TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5

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 an.

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
an = a. rn-1 = 2 × (-3)n-1
∴ r = -3; an = 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
an = a . rn-1 = (-1) × 3n-1
∴ r = 3, an = (-1) × 3n-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 = \(\frac{\mathrm{a}_2}{\mathrm{a}_1}\) = \(=\frac{2}{5}\)
an = a . rn-1 = 5 × \(\left(\frac{2}{5}\right)^{\mathrm{n}-1}\)
∴ r = \(\frac{2}{5}\) ; an = 5\(\left(\frac{2}{5}\right)^{\mathrm{n}-1}\)

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5

Question 2.
Find the 10th and nth 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
an = a. rn-1 = 5 × 5n-1 = 51+n-1 = 5n
∴ a10 = a . r9 = 5 × 59 = 510
∴ a10 = 510; an = 5n

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

i) a1 = 9 ; r = \(\frac{1}{3}\) ; find a7.
Solution:
an = a . rn-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:
an = a. rn-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 nth term of G.P. be 512
an = a . rn-1
512 = 2 × (4)n-1
29 = 2 × (22)n-1
29 = 21 × 22(n-1)
29 = 22n-2+1
29 = 22n-1
[∵ bases are equal, exponents are also equal]
∴ 2n = 9 + 1
n = \(\frac{10}{2}\) = 5
∴ 512 is the 5th 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 11th 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]
∴ 7th term of GP is \(\frac{1}{2187}\)

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.5

Question 5.
Find the 12th term of a G.P whose 8th term is 192 and the common ratio is 2.
Solution:
Given a G.P such that a8 = 192 and
r = 2
an = a . rn-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 4th 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 nth 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 nth terms are equal
an = a. rn-1
⇒ 162 × \(\left(\frac{1}{3}\right)^{\mathrm{n}-1}\) = \(\frac{2}{81}\) × (3)n-1
⇒ 3n-1 × 3n-1 = 162 × \(\frac{81}{2}\)
⇒ 3n-1+n-1 = 81 × 81
⇒ 32n-2 = 38 [am . an = am+n]
⇒ 2n – 2 = 8
[∵ bases are equal, exponents are also equal]
2n = 8 + 2
∴ n = \(\frac{10}{2}\) = 5
The 5th terms of the two G.Ps are equal.

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.4

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
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.4 2
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

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.4

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 1st equilateral triangle = 3 × 24 = 72 cm
Perimeter of the 2nd equilateral triangle = 3 × 12 = 36 cm
Perimeter of the 3rd 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.
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.4 6

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
an =a.rn-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.

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.4

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.rn-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
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.4 8

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
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.4 11

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

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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q1.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q1.2

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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q2.1
∴ Foot of the perpendicular P(h, k) = (4, 2)

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

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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q3.1

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

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

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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q7.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q7.2

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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q9.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q9.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q9.3

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

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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q10.1
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)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Q12.1
L.H.S = 4p2 + q2
= 4(a sin α cos α)2 + (a cos 2α)2
= 4a2 sin2α cos2α + a2 cos2
= a2(2 sin α cos α)2 + a2 cos2
= a2 sin2 2α + a2 cos2
= 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.

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

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

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

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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q10.1

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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q11.1

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.

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

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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q17.1

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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q18.1

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

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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q19.1

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.
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q20.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q20.2

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.
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q21.1
∴ 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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q22.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q22.2

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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q23.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q23.2

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

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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q25.2
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)

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

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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q27.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q27.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q27.3

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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q28.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q28.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q28.3
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q28.4
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q28.5

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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q29.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q29.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q29.3
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q29.4
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q29.5

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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q30.1
∴ Vertex C = (-1, -3)
Let S(α, β) be the circumcentre of the triangle then, SA = SB = SC
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q30.2
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)

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

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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q31.1
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 = \(\frac{1}{\sqrt{5}}\)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type DTP Q32.1
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).

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

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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q3.1

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.
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q4.1

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

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

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

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

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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q13.1
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

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

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)

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

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

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

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

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

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,
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q35.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q35.2
∴ 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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q36.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q36.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q36.3

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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q37.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q37.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q37.3

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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q38.1

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)
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q39.1
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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q39.3

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

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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q40.1
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

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

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,
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q43.1
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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q44.1
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q44.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q44.3
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q44.4
⇒ -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
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q45.2
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q45.3
TS Inter First Year Maths 1B Straight Lines Important Questions Long Answer Type Some more Q45.4

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

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

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

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
sn = \(\frac{\mathrm{n}}{2}\)[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
Sn = \(\frac{n}{2}\)[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

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3

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 = an
But, an = 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, Sn = \(\frac{\mathrm{n}}{2}\)(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‘ = an = 10
But, an = a + (n – 1) d
⇒ 10 = 34 + (n – 1)(-2)
⇒ 10 – 34 = -2n + 2
⇒ -2n = -24 – 2
⇒ n = \(\frac{-26}{-2}\) ∴ n = 13
Also, Sn = \(\frac{\mathrm{n}}{2}\)(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 = an = -230:
d = a2 – a1 = (-8) – (-5)
= -8 + 5 = -3
But, an = 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 Sn = \(\frac{\mathrm{n}}{2}\)(a + l)
S76 = \(\frac{76}{2}\)[(-5) + (-230)]
= 38 × (-235) = -8930

Question 3.
In an AP:

i) Given a = 5, d = 3, an = 50, find n and Sn.
Solution:
Given:
a = 5; d = 3 ; an = a + (n – 1)d = 50
⇒ 50 = 5 + (n – 1)3
⇒ 50 – 5 = 3n – 3
⇒ 3n = 45 + 3
⇒ n = \(\frac{48}{3}\) = 16
Sn = \(\frac{\mathrm{n}}{2}\)(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, Sn = \(\frac{\mathrm{n}}{2}\)(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, Sn = \(\frac{\mathrm{n}}{2}\)(a + l)
S12 = \(\frac{12}{2}\)[4 + 37] = 6 × 41 = 246.

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3

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,
Sn = \(\frac{\mathrm{n}}{2}\)(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, an = a + (n – 1) d
a10 = a + 9d = 13 + 9 × 1
= 13 + 9 = 22
∴ a10 = 22 ; d = 1.

v) Given a = 2, d = 8, Sn = 0, find n and an.
Solution:
Given : a = 2; d = 8 and Sn = 90
But Sn = \(\frac{\mathrm{n}}{2}\)[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 an = a5 = a + 4d = 2 + 4 × 8
= 2 + 32 = 34.

vi) Given an = 4, d = 2, Sn = -14, find n and a.
Solution:
Given an = a + (n – 1) d = 4 —– (1)
d = 2; Sn = -14
From (1); a + (n – 1) 2 = 4
a = 4 – 2n + 2 ⇒ a = 6 – 2n
Sn = \(\frac{\mathrm{n}}{2}\) [a + an]
-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, Sn = \(\frac{9}{2}\)(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, an = a + (n – 1) d
350 = 17 + (n – 1)9
⇒ 350 = 17 + 9n – 9
⇒ 9n = 350 – 8
⇒ n = \(\frac{342}{9}\) = 38
Now, Sn = \(\frac{n}{2}\)(a + l)
S38 = \(\frac{38}{2}\)[17 + 350]
= 19 × 367 = 6973
∴ n = 38
Sn = 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, Sn = \(\frac{n}{2}\)[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,
Sn = \(\frac{n}{2}\)[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, Sn = \(\frac{\mathrm{n}}{2}\)[2a + (n – 1)d]
Sn = \(\frac{\mathrm{n}}{2}\) [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 Sn = n2.
Shortcut : S7 = 49 = 72
S17 = 289 = 172
∴ Sn = n2

Question 7.
Show that a1, a2, ……. an, …. form an AP where an is defined as below :
i) an = 3 + 4n
ii) an = 9 – 5n. Also find the sum of the first 15 terms in each case.
Solution:
i) Given an = 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, …., an … 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.

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3

ii) an = 9 – 5n
Solution:
Given : an = 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 {an} 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 3rd, the 10th and the nth 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 [∴ an = a + (n – 1) d]
= 3 + 9 × (- 2)
= 3 – 18 = -15
an = 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
Sn = \(\frac{\mathrm{n}}{2}\) [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
∴ Sn = \(\frac{\mathrm{n}}{2}\) [a + l]
S12 = \(\frac{12}{2}\) [3 + 36]
= 6 × 39 = 234
∴ Total plants = 234

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3

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:
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.3 5
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, Sn = 200
∴ Sn = \(\frac{n}{2}\) [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.
1st ball: Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 5 = 10 m.
2nd ball : Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 8 = 16 m.
3rd ball : Distance covered by the competitor in picking up and dropping it in the bucket = 2 × 11 = 22 m.
……………………………………………………
10th 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.
Sn = \(\frac{\mathrm{n}}{2}\) [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

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.

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

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

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

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

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

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

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

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
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q12.1

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
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q14.2
which is in the form x cos α + y sin α = p
∴ α = \(\frac{\pi}{6}\), p = 2

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

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
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q17.1

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
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q18.1
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q18.2

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

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

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
TS Inter First Year Maths 1B Straight Lines Important Questions Very Short Answer Type Q23.1
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 \(\frac{-25}{3}\)

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

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

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

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

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)

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

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
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q3.1
∴ 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
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q5.2
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)

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

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
TS Inter First Year Maths 1B Straight Lines Important Questions Short Answer Type Q6.1

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 = \(\left(\frac{3}{4}, \frac{1}{2}\right)\)

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

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

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

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

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

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

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.

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

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 1st 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
TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 7
∴ Sum of roots of a Q.E. is \(\frac{-b}{a}\)

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

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
TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 9
Product of the roots
TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise 10

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)

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

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

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
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q6.1

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

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

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
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q9.1

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
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) I Q10.1

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
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q1.1

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

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
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q4.1

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 \(\int_0^{2 a} f(x) d x=0\)]

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

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
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q6.1

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
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) II Q7.1

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
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q1.1

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

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
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q2.1

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
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q4.1

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

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 Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q5.1
TS Inter 2nd Year Maths 2B Solutions Chapter 7 Definite Integrals Ex 7(c) III Q5.2

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

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.

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

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

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

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

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2

Students can practice TS Class 10 Maths Solutions Chapter 6 Progressions Ex 6.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 6 Progressions Exercise 6.2

Question 1.
Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the A.P :
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2 1
Solution:
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2 2

Question 2.
Find the
i) 30th term of the A.P. : 10, 7, 4, …… (A.P. June 15)
ii) 11th term of the A.P. : -3, \(\frac{-1}{2}\) ,2, …….
Solution:
i) Given A.P. = 10, 7, 4, ….
a1 = 10; d = a2 – a1 = 7 – 10 = – 3
an = a + (n – 1) d
a30 = 10 + (30 – 1)(-3)
= 10 + 29 × (- 3)
= 10 – 87
= -77

ii) Given A.P. = -3, \(\frac{-1}{2}\), 2,……..
a1 = -3; d = a2 – a1 = \(\frac{-1}{2}\) – (-3)
= \(\frac{-1}{2}\) + 3 = = \(\frac{-1+6}{2}\) = \(\frac{5}{2}\)
an = a + (n – 1)d
a11 = -3 + (11 – 1) × (\(\frac{5}{2}\))
= -3 + 10 × \(\frac{5}{2}\)
= -3 + 5 × 5
= -3 + 25 = 22

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2

Question 3.
Find the respective terms for the following APs.

i) a1 = 2; a3 = 26, find a2.
Solution:
Given : a1 = a = 2 —— (1)
a3 = a + 2d = 26 —— (2)
Equation (2) – equation (1)
⇒ (a + 2d) – a = 26 – 2
⇒ 2d = 24
d = \(\frac{24}{2}\) = 12
Now a2 = a + d = 2 + 12 = 14

ii) a2 = 13; a4 = 3, find a1, a3.
Solution:
Given : a2 = a + d = 13 —— (1)
a4 = a + 3d = 3 —- (2)
Solving equations (1) and (2);
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2 3
Substituting, d = – 5 in equation (1) we get
a + (-5) = 13
∴ a = 13 + 5 = 18
i.e., a1 = 18
a3 = a + 2d = 18 + 2(-5)
= 18 – 10 = 8

iii) a1 = 5; a4 = 9\(\frac{1}{2}\), a2, a3.
Solution:
Given : a1 = a = 5 —– (1)
a4 = a + 3d = 9\(\frac{1}{2}\) —– (2)
Solving equations (1) and (2);
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2 4

iv) a1 = – 4; a6 = 6, find a2, a3, a4, a5.
Solution:
Given : a1 = a = – 4 —– (1)
a6 = a + 5d = 6 —– (2)
Solving equations (1) and (2);
(-4) + 5d = 6 ⇒ 5d = 6 + 4
⇒ 5d = 10 ⇒ d = \(\frac{10}{5}\) = 2
Now, a2 = a + d = -4 + 2 = -2;
a3 = a + 2d = -4 + 2 × 2
= -4 + 4 = 0;
a4 = a + 3d = (-4) + 3 × 2
= -4 + 6 = 2;
a5 = a + 4d = -4 + 4 × 2
= -4 + 8 = 4

v) a2 = 38; a6 = -22, find a1, a3, a4, a5.
Solution:
Given : a2 = a + d = 38 —– (1)
a6 = a + 5d = -22 —– (2)
Subtracting (2) from (1) we get
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2 5
Now substituting, d = -15 in equation (1), we get
a + (-15) = 38 ⇒ a = 38 + 15 = 53 Thus,
a1 = a = 53;
a3 = a + 2d = 53 + 2 × (-15) = 53 – 30 = 23;
a4 = a + 3d = 53 + 3 × (-15) = 53 – 45 = 8;
a5 = a + 4d = 53 + 4 × (-15) = 53 – 60 = -7

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2

Question 4.
Which term of the AP :
3, 8, 13, 18, ……. is 78 ?
Solution:
Given : 3, 8, 13, 18,
Here a = 3; d = a2 – a1 = 8 – 3 = 5
Let ’78’ be the nth term of the given A.P.
∴ an = a + (n – 1) d
78 = 3 + (n – 1)5
78 = 3 + 5n – 5
⇒ 5n = 78 + 2 ⇒ n = \(\frac{80}{5}\) = 16
∴ 78 is the 16th term of the given A.P.

Question 5.
Find the number of terms in each of the following APs :
i) 7, 13, 19, ……., 205 (A.P. Mar. ’16)
Solution:
Given : A.P : 7, 13, 19, ………
Here a1 = a = 7; d = a2 – a1 = 13 – 7 = 6
Let 205 be the nth term of the given A.P.
Then, an= a + (n – 1) d
205 = 7 + (n – 1) 6
⇒ 205 = 7 + 6n – 6
⇒ 205 = 6n + 1
⇒ 6n = 205 – 1 = 204
∴ n = \(\frac{204}{6}\) = 34
∴ 34 terms are there.

ii) 18, 15\(\frac{1}{2}\), 13, ………, -47.
Solution:
Given A.P = 18, 15\(\frac{1}{2}\), 13, ….
Here a1 = a = 18
d = a2 – a1 = 15\(\frac{1}{2}\) – 18 = -2\(\frac{1}{2}\) = –\(\frac{5}{2}\)
Let ‘-47’ be the nth term of the given A.P.
an = a + (n – 1) × d
⇒ -47 = \(\frac{18 \times 2+(n-1)(-5)}{2}\)
⇒ – 94 = 36 – 5n + 5
⇒ 5n = 94 + 41 ⇒ n = \(\frac{135}{5}\) = 27
i.e., 27 terms are there.

Question 6.
Check whether, -150 is a term of the AP : 11, 8, 5, 2, ….
Solution:
Given A.P. = 11, 8, 5, 2,
Here a = a2 – a1 = 8 – 11 = -3
If possible, take -150 as the n,sup>th term of the given A.P.
an = a + (n – 1) d
-150 = 11 + (n – 1) × (- 3)
⇒ -150 = 11 – 3n + 3
⇒ 14 – 3n = -150
⇒ 3n = 14 + 150 = 164
∴ n = \(\frac{164}{3}\) = 54\(\frac{2}{3}\)
Here n is not an integer.
∴ -150 is not a term of the given A.P.

Question 7.
Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.
Solution:
Given : An A.P. whose
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2 6
Substituting d = 7 in the equation (1) we get,
a + 10 × 7 = 38
⇒ a + 70 = 38
⇒ a = 38 – 70 = – 32
Now, the 31st term = a + 30d
= (- 32) + 30 × 7
= – 32 + 210 = 178

Question 8.
If the 3rd and the 9th terms of an A.P are 4 and – 8 respectively, which term of this A.P is zero ?
Solution:
Given : An A.P whose
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2 7
Substituting d = – 2 in equation (1) we get a + 2 × (- 2) = 4
⇒ a – 4 = 4
⇒ a = 4 + 4 = 8
Let nth term of the given A.P be equal to zero.
an = a + (n – 1) d
⇒ 0 = 8 + (n – 1) × (-2)
⇒ 0 = 8 – 2n + 2
⇒ 10 – 2n = 0
⇒ 2n = 10 and n = \(\frac{10}{2}\) = 5
∴ The 5th term of the given A.P is zero.

Question 9.
The 17th term of an A.P exceeds its 10th term by 7. Find the common difference.
Solution:
Given an A.P. in which a17 = a10 + 7
⇒ a17 – a10 = 7
We know that an = a + (n – 1) d
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2 8

Question 10.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms ?
Solution:
Let the first A.P be : a, a + d, a + 2d,
Second A.P be : b, b + d, b + 2d, b + 3d,….
Also, general term, an = a + (n – 1) d
Given that, a100 – b100 = 100
⇒ a + 99d – (b + 99d) = 100
⇒ a – b = 100
Now the difference between their 1000 terms,
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2 9
∴ The difference between their 1000th terms is (a – b) = 100.

Note : If the common difference for any two A.Ps are equal then difference nth terms of two A.Ps is same for all natural values of n.

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2

Question 11.
How many three-digit numbers are divisible by 7 ?
Solution:
The list of three digit numbers divisible by 7 , is,
105, 112, 119, ………, 994
Here a = 105; d = 7; an = 994
an = a + (n – 1) d
994 = 105 + (n – 1) 7
(n – 1)7 = 889
n -1 = \(\frac{889}{7}\)
n – 1 = 127
∴ n = 127 + 1 = 128
There are 128 three digit numbers which are divisible by 7.

Question 12.
How many multiples of 4 lie between 10 and 250 ?
Solution:
Given numbers : 10 to 250
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2 10
∴ Multiples of 4 between 10 and 250 are
First term : 10 + (4 – 2) = 12
Last term : 250 – 2 = 248
∴ 12, 16, 20, 24, ………, 248
a = a1 = 12; d = 4: an = 248
an = a + (n – 1) d
248 = 12 + (n – 1) × 4
⇒ (n – 1) 4 = 248 – 12
⇒ n – 1 = \(\frac{236}{4}\) = 59
∴ n = 59 + 1 = 60
There are 60 numbers between 10 and 250 which are divisible by 4.

Question 13.
For what value of n, are the nth terms of two APs : 63, 65, 67, … and 3, 10, 17, …. equal ?
Solution:
Given : The first A.P is 63, 65, 67, ….
where a = 63, d = a2 – a1
⇒ d = 65 – 63 = 2
and the second A.P is 3, 10, 17, ….
where a = 3, d = a2 – a1 = 10 – 3 = 7
Suppose the nth terms of the two A.Ps are equal, where an = a + (n – 1) d
⇒ 63 + (n -1) 2 = 3 + (n – 1) 7
⇒ 63 + 2n – 2 = 3 + 7n – 7
⇒ 61 + 2n = 7n – 4
⇒ 7n – 2n = 61 +4
⇒ 5n = 65
⇒ n = \(\frac{65}{5}\) = 13
∴ 13th terms of the two A.Ps are equal.

Question 14.
Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution:
Given : An A.P in which
a3 = a + 2d = 16 —– (1)
and a7 = a5 + 12
i.e., a + 6d = a + 4d + 12
⇒ 6d – 4d = 12
⇒ 2d = 12
⇒ d = \(\frac{12}{2}\) = 6
Substituting d = 6 in equation (1) we get
a + 2 × 6 = 16
⇒ a = 16 – 12 = 4
∴ The series/A. P is a, a + d, a + 2d, a + 3d, …
⇒ 4, 4 + 6, 4 + 12, 4 + 18, …
⇒ A.P: 4, 10, 16, 22,….

Question 15.
Find the 20th term from the end of the AP : 3, 8, 13,…….., 253.
Solution:
Given : An A.P : 3, 8, 13, …., 253
Here a = a1 = 3
d = a2 – a1 = 8 – 3 = 5
an = 253, where 253 is the last term
an = a + (n – 1) d
∴ 253 = 3 + (n – 1) 5
⇒ 253 = 3 + 5n – 5
⇒ 5n = 253 + 2
⇒ n = \(\frac{255}{5}\) = 51
∴ The 20th term from the other end would be
1 + (51 – 20) = 31 + 1 = 32
∴ a32 = 3 + (32 – 1) × 5
= 3 + 31 × 5
= 3 + 155
= 158

TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2

Question 16.
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the (A.P. Mar. ’15)
Solution:
Given an A.P in which
a4 + a8 = 24 ⇒ a + 3d + a + 7d = 24
⇒ 2a + 10d = 24
⇒ a + 5d = 12 —– (1)
and a6 + a10 = 44
⇒ a + 5d + a + 9d = 44
⇒ 2a + 14d = 44
⇒ a + 7d = 22 —– (2)
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2 11
Also a + 5d = 12
⇒ a + 5(5) = 12
⇒ a + 25 = 12
⇒ a = 12 – 25 = -13
∴ The A.P is a , a + d, a + 2d, ……..
i.e., – 13, (- 13 + 5), (- 13 + 2 × 5)
⇒ -13, -8, -3, ………

Question 17.
Subba Rao started work in 1995 at an annual salary of ₹ 5,000 and received an increment of ₹ 200 each year. In which year did his income reach Rs. 7000?
Solution:
Given : Salary of Subba Rao in 1995 = ₹ 5000
Annual increment = ₹ 200
i.e., His salary increases by Rs. 200 every year.
TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.2 12
Clearly 5000, 5200, 5400, ……… forms an A.P in which a = 5000 and d = 200.
Now suppose that his salary reached ₹ 7000
after x – years,
i.e., an = 7000
But, an = a + (n – 1) d
7000 = 5000 + (n – 1) 200
⇒ 7000 – 5000 = (n – 1) 200
⇒ n – 1 = \(\frac{2000}{200}\)
= 10
⇒ n = 10 + 1 = 11
∴ In 11th year his salary reached ₹ 7000.

TS Inter 1st Year Commerce Study Material Chapter 2 Business Activities

Telangana TSBIE TS Inter 1st Year Commerce Study Material 2nd Lesson Business Activities Textbook Questions and Answers.

TS Inter 1st Year Commerce Study Material 2nd Lesson Business Activities

Long Answer Questions

Question 1.
What is meant by industry? Explain various types of industries with suitable examples.
Answer:
Industry is concerned with the making or manufacturing of goods. It is that part of the production which is involved in changing the form of goods at any stage from raw material to the finished product. E.g.: Weaving woollen yam into cloth. Thus industry imparts form utility in goods.

Classification or types of industries: The industries may be classified as follows.
1) Primary industry: Primary industry is concerned with production of goods with the help of nature. It is nature-oriented industry, which requires very little human effort. E.g: Agriculture, Farming, Fishing, Horticulture etc.

2) Genetic industry: Genetic industry is related to the reproducing and multiplying of certain species of animals and plants with the object of earning profits from their sale. E.g: Nurseries, cattle breeding poultry, fish hatcheries etc.

3) Extractive industry: It is engaged in raising some form of wealth from the soil, cli-mate, air, water or from beneath the surface of the earth. Generally the products of extractive industries comes in raw farm and they are used by manufacturing and construction industries for producing finished products. E.g: Mining, coal, mineral, iron ore, oil industry, extraction of timber and rubber from forests.

4) Construction industry: The industry is engaged in the creation of infrastructure for the smooth development of the economy. It is concerned with the construction, erection or fabrication of products. These industries are engaged in the construction of buildings, roads, dams, bridges and canals.

5) Manufacturing industry: This industry is engaged in the conversion of raw material into semifinished or finished goods. This industry creates form utility in goods by making them suitable for human uses. E.g: Cement industry, Sugar industry, Cotton textile industry, Iron and steel industry, Fertiliser industry etc.

6) Service industry: In modern times, service sector plays an important role in the development of the nation and therefore it is named as service industry. These are engaged in the provision of essential services to the community. E.g: Banking, trans-port, insurance etc.

Question 2.
What is commerce? Describe its branches.
Answer:
Commerce is concerned with exchange of goods. It includes all those activities which are related to transfer of goods from the places of production to the ultimate consumer. Commerce embraces all those processes which help to break the barriers between producers and consumers. It is the sum total of those processes which are engaged in the removal of hindrances of persons, place, time and exchange.

Branches of commerce:
The activities of commerce may be classified into following two broad categories.

  • Trade
  • Aids to trade

I) Trade: Trade is a branch of commerce. It connects with buying and selling of goods and services. An individual who does trade is called a trader. Trader transfers the goods from the producer to the consumer.

Trade may be classified into a) Home trade, b) Foreign trade. Home trade may again sub-divided into (i) Wholesale trade and (ii) Retail trade.

a) Home trade: Home trade is also known as ‘domestic trade’ or ‘Internal trade’ Home trade is carried on within the boundaries of a nation. Home trade again is of two types: wholesale trade and retail trade.

  • Wholesale trade: It implies buying and selling of goods in large quantities. Traders who engage themselves in whole sale trade are called ‘wholesalers’. Wholesale serves as a connecting link between the producers and the retailers.
  • Retail trade: It involves buying and selling of goods in small quantities. Traders engaged in retail trade are called ‘retailers’. They serve as a connecting link between wholesalers and consumers.

b) Foreign trade: It refers to buying and selling of goods and services between two or more countries through international air ports and sea ports. Foreign trade is also known as ‘external trade’ or ‘International trade’.

  • Export trade: It means the sale of goods to foreign countries. For example India exports tea to the U.K.
  • Import trade: It refers to the purchase of goods from foreign countries. For instance India buys petrol from Iran.
  • Entrepot trade: Importing (buying) goods from one country for the purpose of ex-porting (selling) them to another country is called entrepot trade. This type of trade is also known as re-export trade.

II) Aids to trade: Trade or exchange of goods involves several difficulties, which can be removed by auxiliaries are known as aids to trade. It refers to all those activities, which directly or indirectly facilitate smooth exchange of goods and services.

TS Inter 1st Year Commerce Study Material Chapter 2 Business Activities

Aids to trade includes transport, Communication, Warehousing, Banking, Insurance, Advertising. These ensure smooth flow of goods from producers to the consumers. The various aids to trade in commerce are explained in below:

1) Transport: There will be a vast distance between centers of production and centres of consumption. This difficulty is removed by transport. Transport creates place utility. There are several kinds of transport such as air, water and land transport. The geo-graphical distance between producers and consumers is removed with the help of following means of transport.

2) Communication: Communication means transmitting or exchange of information froth one person to another. It can be oral or in writing. It is necessary to communicate information from one to another. Modern means of communication like telephone, email, video conference etc play an important role in establishing contact between businessmen, producers and consumers.

3) Warehousing: There is a time gap between production and consumption. It becomes necessary to make arrangements for storage or warehousing. The goods such as umbrellas and woolen clothes are produced throughout the year but are demanded only during particular seasons like rainy and winter season. Therefore goods need to be stored in warehouses till they are demanded. So, it creates time utility by supplying the goods at right time to consumer.

4) Insurance: Insurance reduces the problem of risks. Business is subject to risks and uncertainities. Risks may be due to fire, theft, accident or any other natural calamity. Insurance companies who act as risk bearer cover risks. Insurance tries to reduce risks by spreading them out over a larger number of people by encouraging them to take insurance policies.

5) Banking: Banking solves the problem of finance. Banking and financial institutions solves the problem of payment and facilitate exchange between buyer and seller. Banks provide such finance to them. Banks also advance loans in the form of overdraft, cash credit and discounting of bills of exchange.

6) Advertising: Advertising fills the knowledge gap. Exchange of goods and services is possible only if producers can bring the products to the consumers. Advertising and publicity are important medias of mass communication. Advertising helps the consumers to know about the various brands manufactured by several manufacturers. The medias used to advertise products are Radio, Newspapers, Magazines, TV, Internet etc.

Question 3.
Discuss the significance of commerce in the present scenario.
Answer:
Commerce can also be defined as The sum total of those processes. Which are engaged in the removal of hindrances of person, place and time in the exchange of commodities’.

Importance of commerce:
The importance of commerce is explained below:
1) Commerce tries to satisfy increasing human wants: Human wants and desires are never ending. Commerce has made distribution and movement of goods possible from one part of the world to the other. Today we can buy anything produced anywhere in the world. Hence commerce facilitates the people to satisfy their needs, desires and wants by distribution and exchange of the goods and services.

2) Commerce helps us to increase our standard of living: Standard of living refers to quality of life enjoyed by the members of a society. When a man consumes more products his standard of living improves. Commerce helps us to get what we want at the right time, right place and at the right price and thus helps in improving our standard of living.

3) Commerce links producers and consumers: Commerce makes a link between producers and consumers through retailers and wholesalers and also through the aids to trade. Thus it creates and facilitates the contact between the centres of production and consumption.

4) Commerce generates employment opportunities: The growth of commerce and trade cause the growth of agencies of trade such as banking, transport, warehousing, insurance, advertising etc. Thus, development of commerce generates more and more employment opportunities.

5) Commerce increases national income and wealth: When production increases, national income also increases. It also helps to earn foreign exchange by way of exports and duties levied on imports.

6) Commerce helps in expansion of aids-to-trade: With the growth in trade and commerce there is a growing need for expansion and modernization of aids to trade. Aids to trade such as banking communication, advertising and publicity, transport, insurance etc. are expanded and modernised for the smooth conduct of commerce.

7) Commerce encourages international trade: With the help of transport and communication development, countries can exchange their surplus. Commodities and earn foreign exchange. Thus, the commerce ensures faster economic growth of the country.

8) Commerce benefit underdeveloped countries: Underdeveloped countries can import skilled labour and technical knowhow from developed countries, while the advanced countries can import raw materials from underdeveloped countries. This helps in laying down the seeds of industrialisation in the underdeveloped countries.

9) Commerce helps during emergencies: During emergencies like floods, earthquakes and wars, commerce helps in reaching the essential requirements like food stuff, medicines and relief measures to the affected area.

TS Inter 1st Year Commerce Study Material Chapter 2 Business Activities

Question 4.
Define trade and explain various types of aids to trade.
Answer:
Trade: Buying and selling of goods and services to earn profit is called trade. The person who undertakes this job is known as trader. It is a branch of commerce.

Aids to trade:
Trade or exchange / distribution of goods involves several difficulties, which can be removed by auxiliaries are known as aids to trade. It refers to all those activities which directly or indirectly facilitate smooth exchange of goods and services.

Aids to trade includes Transport, Communication, Warehousing, Banking, Insurance, Advertising. These ensure smooth flow of goods from producers to the consumers. The various aids to trade in commerce are explained in following points.

1) Transport: There will be a vast distance between centres of production and centres of consumption. This difficulty is removed by transport. Transport creates place utility. There are several kinds of transport such as air, water and land transport. The geo-graphical distance between producers and consumers is removed with the help of following means of transport, (i) Land (ii) Water (iii) Air.

2) Communication: Communication means transmitting or exchange of information from one person to another. It can be oral or in writing. It is necessary to communicate information from one to another. Modem means of communication like telephone, email, teleconference, video-conference etc. play an important role in establishing contact between businessmen, producers and consumers.

3) Warehousing: There is a time gap between production & consumption. It becomes necessary to make arrangements for storage or warehousing. The goods such as umbrellas and woolen clothes are produced throughout the year but are demanded only during particular seasons like rainy and winter season. Therefore goods need to be stored in warehouses till they are demanded. So it creates time utility by supplying the goods at right time to consumers.

4) Insurance: Insurance reduces the problems of risks. Business is subject to risks and uncertainities. Risks may be due to fire, theft, accident or any other natural calamity. Insurance companies who act as risk bearer cover risks. Insurance tries to reduce risks by spreading them out over a large number of people by encouraging them to take insurance policies.

5) Banking: Banking solves the problem of finance. Banking and financial institutions solves the problem of payment and facilitate exchange between buyer and seller. Banks provide such finance to them. Banks also advance loans in the form of overdraft, cash credit and discounting of bills of exchange.

6) Advertising: Exchange of goods and services is possible only if producers can bring the products to the consumers. Advertising and publicity are important medias of mass communication. Advertising helps the consumers to know about the various brands manufactured by several manufacturers. The medias used to advertise products are Radio, Newspaper, Magazines, TV; Internet etc.

Question 5.
Explain inter-relationship between industry, commerce and trade.
Answer:
Business is divided into two categories: Industry and commerce. Commerce is again sub-divided into trade and aids to trade practically all of them are closely related to each other. They are inseparable. All of them are parts of the whole business system. Industry and commerce are closely related to each other. Industry cannot exist without commerce and commerce cannot exist without industry because every producer has to find his market for his product to sell. But the producer has no direct connection with the buyer or consumers. Hence, industry needs commerce.

BUSINESS = INDUSTRY + COMMERCE
Commerce is concerned with the sale, transfer or exchange of goods and services. Hence commerce needs industry for the production of goods and services. Commerce makes the necessary arrangement for linking between producers and ultimate consumers. It includes all those activities that are involved in buying, selling, transporting, banking, warehousing of goods, and insurance of safeguarding the goods.

COMMERCE = TRADE + AIDS TO TRADE
Thus industry, commerce, and trade are closely related to one another and are inter-dependent as shown in the figure below. In conclusion, we can say that industry, trade and commerce are inter-related with each other. Industry is concered with production of goods and services and commerce arranges its sales; but the actual operation of sales is in the hands of trade. So they cannot work independently.

TS Inter 1st Year Commerce Study Material Chapter 2 Business Activities

TS Inter 1st Year Commerce Study Material Chapter 2 Business Activities

Question 6.
Describe the various Hindrances of Commerce.
Answer:
Commerce is concerned with exchange of goods. It includes all those activities which are related to transfer of goods from the place of production to the ultimate consumers. Whereas trade involves buying and selling of goods, commerce has a wider meaning. Commerce include trade and aids to trade. The aids to trade include transport, banking, insurance, warehousing, advertisement and salesmanship.

Hindrances of trade: In the course of exchange of goods various problems are encountered. The hindrances in the way of smooth trade may be place, person, finance, time, knowledge and risk.

1) Hindrances of place: Generally, all the goods are not consumed at the same place where are produced. The goods are to be taken from a place where there is less demand, to the place where there is more demand. The activity of movement of the goods is called transportation. Thus transport eliminates the hindrances of place.

2) Hindrances of persons: In the present day world the consumers are in millions and it is not possible for the producers to know the consumers who are in need of goods produced by them. A chain of middlemen like wholesalers, retailers, dealers etc. Purchase goods from the producers and take them to the customers. Thus, middlemen remove the hindrances of persons.

3) Hindrances of finance: There is always time lag between the production and sale of goods. It takes time to collect money and hence need finance for trade. Commerce makes exchange of goods and services possible by removing these hindrances through the agency of banks.

4) Hindrances of time: As the goods are produced in anticipation of demand, there is a need to store the until they are required for consumption. Warehousing eliminates the hindrances of time and provides time utility to goods.

5) Hindrances of knowledge: The consumers may not be aware of the availability of various goods in the market. The absence of information is another hindrance. This is eliminated through advertising. Advertisement is done through T.V., radio, news papers, magazines, wall posters, hoardings etc.

6) Hindrances of risk: There are risks involved in production, transporting goods from one place to another, warehousing. The businessmen would like to cover these risks. Insurance companies undertake to compensate the loss suffered due to such risks. So, insurances eliminates hindrances of risks.

Question 7.
Distinguish between trade, commerce and industry.
Answer:
Differences between trade, commerce and industry.

Basis of Difference Trade Commerce Industry
1. Meaning It is related to the purchase and sale of goods. It is related to the activities which deals with taking of goods from producers to consumers. All those activities which deal with conversion of raw material into finished goods.
2. Capital More capital is required than commerce. It requires less capital. Capital needs are high.
3. Scope It deals with purchase and sale of goods. It includes trade and aids to trade. Primary manufacturing, processing etc., industries are covered under industry.
4. Risk It involves greater risk. The risk involved is comparatively less. It involves greater risk compared to any other activity.
5. Utility Trade creates possession utility. Commerce creates time and place utility. Industry creates form utility.

TS Inter 1st Year Commerce Study Material Chapter 2 Business Activities

Short Answer Questions

Question 1.
Define Industry.
Answer:
Industry is concerned with the making or manufacturing of goods. It is that part of the production which is involved in changing the form of goods at any stage from raw material to the finished product. E.g.: Weaving woollen yam into cloth. Thus industry imparts form utility in goods.

Question 2.
What is Trade?
Answer:
Trade: Buying and selling of goods and services to earn profit is called trade. The person who undertakes this job is known as trader. It is a branch of commerce.

Trade is a branch of commerce. It connects with buying and selling of goods and services. An individual who does trade is called a trader. Trader transfers the goods from the producer to the consumer.

Trade is classified into two types: (I) Home trade (II) Foreign trade

I) Home trade: Home trade is also known as ‘domestic trade’ or ‘Internal trade’. Home trade is carried on within the boundaries of a nation.
Home trade is again divided into two types:

  • Wholesale trade and
  • Retail trade

(i) Wholesale trade: It implies buying and selling of goods in large quantities. Traders who engage themselves in wholesale trade are called ‘wholesalers’. Wholesale serves as a connecting link between the producers and the retailers.

(ii) Retail trade: It involves buying and selling of goods in small quantities. Traders en-gaged in retail trade are called ‘retailers’. They serve as a connecting link between wholesalers and consumers.

(II) Foreign trade: It refers to buying and selling of goods and services between two or more countries through international air ports and sea ports. Foreign trade is also known as ‘external trade’ or ‘international trade’. Foreign trade is again may be classified into three categories as mentioned below.

  • Export trade: It means the sale of goods to foreign countries. For example India exports tea to the united kingdom.
  • Import trade: It refers to the purchase of goods from foreign countries. For instance India buys petrol from Iran.
  • Entrepot trade: Importing (buying) goods from one country for the purpose of ex-porting (selling) them to another country is called entrepot trade. This type of trade is also known as ‘re-export’ trade.

Question 3.
State the types of foreign trade.
Answer:
When trade takes place between two countries it is called foreign trade or external trade or international trade. Two countries are involved in foreign trade. External trade generally requires permission from the respective countries. The hindrances of place, time, risk, exchange are overcome with the help of various agencies. Foreign trade may be classified into export trade, import trade and entrepot trade.

1) Export trade: This trade refers to sale of goods to foreign countries.
E.g.: India exports tea to U.K.

2) Import trade: The purchase of goods from other countries is known as import trade.
Ex: India buys petrol from Iran.

3) Entrepot trade: When goods are imported (purchased) from one country with a view to exporting (selling) them to other country, it is called entrepot trade or re-export trade.
Ex: India imported petrol from Iran and export the same to Nepal.

Question 4.
Explain the classification of industries.
Answer:
Classification or types of industries: The industries may be classified as follows.
1) Primary industry: Primary industry is concerned with production of goods with the help of nature. It is nature-oriented industry, which requires very little human effort.
E.g: Agriculture, Farming, Fishing, Horticulture etc.

2) Genetic industry: Genetic industry is related to the reproducing and multiplying of certain species of animals and plants with the object of earning profits from their sale.
E.g: Nurseries, cattle breeding poultry, fish hatcheries etc.

3) Extractive industry: It is engaged in raising some form of wealth from the soil, cli-mate, air, water or from beneath the surface of the earth. Generally the products of extractive industries comes in raw farm and they are used by manufacturing and construction industries for producing finished products.
E.g: Mining, coal, mineral, iron ore, oil industry, extraction of timber and rubber from forests.

4) Construction industry: The industry is engaged in the creation of infrastructure for the smooth development of the economy. It is concerned with the construction, erection or fabrication of products. These industries are engaged in the construction of buildings, roads, dams, bridges and canals.

TS Inter 1st Year Commerce Study Material Chapter 2 Business Activities

5) Manufacturing industry: This industry is engaged in the conversion of raw material into semifinished or finished goods. This industry creates form utility in goods by making them suitable for human uses.
E.g: Cement industry, Sugar industry, Cotton textile industry, Iron and Steel industry, Fertiliser industry etc.

6) Service industry: In modern times, service sector plays an important role in the development of the nation and therefore it is named as service industry. These are engaged in the provision of essential services to the community.
E.g: Banking, Trans-port, Insurance etc.

Question 5.
Define ‘Entrepot trade’ with example.
Answer:
When goods are imported from one country and the same are exported to another country, such trade is called entrepot trade.
E.g.:

  • India importing wheat from U.S. and exporting the same to Srilanka.
  • India imports petrol from Iran and export the same to Nepal.

Very Short Answer Questions

Question 1.
Industry.
Answer:
Industry is concerned with the making or manufacturing of goods. It is that part of the production which is involved in changing the form of goods from raw material to the finished product.
E.g.: Weaving woollen yam into cloth. Thus industry creates form utility in goods.

Question 2.
Commerce.
Answer:

  • Commerce is concerned with exchange of goods. It includes all those activities which are related to transfer of goods from the places of production to the ultimate consumer. Commerce embraces all those processes which help to break the barriers between producers and consumers.
  • It is the sum total of those processes which are engaged in the removal of hindrances of persons, place, time and exchange.

Question 3.
Trade.
Answer:

  • All the activities engaged in buying and selling of goods and services a re called trade.
  • Therefore trade includes sale, transfer or exchange of goods and services with the intention of making profit. The object of trade is to make goods available to those who need them and willing to pay for them.
  • Trade is the final stage of business activities and involves transfer of ownership.

Question 4.
Home Trade.
Answer:

  • The trade is carried on within the boundaries of the nation is called Home Trade. Home trade is also called as internal trade or domestic trade.
  • Home trade refers to a trade where buying and selling of goods takes place between the persons who belong to the same country. So, home For movement of goods internal transport system is used.

TS Inter 1st Year Commerce Study Material Chapter 2 Business Activities

Question 5.
Entrepot trade.
Answer:

  • When goods are imported from one country and the same are exported to another country, such trade is called entrepot trade.
  • E.g.: India importing wheat from U.S. and exporting the same to Srilanka.

Question 6.
Transportation.
Answer:
There is a vast distance between centres of production and centres of consumption. Goods are to be moved from the places of production to the place where they are demanded.

  • The activity which is concerned with movement of goods is called transportation. Trans-port create place utility.
  • There are several kinds of transport such as air, water and land transport. The geo-graphical distance between producers and consumers is removed with the help of transport.

Question 7.
Warehousing.
Answer:

  • There is time gap between production and consumption. Hence, it became necessary to make arrangements for storage or warehousing.
  • It is one of the aid to trade, which facilitates to store the goods until they get demand or consumed.
  • Some goods are to be stored in warehouses till they are demanded. Warehousing creates time utility.

Question 8.
Genetic Industries.
Answer:

  • Genetic industry is related to the reproducing and multiplying certain species of animals and plants with the object of earning profit from their sale.
  • E.g.: Nurseries, cattle breeding, poultry farm, fish hatcheries etc.

Question 9.
Extractive industries.
Answer:

  • These industries are engaged in raising some form of wealth from the soil, climate, air, water or from beneath the surface of the earth. Generally the products of extractive industry comes in raw form and they are used by manufacturing and construction industries for producing finished products.
  • E.g.: Mining, coal, mineral, iron ore oil, extraction of timber and rubber from forests.

Question 10.
Banking.
Answer:

  • Banking solve the problem of finance.
  • Producers and traders require money for carrying on production and trade. Banks are the institutions which supply funds for industries and trade.
  • They pool savings from the public and make them available to industry. So, banking is an important function of commerce.