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

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

Question 1.

Find the condition for the points (a, 0), (h, k), and (0, b), where ab ≠ 0, to be collinear. [Mar. ’10]

Solution:

Let A(a, 0), B(h, k), C (0, b) be the given points.

Slope of \(\overline{\mathrm{AB}}=\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}=\frac{\mathrm{k}-0}{\mathrm{~h}-\mathrm{a}}=\frac{\mathrm{k}}{\mathrm{h}-\mathrm{a}}\)

Slope of \(\overline{\mathrm{BC}}=\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}=\frac{\mathrm{b}-\mathrm{k}}{0-\mathrm{h}}=\frac{\mathrm{b}-\mathrm{k}}{-\mathrm{h}}\)

Since points A, B, C are collinear, then

Slope of \(\overline{\mathrm{AB}}\) = Slope of \(\overline{\mathrm{BC}}\)

\(\frac{k}{h-a}=\frac{b-k}{-h}\)

⇒ -hk = (h – a) (b – k)

⇒ -hk = bh – hk – ab + ak

⇒ bh + ak = ab

⇒ \(\frac{b h}{a b}+\frac{a k}{a b}=1\)

\(\frac{h}{a}+\frac{k}{b}=1\), which is the required condition.

Question 2.

Find the value of x, if the slope of the line passing through (2, 5) and (x, 3) is 2. [Mar. ’18 (AP & TS); May ’12; B.P.]

Solution:

Let A = (2, 5), B = (x, 3) are the given points.

Given, Slope of \(\overline{\mathrm{AB}}\) = 2

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,

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(x_{1}, y_{1}) is (-2, -1).

∴ The equation of the straight line passing through A(-2, -1) and having slope \(\frac{-1}{\sqrt{3}}\) is y – y_{1} = m(x – x_{1})

⇒ y + 1 = latex]\frac{-1}{\sqrt{3}}[/latex] (x + 2)

⇒ √3(y + 1) = -1(x + 2)

⇒ √3y + √3 = -x – 2

⇒ x + √3y + √3 = -2

⇒ x + √3y + √3 + 2 = 0

Question 5.

Find the equations of the straight lines passing through the origin and making equal angles with the coordinate axes. [May ’05]

Solution:

Let l_{1}, l_{2} 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 l_{1} is θ = 45°

Slope of a line l_{1} is, m = tan θ = tan 45° = 1

Let the given point O = (0, 0)

∴ The equation of a straight line l_{1} passing through O(0, 0) and having slope ‘1’ is y – y_{1} = m(x – x_{1})

y – 0 = 1(x – 0)

⇒ y = x

⇒ x – y = 0

Case II: Inclination of a straight line l_{2} is θ = 135°

The slope of a line l_{2} 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 l_{2} passing through O(0, 0) and having slope ‘-1’ is y – y_{1} = m(x – x_{1})

⇒ y – 0 = -1(x – 0)

⇒ y = -x

⇒ x + y = 0

∴ Required equations of the straight lines are x – y = 0; x + y = 0

Question 6.

Find the equation of the straight line passing through (-4, 5) and cutting off equal and non-zero intercepts on the coordinate axes. [Mar. ’15 (TS) ’13 (Old), ’07, ’00; May ’10, ’08]

Solution:

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

Given that, the straight line making equal intercepts on the co-ordinate axis, then a = b

From (1),

\(\frac{x}{a}+\frac{y}{a}=1\)

x + y = a ……….(2)

Since equation (2) passes through the point (-4, 5) then,

-4 + 5 = a

∴ a = 1

Substitute the value of ’a’ in equation (2)

∴ x + y = 1

Question 7.

Find the equation of the straight line passing through (-2, 4) and making non-zero intercepts whose sum is zero. [Mar. ’15 (AP), ’13; May ’15 (TS), ’02]

Solution:

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

Given that, the straight line-making intercepts whose sum is ‘0’

i.e., a + b = 0

b = -a

From (1)

\(\frac{x}{a}+\frac{y}{-a}=1\)

x – y = -a …….(2)

Since equation (2) passes through the point (-2, 4) then,

-2 – 4 = a

∴ a = -6

Substitute the value of ‘a’ in equation (2)

x – y = -6

x – y + 6 = 0

Question 8.

Find the equation of the straight line passing through the point (3, -4) and making X and Y-intercepts which are in the ratio 2 : 3. [Mar. ’08]

Solution:

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

Given that, the ratio of intercepts = 2 : 3

X-intercept = 2a

Y-intercept = 3a

From (1),

\(\frac{x}{2 a}+\frac{y}{3 a}=1\)

\(\frac{3 x+2 y}{6 a}=1\)

3x + 2y = 6a ……..(2)

Since equation (2) passes through point (3, -4) then,

3(3) + 2(-4)= 6a

⇒ 9 – 8 = 6a

⇒ 6a = 1

⇒ a = \(\frac{1}{6}\)

Substitute the value of ‘a’ in equation (2)

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

∴ 3x + 2y = 1

Question 9.

Find the equation of the straight line passing through the points \(\left(a t_1^2, 2 at_1\right)\) and \(\left(a t_2^2, 2 at_2\right)\). [Mar. ’14. ’04; May ’15 (AP), ’00]

Solution:

Let A\(\left(a t_1^2, 2 at_1\right)\) and B\(\left(a t_2^2, 2 at_2\right)\) are the given points

The equation of the straight line passing through the points A\(\left(a t_1^2, 2 at_1\right)\) and B\(\left(a t_2^2, 2 at_2\right)\) is

(y – y_{1}) (x_{2} – x_{1}) = (x – x_{1}) (y_{2} – y_{1})

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.

(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 – y_{1} = m(x – x_{1})

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

y – 3 = \(\frac{-2}{11}\) (x + 1)

⇒ 11y – 33 = -2x – 2

⇒ 2x + 11y – 31 = 0

Question 11.

A(10, 4), B(-4, 9) and C(-2, -1) are the vertices of a triangle. Find the equation of the altitude through B. [May ’13 (Old)]

Solution:

Slope of \(\overline{\mathrm{AC}}\) is,

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 – y_{1} = m(x – x_{1})

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,

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,

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

which is in the form x cos α + y sin α = p

∴ α = \(\frac{\pi}{6}\), p = 2

Question 15.

Transform the equation x + y + 1 = 0 into normal form. [Mar. ’17 (AP), ’08; May ’10; B.P.; Mar. ’18 (TS)]

Solution:

Given, equation of the straight line is x + y + 1 = 0

x + y = -1

– x – y = 1

On dividing both sides with

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 \(\sqrt{a^2+b^2}\)

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

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

(c) Normal form:

Given equation of the straight line is, √3x + y + 10 = 0

√3x + y = -10

-√3x – y = 10

Question 19.

If the area of the triangle is formed by the straight lines, x = 0, y = 0, and 3x + 4y = a [a > 0] is ‘6’. Find the value of ‘a’. [May ’11, Mar. ’09, ’07]

Solution:

Given equations of the straight lines are (a > 0) 3x + 4y = a, x = 0 and y = 0

Comparing with ax + by + c = 0, we get

a = 3, b = 4, c = -a

The area of the triangle formed by this line and the co-ordinate axis is equal to \(\frac{c^2}{2|a b|}\)

Given that area of the triangle = 6

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)

∴ 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(x_{1}, y_{1}) = (0, 0) and B(x_{2}, y_{2}) = (-2, 1)

Required ratio = \(\frac{-\left(a x_1+b {y}_1+c\right)}{a x_2+b y_2+c}\)

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 + c_{1} = 0, we get

a = 6, b = 8, c_{1} = -6

Comparing (2) with ax + by + c_{2} = 0, we get

a = 6, b = 8, c_{2} = -1

Distance between the parallel lines =

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

a_{1} = 4, b_{1} = -1, c_{1} = 7

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

a_{2} = k, b_{2} = -5, c_{2} = -9

Given that, θ = 45°

If ‘θ’ is the angle between the given lines then,

Squaring on both sides

⇒ 17(k^{2} + 25) = 2(4k + 5)^{2}

⇒ 17k^{2} + 425 = 2(16k^{2} + 25 + 40k)

⇒ 17k^{2} + 425 = 32k^{2} + 50 + 80k

⇒ 15k^{2} + 80k – 375 = 0

⇒ 3k^{2} + 16k – 375 = 0

⇒ 3k^{2} + 25k – 9k – 75 = 0

⇒ k(3k + 25) – 3(3k + 25) = 0

⇒ (3k + 25) (k – 3) = 0

⇒ 3k + 25 = 0; k – 3 = 0

⇒ 3k = -25; k = 3

⇒ k = 3 or \(\frac{-25}{3}\)

Question 24.

Find the equation of the straight line parallel to the line 2x + 3y + 7 = 0 and pass through the point (5, 4). [Mar. ’13, ’03]

Solution:

Given, the equation of the straight line is 2x + 3y + 7 = 0

Given points (5, 4)

The equation of the straight line parallel to 2x + 3y + 7 = 0 is 2x + 3y + k = 0

Since equation (1) passes through the point (5, 4) then,

2(5) + 3(4) + k = 0

⇒ 10 + 12 + k = 0

⇒ 22 + k = 0

⇒ k = -22

∴ The required equation of the straight line is 2x + 3y – 22 = 0

Question 25.

Find the value of k, if the straight lines y – 3kx + 4 = 0 and (2k – 1)x – (8k – 1)y – 6 = 0 are perpendicular. [Mar. ’10]

Solution:

Given, the equations of the straight lines are

y – 3kx + 4 = 0 ………(1)

(2k – 1)x – (8k – 1)y – 6 = 0 ……….(2)

Slope of the line (1) is m_{1} = \(\frac{-(-3 k)}{1}\) = 3k

Slope of the line (2) is m_{2} = \(\frac{-(2 k-1)}{-(8 k-1)}=\frac{(2 k-1)}{(8 k-1)}\)

Since the given lines are perpendicular then m_{1} × m_{2} = -1

\(3 \mathrm{k}\left(\frac{2 \mathrm{k}-1}{8 \mathrm{k}-1}\right)\) = -1

⇒ 3k(2k – 1) = -1(8k – 1)

⇒ 6k^{2} – 3k = -8k + 1

⇒ 6k^{2} + 5k – 1 = 0

⇒ 6k^{2} + 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(x_{1}, y_{1}) = (3, 4)

The perpendicular distance from the point (3, 4) to the line 3x – 4y + 10 = 0 is

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