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:

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(x_{1}, y_{1}) = (3, 2)

Given the equation of the straight line is 3x – 4y – 1 = 0

Distance |r| = 5

Required points = (x_{1} + |r| cos θ, y1 + |r| sin θ)

= (x_{1} ± r cos θ, y_{1} + r sin θ)

= (3 ± 5 . \(\frac{4}{5}\), 2 ± 5 . \(\frac{3}{5}\))

= (3 ± 4, 2 ± 3)

= (3 + 4, 2 + 3); (3 – 4, 2 – 3)

= (7, 5), (-1, 1)

Question 3.

Find the value of k, if the lines 2x – 3y + k = 0, 3x – 4y – 13 = 0 and 8x – 11y – 33 = 0 are concurrent. [May ’07, ’95; Mar. ’05, ’80; Mar. ’18 (TS)]

Solution:

Given, the equations of the straight lines are,

2x – 3y + k = 0 ……..(1)

3x – 4y – 13 = 0 ……..(2)

8x – 11y – 33 = 0 ………(3)

Solving (2) & (3)

∴ 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 a^{3} + b^{3} + c^{3} = 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)

\(\frac{y}{b c-a^2}=\frac{1}{a c-b^2}\) ⇒ 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)

⇒ c(ab – c^{2}) + a(bc – a^{2}) + b(ac – b^{2}) = 0

⇒ abc – c^{3} + abc – a^{3} + abc – b^{3} = 0

⇒ 3abc – a^{3} – b^{3} – c^{3} = 0

⇒ a^{3} + b^{3} + c^{3} = 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,

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(x_{1}, y_{1}) be any point on the locus.

Since Q(x_{1}, y_{1}) is the midpoint of \(\overline{\mathbf{A B}}\)

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(x_{1} + y_{1}) = (a + b) 2x_{1}y_{1}

∴ The locus of the midpoint ‘Q’ of AB is 2(a + b) xy = ab(x + y)

Question 6.

A straight line meets the coordinate axes in A and B. Find the equation of the straight line when (p, q) bisects \(\overline{\mathbf{A B}}\). [May ’90]

Solution:

The equation of the straight line in the intercept form is \(\frac{x}{a}+\frac{y}{b}=1\) …….(1)

The straight line (1) meets the X-axis at A(a, 0), Y-axis at B(0, b).

Now, C(p, q) bisects \(\overline{\mathbf{A B}}\), then C is the midpoint of \(\overline{\mathbf{A B}}\).

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)

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

\(a\left(\frac{4 a}{a-3}\right)=48\)

a^{2} = 12(a – 3)

a = 12a – 36

a^{2} – 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 L_{1} + λL_{2} = 0

Here, ax + by + c = 0, represents a set of lines passing through the point of intersection of the lines

L_{1} = 4x – 3 = 0 …….(1)

L_{2} = 4y – 2 = 0 ………(2)

Solving (1) & (2)

From (1), 4x – 3 = 0

4x = 3

x = \(\frac{3}{4}\)

from (2), 4y – 2 = 0

4y = 2

y = \(\frac{1}{2}\)

∴ The point of concurrence = \(\left(\frac{3}{4}, \frac{1}{2}\right)\)

∴ ax + by + c = 0 represents a set of concurrent lines.

The point of concurrence = \(\left(\frac{3}{4}, \frac{1}{2}\right)\)

Question 9.

Find the point on the straight line 3x + y + 4 = 0, which is equidistant from the points (-5, 6) and (3, 2). [Mar. ’13; Nov. ’98]

Solution:

Given, equation of the straight line is 3x + y + 4 = 0 …….(1)

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}

x^{2} + 25 + 10x + y^{2} + 36 – 12y = x^{2} + 9 – 6x + y^{2} – 4y + 4

10x – 12y + 61 = -6x – 4y + 13

16x – 8y + 48 = 0

2x – y + 6 = 0 ………(2)

Solving (1) & (2)

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)

Given point Q(x_{1}, y_{1}) = (√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 – y_{1} = m(x – x_{1})

y – 2 = \(\frac{1}{\sqrt{3}}\) (x – √3)

√3y – 2√3 = x – √3

x – √3y + √3 = 0 …….(2)

Solving (1) & (2)

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)

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 – y_{1} = \(\frac{-1}{m}\) (x – x_{1})

y – 0 = \(\frac{-a}{b}\) (x – a)

y = \(\frac{-a}{b}\) (x – a)

by = -ax + a^{2}

ax + by – a^{2} = 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)

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 – y_{1}) = \(\frac{-1}{m}\) (x – x_{1})

y – 0 = \(\frac{4}{3}\) (x + 4)

3y = 4x + 16

4x – 3y + 16 = 0

Question 13.

Find the equation of the straight line making non-zero equal intercepts on the coordinate axes and passing through the point of intersection of the lines 2x – 5y + 1 = 0 and x – 3y – 4 = 0. [Mar. ’06, ’00]

Solution:

Given the equation of the lines are

2x – 5y + 1 = 0 …….(1)

x – 3y – 4 = 0 …….(2)

\(\frac{x}{23}=\frac{y}{9}=\frac{1}{-1}\)

\(\frac{x}{23}\) = -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)

∴ 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(x_{1}, y_{1}) = (-2, 1)

The perpendicular distance from P(-2, 1) to the straight line

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 a_{1}x + b_{1}y + c_{1} = 0, we get

a_{1} = b, b_{1} = a, c_{1} = -ab

Comparing (2) with a_{2}x + b_{2}y + c_{2} = 0, we get

a_{2} = a, b_{2} = b, c_{2} = -ab

If ‘θ’ is the angle between lines (1) & (2) then,

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 – y_{1} = m(x – x_{1})

⇒ 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 – y_{1} = \(\frac{-1}{m}\) (x – x_{1})

⇒ y – 3 = \(\frac{1}{2}\) (x – 1)

⇒ 2y – 6 = 3x – 3

⇒ 3x – 2y + 3 = 0