) = (3, 4)
The perpendicular distance from the point (3, 4) to the line 3x – 4y + 10 = 0 is
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q26

Question 27.
Find the slopes of the lines x + y = 0 and x – y = 0. [Mar. ’17 (AP)]
Solution:
Slope of x + y = 0 is \(\frac{-a}{b}=\frac{-1}{1}\) = -1
Slope of x – y = 0 is \(\frac{-a}{b}=\frac{-1}{-1}\) = 1

TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type

February 29, 2024February 28, 2024 by Mahesh

Students must practice these Maths 1B Important Questions TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type to help strengthen their preparations for exams.

TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type

Question 1.
Transform the equation \(\frac{x}{a}+\frac{y}{b}=1\) into the normal form when a > 0 and b > 0. If the perpendicular distance of the straight line from the origin is p, deduce that \(\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}\). [May ’08, ’04, ’02, ’97, ’95, ’90; Mar. ’07, ’02, ’00]
Solution:
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q1

Question 2.
Find the points on the line 3x – 4y – 1 = 0 which are at a distance of 5 units from the point (3, 2). [Mar. ’16 (AP); ’15 (AP); B.P.]
Solution:
Let the given point A(x₁, y₁) = (3, 2)
Given the equation of the straight line is 3x – 4y – 1 = 0
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q2
Distance |r| = 5
Required points = (x₁ + |r| cos θ, y1 + |r| sin θ)
= (x₁ ± r cos θ, y₁ + r sin θ)
= (3 ± 5 . \(\frac{4}{5}\), 2 ± 5 . \(\frac{3}{5}\))
= (3 ± 4, 2 ± 3)
= (3 + 4, 2 + 3); (3 – 4, 2 – 3)
= (7, 5), (-1, 1)

Question 3.
Find the value of k, if the lines 2x – 3y + k = 0, 3x – 4y – 13 = 0 and 8x – 11y – 33 = 0 are concurrent. [May ’07, ’95; Mar. ’05, ’80; Mar. ’18 (TS)]
Solution:
Given, the equations of the straight lines are,
2x – 3y + k = 0 ……..(1)
3x – 4y – 13 = 0 ……..(2)
8x – 11y – 33 = 0 ………(3)
Solving (2) & (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q3

∴ The point of intersection (11, 5) lies on (1)
2x – 3y + k = 0
⇒ 2(11) – 3(5) + k = 0
⇒ 22 – 15 + k = 0
⇒ 7 + k = 0
⇒ k = -7

Question 4.
If the straight lines ax + by + c = 0, bx + cy + a = 0 and cx + ay + b = 0 are concurrent, then prove that a³ + b³ + c³ = 3abc. [Mar. ’19 (AP); Mar. ’08; May. ’00]
Solution:
Given, the equations of the straight lines are,
ax + by + c = 0 ……(1)
bx + cy + a = 0 ………(2)
cx + ay + b = 0 ……..(3)
Solving (2) & (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q4
\(\frac{y}{b c-a^2}=\frac{1}{a c-b^2}\) ⇒ y = \(\frac{b c-a^2}{a c-b^2}\)
∴ Point of intersection of the straight lines (1) & (2) is \(\left(\frac{a b-c^2}{a c-b^2}, \frac{b c-a^2}{a c-b^2}\right)\)
Since the given lines are concurrent, then the point of intersection lies on line (3)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q4.1
⇒ c(ab – c²) + a(bc – a²) + b(ac – b²) = 0
⇒ abc – c³ + abc – a³ + abc – b³ = 0
⇒ 3abc – a³ – b³ – c³ = 0
⇒ a³ + b³ + c³ = 3abc

Question 5.
A variable straight line drawn through the point of intersection of the straight lines \(\frac{x}{a}+\frac{y}{b}=1\) and \(\frac{x}{a}+\frac{y}{b}=1\) meets the coordinate axes at A and B. Show that the locus of the midpoint of \(\overline{\mathbf{A B}}\) is 2(a + b) xy = ab(x + y). [May ’05]
Solution:
Given equations of the straight lines are,
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q5
The point of intersection of lines (1) & (2) is
C = \(\left(\frac{a b}{a+b}, \frac{a b}{a+b}\right)\)
The equation of the straight line \(\overline{\mathbf{A B}}\) is the intercept from \(\frac{x}{p}+\frac{y}{q}=1\) ……(3)
The straight line (3) meets the X-axis at A(p, 0), Y-axis at B(0, q).
Let Q(x₁, y₁) be any point on the locus.
Since Q(x₁, y₁) is the midpoint of \(\overline{\mathbf{A B}}\)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q5.1

then, \(y_1\left(\frac{a b}{a+b}\right)+\left(\frac{a b}{a+b}\right) x_1=2 x_1 y_1\)
\(\frac{a b y_1+a b x_1}{a+b}=2 x_1 y_1\)
ab(x₁ + y₁) = (a + b) 2x₁y₁
∴ The locus of the midpoint ‘Q’ of AB is 2(a + b) xy = ab(x + y)

Question 6.
A straight line meets the coordinate axes in A and B. Find the equation of the straight line when (p, q) bisects \(\overline{\mathbf{A B}}\). [May ’90]
Solution:
The equation of the straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) …….(1)
The straight line (1) meets the X-axis at A(a, 0), Y-axis at B(0, b).
Now, C(p, q) bisects \(\overline{\mathbf{A B}}\), then C is the midpoint of \(\overline{\mathbf{A B}}\).
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q6

Question 7.
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). [May ’07]
Solution:
Equation of the straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) …….(1)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q7
Since equation (1) passes through the point (3, 4) then,
\(\frac{3}{a}+\frac{4}{b}=1\)
\(\frac{3 b+4 a}{a b}\) = 1
3b + 4a = ab
4a = ab – 3b
4a = b(a – 3)
b = \(\frac{4 a}{a-3}\) ……(2)
Given that, area of ΔOAB = 24 sq.units
\(\frac{1}{2}\)ab = 24
ab = 48
\(a\left(\frac{4 a}{a-3}\right)=48\)
a² = 12(a – 3)
a = 12a – 36
a² – 12a + 36 = 0
(a – 6)² = 0
a = 6
from (2), b = \(\frac{4(6)}{6-3}\) = 8
The equation of the straight line is, from (1)
\(\frac{x}{6}+\frac{y}{8}\) = 1
\(\frac{4 x+3 y}{24}\) = 1
4x + 3y = 24

Question 8.
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. [May ’10]
Solution:
Given that,
3a + 2b + 4c = 0
4c = -3a – 2b
c = \(\frac{-3 a-2 b}{4}\)
Now, ax + by + c = 0
ax + by + \(\left(\frac{-3 a-2 b}{4}\right)\) = 0
\(\frac{4 a x+4 b y-3 a-2 b}{4}\) = 0
4ax + 4by – 3a – 2b = 0
a(4x – 3) + b(4y – 2) = 0
(4x – 3) + \(\frac{b}{a}\) (4y – 2) = 0
This is of the form L₁ + λL₂ = 0
Here, ax + by + c = 0, represents a set of lines passing through the point of intersection of the lines
L₁ = 4x – 3 = 0 …….(1)
L₂ = 4y – 2 = 0 ………(2)
Solving (1) & (2)
From (1), 4x – 3 = 0
4x = 3
x = \(\frac{3}{4}\)
from (2), 4y – 2 = 0
4y = 2
y = \(\frac{1}{2}\)
∴ The point of concurrence = \(\left(\frac{3}{4}, \frac{1}{2}\right)\)
∴ ax + by + c = 0 represents a set of concurrent lines.
The point of concurrence = \(\left(\frac{3}{4}, \frac{1}{2}\right)\)

Question 9.
Find the point on the straight line 3x + y + 4 = 0, which is equidistant from the points (-5, 6) and (3, 2). [Mar. ’13; Nov. ’98]
Solution:
Given, equation of the straight line is 3x + y + 4 = 0 …….(1)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q9
Let the given points are A(-5, 6) & B(3, 2)
Let P(x, y) be a point on the straight line 3x + y + 4 = 0,
Given that, PA = PB
\(\sqrt{(x+5)^2+(y-6)^2}=\sqrt{(x-3)^2+(y-2)^2}\)
Squaring on both sides
(x + 5)² + (y – 6)² = (x – 3)² + (y – 2)²
x² + 25 + 10x + y² + 36 – 12y = x² + 9 – 6x + y² – 4y + 4
10x – 12y + 61 = -6x – 4y + 13
16x – 8y + 48 = 0
2x – y + 6 = 0 ………(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q9.1

Question 10.
A straight line through Q(√3, 2) makes an angle \(\frac{\pi}{6}\) with the positive direction of the X-axis. If the straight line intersects the line √3x – 4y + 8 = 0 at P, find the distance PQ. [Mar. ’19 (TS); Mar. ’04]
Solution:
Given equation of the straight line is √3x – 4y + 8 = 0 ……..(1)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q10
Given point Q(x₁, y₁) = (√3, 2)
Inclination of a straight line, θ = \(\frac{\pi}{6}\) = 30°
Slope of a straight line, m = tan 30° = \(\frac{1}{\sqrt{3}}\)
∴ The equation of the straight line \(\overline{\mathrm{PQ}}\) having slope \(\frac{1}{\sqrt{3}}\) and passing through the
point Q(√3, 2) is, y – y₁ = m(x – x₁)
y – 2 = \(\frac{1}{\sqrt{3}}\) (x – √3)
√3y – 2√3 = x – √3
x – √3y + √3 = 0 …….(2)
Solving (1) & (2)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q10.1

Question 11.
The line \(\frac{x}{a}-\frac{y}{b}=1\) meets the X-axis at P. Find the equation of the line perpendicular to the line at P. [May ’03]
Solution:
Given, the equation of the straight line is \(\frac{x}{a}-\frac{y}{b}=1\) …..(1)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q11
Since line (1) meets the X-axis at P.
Then y-coordinate = 0
\(\frac{x}{a}-\frac{0}{b}=1\)
x = a
∴ The coordinates of P = (a, 0)
The slope of the line (1) is m = \(\frac{\frac{-1}{a}}{\frac{-1}{b}}=\frac{b}{a}\)
Slope of the perpendicular line = \(\frac{-1}{b/a}=\frac{-a}{b}\)
∴ The equation of the line passing through P(a, 0) and having slope \(\frac{-a}{b}\) is,
y – y₁ = \(\frac{-1}{m}\) (x – x₁)
y – 0 = \(\frac{-a}{b}\) (x – a)
y = \(\frac{-a}{b}\) (x – a)
by = -ax + a²
ax + by – a² = 0
which is the required equation of a straight line.

Question 12.
Find the equation of the line perpendicular to the line 3x + 4y + 6 = 0 and make an intercept -4 on the X-axis. [Mar. ’10]
Solution:
Given, equation of the straight line is 3x + 4y + 6 = 0 …….(1)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q12
Slope of the line (1) is m = \(\frac{-3}{4}\)
Slope of the perpendicular line = \(\frac{-1}{m}=\frac{-1}{\left(\frac{-3}{4}\right)}=\frac{4}{3}\)
Given that, the required a straight line making an intercept -4 on X-axis. Then P = (-4, 0).
Equation of the straight line passing through P(-4, 0) and having slope \(\frac{4}{3}\) is
(y – y₁) = \(\frac{-1}{m}\) (x – x₁)
y – 0 = \(\frac{4}{3}\) (x + 4)
3y = 4x + 16
4x – 3y + 16 = 0

Question 13.
Find the equation of the straight line making non-zero equal intercepts on the coordinate axes and passing through the point of intersection of the lines 2x – 5y + 1 = 0 and x – 3y – 4 = 0. [Mar. ’06, ’00]
Solution:
Given the equation of the lines are
2x – 5y + 1 = 0 …….(1)
x – 3y – 4 = 0 …….(2)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q13
\(\frac{x}{23}=\frac{y}{9}=\frac{1}{-1}\)
\(\frac{x}{23}\) = -1; \(\frac{y}{9}\) = -1
x = -23; y = -9
∴ Point of intersection of lines (1) & (2) is, P = (-23, -9)
The equation of the straight line in the intercept form is, \(\frac{x}{a}+\frac{y}{b}=1\) = 1 ……..(3)
The straight line (3) makes equal intercepts on the coordinate axes
from (3),
\(\frac{x}{a}+\frac{y}{a}\)
x + y = a ……..(4)
Since equation (4) passes through the point P(-23, -9) then,
-23 – 9 = a
a = -32
∴ The equation of the straight line is x + y = -32
x + y + 32 = 0

Question 14.
Find the length of the perpendicular drawn from the point of intersection of the lines 3x + 2y + 4 = 0 and 2x + 5y – 1 = 0 to the straight line 7x + 24y – 15 = 0. [May ’01; Mar. ’91]
Solution:
Given, the equation of the straight lines are
3x + 2y + 4 = 0 ……(1)
2x + 5y – 1 = 0 ……..(2)
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q14
∴ The point of intersection of lines (1) & (2) is, P = (-2, 1)
Given the equation of the straight line is 7x + 24y – 15 = 0
Comparing with ax + by + c = 0 then a = 7, b = 24, c = -15
Point of intersection P(x₁, y₁) = (-2, 1)
The perpendicular distance from P(-2, 1) to the straight line
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q14.1

Question 15.
If θ is the angle between the lines \(\frac{x}{a}+\frac{y}{b}=1\) and \(\frac{x}{b}+\frac{y}{a}=1\), find the value of sin θ, where a > b. [May ’09]
Solution:
Given, the equation of the straight lines are
\(\frac{x}{a}+\frac{y}{b}=1\)
\(\frac{b x+a y}{a b}\) = 1
bx + ay = ab
bx + ay – ab = 0 ……..(1)
\(\frac{x}{b}+\frac{y}{a}=1\)
ax + by = ab
ax + by – ab = 0 ………(2)
Comparing (1) with a₁x + b₁y + c₁ = 0, we get
a₁ = b, b₁ = a, c₁ = -ab
Comparing (2) with a₂x + b₂y + c₂ = 0, we get
a₂ = a, b₂ = b, c₂ = -ab
If ‘θ’ is the angle between lines (1) & (2) then,
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q15

Question 16.
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). [May ’15 (AP)]
Solution:
The slope of the line passing through the points (3, -5) and (-6, 1) is
m = \(\frac{y_2-y_1}{x_2-x_1}=\frac{1+5}{-6-3}=\frac{-6}{9}=\frac{-2}{3}\)
(i) Equation of the line passing through (1, 3) and parallel to the line passing through the points (3, -5) and (-6, 1) is y – y₁ = m(x – x₁)
⇒ y – 3 = \(\frac{-2}{3}\) (x – 1)
⇒ 3y – 9 = -2x + 2
⇒ 2x + 3y – 11 = 0
(ii) Equation of the line passing through (1, 3) and perpendicular to the line passing through the points (3, -5) and (-6, 1) is y – y₁ = \(\frac{-1}{m}\) (x – x₁)
⇒ y – 3 = \(\frac{1}{2}\) (x – 1)
⇒ 2y – 6 = 3x – 3
⇒ 3x – 2y + 3 = 0

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise

February 28, 2024February 27, 2024 by Mahesh

Students can practice TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise

Question 1.
Some points are plotted on a plane. Each point is joined with all remaining points by line segments. Find the number of points if the number of line segments are 10.
Solution:
Number of distinct line segments that can be formed out of n-points = \(\frac{\mathrm{n}(\mathrm{n}-1)}{2}\)
Given : No. of line segments \(\frac{\mathrm{n}(\mathrm{n}-1)}{2}\) = 10
⇒ n² – n = 20
⇒ n² – n – 20 = 0
⇒ n² – 5n + 4n – 20 = 0
⇒ n(n- 5) + 4(n – 5) = 0
⇒ (n – 5) (n + 4) = 0
⇒ n – 5 = 0 (or) n + 4 = 0
⇒ n = 5 (or) – 4
∴ n = 5 [n – can’t be negative]

Question 2.
A two digit number is such that the product of the digits is 8. When 18 is added to the number, they interchange their places. Determine the number.
Solution:
Let the digit in the units place = x
Let the digit in the tens place = y
∴ The number = 10y + x
By interchanging the digits the number becomes 10x + y
By problem (10x + y) – (10y + x) = 18
⇒ 9x + 9y = 18
⇒ 9(x – y) = 18
⇒ x – y = \(\frac{18}{9}\) = 2
⇒ y = x – 2
(i.e.) digit in the tens place = x – 2
digit in the units place = x
Product of the digits = (x – 2) x
By problem x² – 2x = 8
x² – 2x – 8 = 0
⇒ x² – 4x + 2x – 8 = 0
⇒ x(x – 4) + 2(x – 4) = 0
⇒ (x – 4) (x + 2) = 0
⇒ x – 4 = 0 (or) x + 2 = 0
⇒ x = 4 (or) x = -2
x = 4 [ ∵ x can’t be negative]
∴ The number is 24.

Question 3.
A piece of wire 8m in length, cut into two pieces and each piece is bent into a square. Where should the cut in the wire be made if the sum of the areas of these squares is to be 2 m² ?
[Hint: x + y = 8, (\(\frac{x}{4}\))² + (\(\frac{y}{4}\))² = 2
⇒ (\(\frac{x}{4}\))² + \(\left(\frac{8-x}{4}\right)\) = 2
Solution:
Let the length of the first piece = x m
Then length of the second piece = (8 – x) m
∴ Side of the 1^st square = \(\frac{x}{4}\) m and
Side of the second square = \(\frac{8-x}{4}\) m
Sum of the areas = 2 m²
\(\left(\frac{x}{4}\right)^2\) + \(\left(\frac{8-x}{4}\right)^2\) = 2
⇒ x² + 64 + x² – 16x = 16 × 2 = 32
⇒ 2x² – 16x + 64 = 32
⇒ 2x² – 16x + 32 = 0
⇒ 2(x² – 8x + 16) = 0
⇒ x² – 4x – 4x + 16 = 0
⇒ x(x – 4) – 4(x – 4) = 0
⇒ (x – 4) (x – 4) = 0
∴ x = 4
The cut should be made at the centre making two equal pieces of length 4m, 4m.

Question 4.
Vinay and Praveen working together can paint the exterior of a house in 6 days. Vinay by himself can complete the job in 5 days less than Praveen. How long will it take Vinay to complete the job by himself ?
Solution:
Let the time taken by Vinay to complete the job = x days
Then the time taken by Praveen to complete the job = x + 5 days
Both worked for 6 days to complete a job.
∴ Total work done by them is
\(\frac{6}{x}\) + \(\frac{6}{x+5}\) = 1
⇒ 6(2x + 5) = x² + 5x
⇒ x² – 7x – 30 = 0
⇒ x² – 10x + 3x – 30 = 0
⇒ x(x – 10) + 3(x – 10) = 0
⇒ x – 10 = 0 (or) x + 3 = 0
⇒ x = 10(or) x = -3
x = 10 (∵ x can’t be negative)
∴ Time taken by Vinay = x = 10 days
Time taken by Praveen = x + 5 = 15 days.

Question 5.
Show that the sum of roots of a quadratic equation is \(\frac{-\mathbf{b}}{\mathbf{a}}\).
Solution:
Let the Q.E. = ax² + bx + c = 0 (a ≠ 0)
⇒ ax² + bx = -c
⇒ x² + \(\frac{\mathrm{b}}{\mathrm{a}}\)x = \(\frac{-\mathrm{c}}{\mathrm{a}}\)
⇒ x² + 2.\(x \cdot \frac{b}{2 a}\) = \(-\frac{c}{a}\)
TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 6
sum of the roots

∴ Sum of roots of a Q.E. is \(\frac{-b}{a}\)

Question 6.
Show that the product of the roots of a Q.E is \(\frac{\mathbf{c}}{\mathrm{a}}\).
Solution:
Let the Q.E. = ax² + bx + c = 0(a ≠ 0)
⇒ ax² + bx = -c
TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 8

Product of the roots

Question 7.
The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 2\(\frac{16}{21}\), find the fraction.
Solution:
Let the numerator = x
then denominator = 2x + 1
Then the fraction = \(\frac{x}{2 x+1}\)
Its reciprocal = \(\frac{2 x+1}{x}\)
TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 11
⇒ 105x² + 84x + 21 = 116x² + 58x
⇒ 11x² – 26x – 21 = 0
⇒ 11x² – 33x + 7x – 21 = 0
⇒ 11x (x – 3) + 7(x – 3) = 0
⇒ (x – 3)(11x + 7) = 0
⇒ x – 3 = 0 (or) 11x + 3 = 0
⇒ x = 3 (or) \(\frac{-3}{11}\)
∴x = 3
Numerator = 3;
Denominator = 2 × 3 + 1 = 7
Fraction = \(\frac{3}{7}\)

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c)

February 27, 2024February 26, 2024 by Mahesh

Students must practice this TS Intermediate Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Exercise 7(c)

I. Find the values of the following integrals.

Question 1.
\(\int_0^{\frac{\pi}{2}} \sin ^{10} x d x\) (May ’06; Mar. ’03)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q1

Question 2.
\(\int_0^{\frac{\pi}{2}} \cos ^{11} x d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q2

Question 3.
\(\int_0^{\frac{\pi}{2}} \cos ^7 x \sin ^2 x d x\)
Solution:
\(\int_0^{\frac{\pi}{2}} \sin ^m x \cos ^n x d x\)
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q3

Question 4.
\(\int_0^{\frac{\pi}{2}} \sin ^4 x \cos ^4 x d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q4

Question 5.
\(\int_0^\pi \sin ^3 x \cos ^6 x d x\)
Solution:
We have \(\int_0^{2 a} f(x) d x=2 \int_0^a f(x) d x\)
if f(2a – x) = f(x)
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q5

Question 6.
\(\int_0^{2 \pi} \sin ^2 x \cos ^4 x d x\)
Solution:
Take f(x) = sin²x cos⁴x
Then f(π – x) = sin²(π – x) cos⁴(π – x)
= sin²x cos⁴x
= f(x)
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q6

Question 7.
\(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^2 \theta \cos ^7 \theta d \theta\)
Solution:
f(θ) = sin²θ cos⁷θ
f(-θ) = sin²(-θ) cos⁷(-θ)
= sin²θ cos⁷θ
= f(θ); and f is even
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q7

Question 8.
\(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^3 \theta \cos ^3 \theta d \theta\)
Solution:
Let f(θ) = sin³θ cos³θ
∴ f(-θ) = sin³(-θ) cos³(-θ)
= -sin³θ cos³θ
= -f(θ)
∴ f is an odd function.
Hence \(\int_{-a}^a f(x) d x=0\) when f is odd.
∴ \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^3 \theta \cos ^3 \theta d \theta=0\)

Question 9.
\(\int_0^a x\left(a^2-x^2\right)^{\frac{7}{2}} d x\)
Solution:
Take x = a sin θ then dx = a cos θ dθ
Upper limit when x = a is θ = \(\frac{\pi}{2}\)
Lower limit when x = 0 is θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q9
Let cos θ = t then -sin θ dθ = dt
Upper limit when θ = \(\frac{\pi}{2}\) is t = 0
Lower limit when θ = 0 is t = 1

Question 10.
\(\int_0^2 x^{\frac{3}{2}} \sqrt{2-x} d x\)
Solution:
Take x = 2 sin²θ, then dx = 4 sin θ cos θ dθ
Upper limit when x = 2 is sin²θ = 1 ⇒ θ = \(\frac{\pi}{2}\)
Lower limit when x = 0 is sin²θ = 0 ⇒ θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q10

II. Evaluate the following integrals.

Question 1.
\(\int_0^1 x^5(1-x)^{\frac{5}{2}} d x\)
Solution:
Let x = sin²θ then dx = 2 sin θ cos θ dθ
Upper limit when x = 1 is θ = \(\frac{\pi}{2}\)
Lower limit when x = 0 is θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q1

Question 2.
\(\int_0^4\left(16-x^2\right)^{\frac{5}{2}} d x\)
Solution:
Let x = 4 sin θ then dx = 4 cos θ dθ
Upper limit when x = 4 is θ = \(\frac{\pi}{2}\)
and Lower limit when x = 0 is θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q2

Question 3.
\(\int_{-3}^3\left(9-x^2\right)^{\frac{3}{2}} x d x\)
Solution:
Let x = 3 sin θ then dx = 3 cos θ dθ
Upper limit when x = 3 is θ = \(\frac{\pi}{2}\)
and Lower limit when x = -3 is θ = \(-\frac{\pi}{2}\)
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q3

Question 4.
\(\int_0^5 x^3\left(25-x^2\right)^{\frac{7}{2}} d x\)
Solution:
Let x = 5 sin θ then
Upper limit when x = 5 is θ = \(\frac{\pi}{2}\)
and Lower limit when x = 0 is θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q4

Question 5.
\(\int_{-\pi}^\pi \sin ^8 x \cos ^7 x d x\)
Solution:
Take f(x) = sin⁸x cos⁷x
then f(-x) = sin⁸(-x) cos⁷(-x)
= sin⁸x cos⁷x
= f(x)
f is an even function of x.
∴ \(\int_{-\pi}^\pi \sin ^8 x \cos ^7 x d x=2 \int_0^\pi \sin ^8 x \cos ^7 x d x\)
Now f(x) = sin⁸x cos⁷x
and f(π – x) = sin⁸(π – x) cos⁷(π – x)
= -sin⁸x cos⁷x
= -f(x)
Hence \(\int_0^\pi \sin ^8 x \cos ^7 x d x=0\)
∴ \(\int_{-\pi}^\pi \sin ^8 x \cos ^7 x d x=0\)
[By the result that f = [0, 2a] → R is integrable on [0, a] and if f(2a – x) = -f(x) ∀ x ∈ [a, 2a] then \(\int_0^{2 a} f(x) d x=0\)]

Question 6.
\(\int_3^7 \sqrt{\frac{7-x}{x-3}} d x\)
Solution:
Let x = 3 cos²θ + 7 sin²θ then
dx = -6 cos θ sin θ + 14 sin θ cos θ = 8 cos θ sin θ
Upper limit when x = 7 is
7 = 3 cos²θ + 7 sin²θ
⇒ 7(1 – sin²θ) = 3 cos²θ
⇒ cos θ = 0
⇒ θ = \(\frac{\pi}{2}\)
The lower limit when x = 3 is
3 = 3 cos²θ + 7 sin²θ
⇒ 3 sin²θ = 7 sin²θ
⇒ 4 sin²θ = 0
⇒ sin θ = 0
⇒ θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q6

Question 7.
\(\int_2^6 \sqrt{(6-x)(x-2)} d x\)
Solution:
Put x = 2 cos²θ + 6 sin²θ
then dx = (-4 cos θ sin θ + 12 sin θ cos θ) dθ = 8 sin θ cos θ dθ
Upper limit when x = 6 is 6 = 2 cos²θ + 6 sin²θ
⇒ 6 cos²θ = 2 cos²θ
⇒ 4 cos²θ = 0
⇒ cos θ = 0
⇒ θ = \(\frac{\pi}{2}\)
Lower limit when x = 2 is 2 = 2 cos²θ + 6 sin²θ
⇒ 2 sin²θ = 6 sin²θ
⇒ 4 sin²θ = 0
⇒ sin θ = 0
⇒ θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q7

Question 8.
\(\int_0^{\frac{\pi}{2}} \tan ^5 x \cos ^8 x d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q8

III. Evaluate the following integrals.

Question 1.
\(\int_0^1 x^{7 / 2}(1-x)^{5 / 2} d x\)
Solution:
Let x = sin²θ then dx = 2 sin θ cos θ dθ
Upper limit when x = 1 is sin²θ = 1 ⇒ θ = \(\frac{\pi}{2}\)
Lower limit when x = 0 is sin²θ = 0 ⇒ θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q1

Question 2.
\(\int_0^\pi(1+\cos x)^3 d x\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q2

Question 3.
\(\int_4^9 \frac{d x}{\sqrt{(9-x)(x-4)}}\)
Solution:
Take x = 4 cos²θ + 9 sin²θ then
dx = (-8 cos θ sin θ + 18 sin θ cos θ) dθ = 10 cos θ sin θ
Upper limit when x = 9 is 9 = 4 cos²θ + 9 sin²θ
⇒ 9(1 – sin²θ) = 4 cos²θ
⇒ 5 cos²θ = 0
⇒ cos θ = 0
⇒ θ = \(\frac{\pi}{2}\)
Lower limit when x = 4 is 4 = 4 cos²θ + 9 sin²θ
⇒ 4(1 – cos²θ) = 9 sin²θ
⇒ 5 sin²θ = 0
⇒ sin θ = 0
⇒ θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q3

Question 4.
\(\int_0^5 x^2(\sqrt{5-x})^7 d x\)
Solution:
Let x = 5 sin²θ then dx = 10 sin θ cos θ dθ
Upper limit when x = 5 is sin²θ = 1 ⇒ θ = \(\frac{\pi}{2}\)
Lower limit when x = 0 is sin²θ = 0 ⇒ θ = 0
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q4

Question 5.
\(\int_0^{2 \pi}(1+\cos x)^5(1-\cos x)^3 d \dot{x}\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q5

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations InText Questions

February 27, 2024February 26, 2024 by Mahesh

Students can practice 10th Class Maths Solutions Telangana Chapter 5 Quadratic Equations InText Questions to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 5 Quadratic Equations InText Questions

Try This

Question 1.
Check whether the following equations are quadratic or not.
i) x² – 6x – 4 = 0
ii) x³ – 6x² + 2x – 1 = 0
iii) 7x = 2x²
iv) x² + \(\frac{1}{\mathrm{x}^2}\) = 2
v) (2x + 1) (3x + 1) = 6(x – 1)(x – 2)
vi) 3y² = 192 (Page No. 102)
Solution:
i) x² – 6x – 4 = 0
Yes, It’s a quadratic equation.

ii) x³ – 6x² + 2x – 1 = 0
No, It is not a quadratic equation.
[∵ degree is 3]

iii) 7x = 2x²
Yes, It’s a quadratic equation.

iv) x² + \(\frac{1}{x^2}\) = 2
No, It is not a quadratic equation.
[∵ degree is 4]

v) (2x + 1) (3x + 1) = 6(x – 1) (x – 2)
No, It is not a quadratic equation.
[coefficient of x² on both sides is same i.e., 6]

vi) 3y² = 192
Yes, It’s a quadratic equation.

Try This

Question 1.
Verify whether 1 and \(\frac{3}{2}\) are the roots of the equation 2x² – 5x + 3 = 0. (Page No. 107)
Solution:
Let the given
Q.E. be p(x) = 2x² – 5x + 3
Now p(1) = 2(1)² – 5(1) + 3
= 2 – 5 + 3 = 0
∴ 1 is a root of 2x² – 5x + 3 = 0
also p(\(\frac{3}{2}\)) = 2(\(\frac{3}{2}\))² – 5(\(\frac{3}{2}\)) + 3
= 2 × \(\frac{9}{4}\) – \(\frac{15}{2}\) + 3
= \(\frac{9}{2}\) + 3 – \(\frac{15}{2}\) = \(\frac{9+6-15}{2}\) = 0
∴ \(\frac{3}{2}\) is also a root of 2x² – 5x + 3 = 0

Do This

Question 1.
Solve the equations by completing the square. (Page No. 113)

i) x² – 10x + 9 = 0
Solution:
Given x² – 10x + 9 = 0
⇒ x² – 10x = – 9
⇒ x² – 2. x. 5 = – 9
⇒ x² – 2.x. 5 + 5² = -9 + 5²
⇒ (x – 5)² = 16
∴ x – 5 = ± 4
∴ x – 5 = 4 (or) x – 5 = – 4
⇒ x = 9 (or) x = 1
⇒ x = 9 (or) 1

ii) x² – 5x + 5 = 0
Solution:
Given : x² – 5x + 5 = 0
⇒ x² – 5x = -5
TS 10th Class Maths Solutions Chapter 5 Quadratic Equations InText Questions 1

iii) x² + 7x – 6 = 0
Solution:
x² + 7x – 6 = 0
x² + 7x = 6
x² + 2.x.\(\frac{7}{2}\) = 6
x² + 2.x.\(\frac{7}{2}\) + (\(\frac{7}{2}\))² = 6 + (\(\frac{7}{2}\))²
TS 10th Class Maths Solutions Chapter 5 Quadratic Equations InText Questions 2

Think – Discuss

Question 1.
We have three methods to solve a quadratic equation. Among these three, which method would you like to use ? Why ? (Page No. 115)
Solution:
If the Q.E. has district and real roots, we use factorisation. If Q.E. has no real roots, we use quadratic formula.

Try This

Question 1.
Explain the benefits of evaluating the discriminate of a quadratic equation before attempting to solve it. What does its value signifies ? (Page No. 122)
Solution:
The discriminant b² – 4ac of a Q.E. ax² + bx + c = 0 give the clear idea about the nature of Q the roots of the Q.E. If the discriminant D = b² – 4ac > 0, the Q.E. has distinct and real roots.

If b² – 4ac = 0, the Q.E. had equal roots. If b² – 4ac < 0, the Q.E. has no real roots. By the 3 value of the discriminant, we can state the nature of the roots of a Q.E. without actually finding them.

Question 2.
Write three quadratic equations one having two distinct real solutions, one having no real solution and one having exactly one real solution. (Page No. 122)
Solution:
TS 10th Class Maths Solutions Chapter 5 Quadratic Equations InText Questions 3