Students must practice these TS Intermediate Maths 1B Solutions Chapter 3 Straight Lines Ex 3(b) to find a better approach to solving the problems.

## TS Inter 1st Year Maths 1B Straight Lines Solutions Exercise 3(b)

I.

Question 1.

Find the sum of squares of the intercepts of the line 4x – 3y = 12 on the axes of coordinates. (VS.A.Q.)

Answer:

Equation of the given line is 4x – 3y = 12

Writing in the intercepts form \(\frac{x}{a}+\frac{y}{b}\) = 1

∴ a = 3, b = – 4

∴ Sum of the squares = a^{2} + b^{2} = 9 + 16 – 25

Question 2.

If the portion of a straight line intercepted between the axes of coordinates is bisected at (2p, 2q), write the equation of the straight (V.S.A.Q.)

Answer:

Coordinates of A = (a, 0) and B = (0, b)

M is the midpoint of AB

∴ M = \(\left(\frac{\mathrm{a}}{2}, \frac{\mathrm{b}}{2}\right)\) = (2p, 2q)

∴ \(\frac{a}{2}\) = 2p and \(\frac{b}{2}\) = 2q ⇒ a = 4p, b = 4q

Question 3.

If the linear equations ax + by + c = 0 (abc ≠ 0) and lx + my + n = 0 represents the same line and r = \(\frac{l}{a}\) = \(\frac{n}{c}\), write the value of r in terms of m and b. (V.S.A.Q.)

Answer:

ax + by + c = 0 and lx + my + n = 0 represents the same line.

Hence coefficients are proportional

∴ \(\frac{l}{a}=\frac{m}{b}=\frac{n}{c}\) = r

r = \(\frac{\mathrm{m}}{\mathrm{b}}\) is the value of ’r’ interms of m and b.

Question 4.

Find the angle made by the straight line y = – √3 x + 3 with the positive direction of the X – axis measured in the counter clock-wise direction. (V.S.A.Q.)

Answer:

Equation of the given line is y = – √3x + 3 which is of the form y = mx + c where

m = tan α = √3 = tan \(\frac{2 \pi}{3}\) (angle made by the line in counter clock-wise direction)

∴ α = \(\frac{2 \pi}{3}\)

Question 5.

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. (V.S.A.Q.)

Answer:

Equation of the line in the intercepts form is

\(\frac{x}{a}+\frac{y}{b}\) – 1 = 0

p = length of the perpendicular drawn from origin to the line then

II.

Question 1.

In what follows, p denotes the distance of the straight line from the origin and a denotes the angle made by the normal ray drawn from the origin to the straight line with \(\overleftrightarrow{O x}\) measured in the anti-clockwise sense. Find the equations of the straight lines with the following values of p and α. (V.S.A.Q.)

(i) p = 5, α = 60°

(ii) p = 6, α = 150°

(iii) p = 1, α = \(\frac{7 \pi}{4}\)

(iv) p = 4, α = 90°

(v) p = 0, α = 0

vi) P = 2√2, α = \(\frac{5 \pi}{4}\)

Answer:

Equation of line in the normal form is x cos α + y sin α = p

(i) p = 5, α = 60°

Equation of line is x cos 60°+ y sin 60°= 5

⇒ \(\frac{x}{2}\) + y\(\frac{\sqrt{3}}{2}\) = 5

⇒ x + √3y = 10

(ii) p = 6, α = 150°

Equation of line is x cos 150° + y sin 150°= 6

⇒ x cos ( 180° – 30°) + y sin (180° -30°) = 6

⇒ – x cos 30° + y sin 30° = 6

⇒ – x \(\frac{\sqrt{3}}{2}\) + y \(\frac{1}{2}\) = 6

⇒ √3x + y = 12

⇒ √3x – y + 12 = 0

(iii) P = 1, α = \(\frac{7 \pi}{4}\)

Equation of the line is x cos \(\frac{7 \pi}{4}\) + y sin \(\frac{7 \pi}{4}\) = 1

⇒ x cos 315° + y sin 315° = 1

⇒ x cos (360° – 45°) + y sin (360° – 45°) = 1

⇒ x cos 45° – y sin 45° = 1

⇒ \(\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}\) = 1

⇒ x – y – √2 = 0

(iv) p = 4, α = 90°

Equation of the line is

x cos 90° + y sin 90° = 4

⇒ x(0) + y(1) = 4

⇒ y – 4 = 0

(v) p = 0, α = 0

Equation of the line is

x cos 0 + y sin 0 = 0

⇒ x = 0

(vi) p = 2√2 , α = \(\frac{5 \pi}{4}\)

Equation of the line is

x cos \(\frac{5 \pi}{4}\) + y sin \(\frac{5 \pi}{4}\) = 2√2

⇒ x cos 225° + y sin 225° = 2√2

⇒ x cos (180° + 45°) + y sin( 180° + 45°) = 2√2

⇒ – x cos 45° – y sin 45° = 2√2

⇒ – \(\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}\) = 2√2

⇒ x + y + 4 = 0

Question 2.

Find the equations of the straight lines in the symmetric form, given the slope and a point on the line in each part of the question. (S.A.Q.)

(i) √3, (2, 3)

(ii) – \(\frac{1}{\sqrt{3}}\), (- 2, 0)

(iii) – 1, (1, 1)

Answer:

(i) Equation of the line in symmetric form is

\(\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}\) = r

(x_{1}, y_{1}) = (2, 3), slope m = tan θ = √3 ⇒ θ = 60°

∴ Equation of the line is

\(\frac{x-2}{\cos \theta}=\frac{y-3}{\sin \theta}\) = √3

⇒ \(\frac{x-2}{\cos 60^{\circ}}=\frac{y-3}{\sin 60^{\circ}}\)

(ii) (x_{1}, y_{1}) = (-2, 0)

Answer:

Slope m = tan θ = – \(\frac{1}{\sqrt{3}}\)

⇒ θ = 180° – 30° = 150°

∴ Equation of the line is \(\frac{x+2}{\cos 150^{\circ}}=\frac{y}{\sin 150^{\circ}}\)

(iii) tan α = – 1, α = 180 – 45°

(x_{1}, y_{1}) = (1, 1)

∴ Equation of the line is

\(\frac{x-1}{\cos \left(\frac{3 \pi}{4}\right)}\) = \(\frac{y-1}{\sin \left(\frac{3 \pi}{4}\right)}\)

Question 3.

Transform the following equations into

(a) slope – intercept form (b) intercept form and (c) normal form. (S.A.Q.)

(i) 3x + 4y = 5

(ii) 4x – 3y + 12 = 0,

(iii) √3x + y = 4

(iv) x + y + 2 = 0

(v) x + y – 2 = 0

(vi) √3x + y + 10 = 0

Answer:

(i) 3x + 4y = 5

(a) Slope – intercept form : 4y = 5 – 3x

⇒ y = \(\frac{5}{4}-\frac{3}{4}\)

Slope m = – \(\frac{3}{4}\) and Y- intercept = \(\frac{5}{4}\)

(b) Intercept form : 3x + 4y = 5

⇒ \(\frac{x}{\left(\frac{5}{3}\right)}+\frac{y}{\left(\frac{5}{4}\right)}\) = 1

This is of the form \(\frac{x}{a}+\frac{y}{b}\) = 1

(c) Normal form: 3x + 4y = 5,

Dividing by \(\sqrt{9+16}\) = 5 on both sides

(ii) 4x – 3y + 12 = 0,

(a) Slope – intercept form: 3y = 4x + 12

⇒ y = \(\left(\frac{4}{3}\right)\)x + 4

Slope = \(\frac{4}{3}\) and y-intercept = 4

(b) Intercept form: 4x – 3y = – 12

⇒ \(\frac{4}{-12} x+\frac{3}{12} y\) = 1

⇒ \(\frac{x}{(-3)}+\frac{y}{(4)}\) = 1

(c) Normal form : 4x – 3y + 12 = 0

⇒ – 4x + 3y = 12

Dividing by \(\sqrt{16+9}\) = 5 on both sides

(iii) √3x + y = 4

Answer:

(a) Slope – intercept form : y = -√3x + 4

Slope = – √3 , Y-intercept = 4

(b) Intercept form: √3 x + y = 4

(c) Normal form: √3x + y = 4

Dividing by √3 + 1 = 2 on both sides

(iv) x + y + 2 = 0

(a) Slope – intercept form: y = – x – 2

Slope = – 1, Y – intercept = – 2,

(b) Intercept form: x + y = – 2

⇒ \(\frac{x}{-2}+\frac{y}{-2}\) = 1

(c) Normal form : x + y + 2 = 0

⇒ – x – y = 2

Dividing by \(\sqrt{1+1}\) = √2 on both sides

(v) x + y – 2 = 0

(a) Slope – intercept form: y = – x + 2

Slope = -1, Y – intercept = 2

(b) Intercept form: x + y -2 = 0

⇒ x + y = 2

⇒ \(\frac{x}{2}+\frac{y}{2}\) = 1

(c) Normal form: x + y -2 = 0

⇒ x + y = 2

Dividing by \(\sqrt{1+1}\) = √2 on both sides

(vi) √3x + y + 10 = 0

Answer:

(a) Slope – intercept form : y = -√3x – 10

Slope = – √3 , Y – intercept = – 10

(b) Intercept form: √3 x + y = – 10

(c) Normal form: √3 x + y = – 10

Dividing by \(\sqrt{3+1}\) = 2 both sides

Question 4.

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. (S.A.Q.)

Answer:

Equation of the lines x tan α + y sec α = 1

⇒ \(\frac{x}{\cot \alpha}+\frac{y}{\cos \alpha}\) = 1

This is in intercepts form and

a = cot α, b = cos α

given ab = product of intercepts on the axes = sin α

∴ ab = sin α

⇒ cot α . cos α = sin α

⇒ \(\frac{\cos \alpha}{\sin \alpha}\) .cos α = sin α

⇒ cos^{2} α = sin^{2} α

⇒ tan^{2} α = 1 ⇒ tan α = ± 1

(∵ 0 ≤ α ≤ \(\frac{\pi}{2}\)) we take α = \(\frac{\pi}{4}\) = 45°

Question 5.

If the sum of the reciprocals of the intercepts made by a variable straight line on the axes of coordinates is a constant, then prove that the line always passes through a fixed point. (S.A.Q.)

Answer:

Equation of the line in the intercept form is

\(\frac{x}{a}+\frac{x}{b}+\frac{x}{c}\) = 1 ……………. (1)

Sum of the reciprocals of the intercepts

Question 6.

Line L has intercepts a and b on the axes of the coordinates. When the axes are rotated through a given angle, keeping the origin fixed, the same line L has intercepts p and q on the transformed axes. Prove that \(\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2}\). (S.A.Q.)

Answer:

Equation of the line in the original system in the intercepts form is \(\frac{x}{a}+\frac{y}{b}\) = 1

⇒ \(\frac{x}{a}+\frac{y}{b}\) – 1 = 0

Length of the perpendicular from origin

= \(\frac{|0+0-1|}{\sqrt{\frac{1}{\mathrm{a}^2}+\frac{1}{\mathrm{~b}^2}}}\) ………………. (1)

Equation of the line in the second system in the intercepts form is

Question 7.

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 \(\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}\). (S.A.Q)

Answer:

III.

Question 1.

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. (S.A.Q.)

Answer:

We have symmetric form of the line as

\(\frac{x-x_1}{\cos \alpha}=\frac{y-y_1}{\sin \alpha}\) = r

∴ Coordinates of any point on the given line are (x_{1} + r cos α , y_{1} + r sin α)

Given α = 30°

⇒ cos α = cos 30° = \(\frac{\sqrt{3}}{2}\)

⇒ sin α = sin 30° = \(\frac{1}{2}\)

∴ (x_{1}, y_{1}) = (- 2, 1)

⇒ x_{1} = – 2, y_{1} = 1

Taking r = 4, coordinates of P

= [- 2 + 4\(\left(\frac{\sqrt{3}}{2}\right)\), 1 + 4\(\left(\frac{1}{2}\right)\)]

= [- 2 + 2√3, 3]

Taking r = – 4 coordinates of P’ are

= [- 2 + 4\(\left(\frac{\sqrt{3}}{2}\right)\), 1 – 4\(\left(\frac{1}{2}\right)\)]

= (- 2 + 2√3, – 1)

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 ). (S.A.Q.)

Answer:

Equation of the line in symmetric form is

\(\frac{x-3}{\cos \alpha}=\frac{y-2}{\sin \alpha}\) = r

Coordinates of the point P are

= ( 3 + r cos α, 2 + r sin α)

= ( 3 + 5 cos α, 2 + 5 sin α)

P is a point on the line 3x – 4y – 1 = 0

⇒ 3 (3 + 5 cos α) – 4(2 + 5 sin α) – 1 = 0

⇒ 15 cos α – 20 sin α = 0

⇒ 15 cos α = 20 sin α

⇒ tan α = \(\frac{15}{20}\) = \(\frac{3}{4}\)

Case (i): When α is in the first quadrant

cos α = \(\frac{-4}{5}\), sin α = – \(\frac{3}{5}\)

Case (ii): When α is in the third quadrant then

cos α = \(\frac{-4}{5}\), sin α = – \(\frac{3}{5}\)

From case (i) coordinates of P are

[3 + 5\(\left(\frac{4}{5}\right)\), 2 + 5\(\left(\frac{3}{5}\right)\)]

Case (iii): Coordinates of P are

[3 – 5\(\left(\frac{4}{5}\right)\), 2 – 5\(\left(\frac{3}{5}\right)\)] = [- 1, 1]

Question 3.

A straight line whose inclination with the positive direction of the X-axis measured in the anti clockwise sense is \(\frac{\pi}{3}\) makes positive intercept on the Y – axis. If the straight line is at a distance of 4 from the origin, find its equation. (S.A.Q.)

Answer:

Given α = \(\frac{\pi}{3}\), p = 4

We have m = tan α = tan 60° = √3

∴ Equation of the line in slope – intercept form is y = √3 x + c

⇒ √3x – y + c = 0

Distance from the origin = 4

∴ \(\frac{|0-0+c|}{\sqrt{3+1}}\) = 4

⇒ |c| = 2 × 4 = 8

⇒ c = ±8

When c > 0 we have c = 8

Hence the equation of the line is

√3x – y + 8 = 0

Question 4.

A straight line L is drawn through the point A (2, 1) such that its point of intersection with the straight line x + y = 9 is at a distance

of 3√2 from A. Find the angle which the line L makes with the positive direction of the X-axis. (S.A.Q.)

Answer:

Suppose a is the angle made by L with the positive X-axis.

Any point on the line is (x_{1} + r cos α, y_{1} + r sin α)

= (2 + 3√2 cos α, 1 + 3√2 sin α)

This is a point over the line x + y = 9

∴ (2 + 3√2 cos α) + ( 1 + 3√2 sin α) = 9

⇒ 3 √2 ( cos α + sin α) = 6

⇒ cos α + sin α = \(\frac{6}{3 \sqrt{2}}\) = √2

⇒ \(\frac{1}{\sqrt{2}}\) cos α + \(\frac{1}{\sqrt{2}}\) sin α = 1

⇒ cos α . cos 45° + sin α . sin 45° = 1

⇒ cos (α – 45°) = cos 0°

⇒ α – 45° = 0 ⇒ α = 45° = \(\frac{\pi}{4}\)

Question 5.

A straight line L with negative slope passes through the point ( 8, 2 ) and cuts positive coordinate axes at the points P and Q. Find the minimum value of OP + OQ as L varies, where O is the origin. (S.A.Q.)

Answer:

Equation of the line passing through A (8, 2) with negative slope – m is

y – 2 = – m(x – 8)

⇒ mx + y – (2 + 8m) = 0

⇒ mx + y = 2 + 8m

∴ Minimum value of OP + OQ as L varies where O is the origin is 18.