Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B The Straight Line Solutions Exercise 3(a) will help students to clear their doubts quickly.

## Intermediate 1st Year Maths 1B The Straight Line Solutions Exercise 3(a)

I.

Question 1.

Find the slope of the line x + y = 0 and x – y = 0.

Solution:

Slope of x + y = 0 is – \(\frac{a}{b}\) = -1

Slope of x – y = 0 is 1

Question 2.

Find the equation of the line containing the points (2, -3) and (0, -3).

Solution:

Equation of the line is

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

(y + 3)(2 – 0) = (x – 2)(-3 + 3)

2(y + 3) = 0

⇒ y + 3 = 0

Question 3.

Find the equation of the line containing the points (1, 2) and (1, -2).

Solution:

Equation of the line is

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

(y – 2)(1 – 1) = (x – 1) (2 + 2)

0 = 4(x – 1) ⇒ x – 1 =0

Question 4.

Find the angle which the straight line y = √3x – 4 makes with the Y-axis.

Solution:

Equation of the line is y = √3x – 4

Slope = m = √3 = tan \(\frac{\pi}{6}\)

Angle made with X-axis = \(\frac{\pi}{6}\)

Angle made with Y – axis = \(\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}\)

Question 5.

Write the equation of the reflection of the line x = 1 in the Y-axis.

Solution:

Equation of PQ is x = 1

Reflection about Y – axis is x = – 1

i.e., x + 1 = 0

Question 6.

Find the condition for the points (a, 0), (h, k) and (0, b) when ab ≠ 0 to be collinear.

Solution:

A(a, 0), B(h, k), C(0, b) are collinear.

⇒ Slope of AB = Slope of AC

\(\frac{k-0}{h-a}=\frac{-b}{a}\)

ak = -bh + ab

bh + ak = ab

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

Question 7.

Write the equations of the straight lines parallel to X-axis is

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)

Equation of the required line AB is y = 3

ii)

Equation of A’B’ is y = — 4 ; y + 4 = 0

8. Write the equations of the straight line parallel to 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)

Equation of the required line AB is x = 2

ii)

Equation of the required line A’B’

x = -5

x + 5 = 0

II.

Question 1.

Find the slopes of the straight line passing through the following pairs of points.

i) (-3, 8) (10, 5)

ii) (3, 4) (7, -6)

iii) (8,1), (-1, 7)

iv) (-P, q) (q, -p) (pq ≠ 0)

Solution:

Question 2.

Find the value of x, if the slope of the line passing through (2, 5) and (x, 3) is 2.

Solution:

Slope = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{5-3}{2-x}=2\)

2 = 2(2 – x)

x = 2 – 1 = 1

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).

Solution:

A(3, y), B(2, 7), P(-1, 4) and Q(0, 6) are the given points.

m_{1} = Slope of AB = \(\frac{y-7}{3-2}\) = y – 7

m_{2} = Slope of PQ = \(\frac{4-6}{-1-0}=\frac{-2}{-1}=2\)

AB and PQ are parallel

m_{1} = m_{2} ⇒ y – 7 = 2

y = 2 + 7 = 9

Question 4.

Find the slopes of the lines i) parallel to and ii) perpendicular to the line passing through (6, 3) and (- 4,5).

Solution:

A(6, 3) and B(-4, 5) are the given points.

m = Slope of AB = \(\frac{3-5}{6+4}=\frac{-2}{10}=-\frac{1}{5}\)

PQ is parallel to AB

(i) Slope of PQ = m = – \(\frac{1}{5}\)

RS is perpendicular to AB

(ii)Slope of RS = – \(\frac{1}{m}\) =5 m

Question 5.

Find the equation of the straight line which makes the following angles with the positive X-axis in the positive direction and which pass through the points given below

i) \(\frac{\pi}{4}\) and (0,0)

ii) \(\frac{\pi}{3}\) and(1, 2)

iii) 135° and (3, -2)

iv) 150° and (-2, -1)

Solution:

i) m = Slope = tan 45° = 1

Equation of the line is y – y_{1} = m(x – x_{1})

y – 0 = 1(x – 0)

i.e., y = x

or x – y = 0

ii) m = tan 60° – √3

Equation of the line is

y – 2 = √3 (x – 1)

= √3x – √3

√3x – y+(2 – √3) = 0

iii) m = tan 135° = tan (180° -45°)

= – tan 45° = -1

Equation of the line is y + 2 = -1(x – 3)

= – x + 3

i.e., x + y – 1 = 0

iv) m = tan 150° = tan (180° – 30°)

= – tan 30° = – \(\frac{1}{\sqrt{3}}\)

Equation of the line is

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

√3y + √3 = – x – 2

x + √3y + (2 + √3) = 0

Question 6.

Find the equations of the straight lines passing through the origin and making equal angles with the co-ordinate axes.

Solution:

Case (i) : PP1 makes an angle 45° with positive X-axis

m = tan 45° = 1

PP’ passes through 0(0, 0)

Equation of PP’ is y – 0 = 1 (x – 0)

Case ii) : QQ’ makes an angle 135° with positive X-axis

m = tan 135° = tan (180° – 45°) = -tan 45°

Equation of QQ’ is y – 0 = -1 (x – 0)

y = -x

Question 7.

The angle made by a straight line with the positive X-axis in the positive direction and the Y-intercept cut off by it are given below. Find the equation of the straight line.

i) 60°, 3

ii) 150°, 2

iii) 45°, -2

iv) Tan^{-1}(\(\frac{2}{3}\)), 3

Solution:

i) Equation of the line is y = mx + c

m = tan 60° = √3, c = 3

Equation of the line is y = √3x + 3

√3x – y + 3 = 0

ii) m = tan 150° = tan (180°-30°)

= – tan 30° = \(\frac{-1}{\sqrt{3}}\), c = 2

Equation of the line is y =- \(\frac{1}{\sqrt{3}}\)x + 2

√3 y = -x + 2 √3x

x + √3y – 2 – √3 =0

iii) m = tan 45° = 1

c = -2

Equation of the line is

y = x – 2

x – y – 2=0

iv) θ = tan^{-1}(\(\frac{2}{3}\)) ⇒ m = tan θ = \(\frac{2}{3}\),c = 3

Equatidn of the line is y = \(\frac{2}{3}\) x + 3

3y = 2x + 9

2x – 3y + 9 = 0

Question 8.

Find the equation of the straight line passing through (-4, 5) and cutting off equal and non-zero intercepts on the coordinate axes.

Solution:

Equation of the line in the intercept form is

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

Given a = b

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

⇒ x + y = a

This line passes through P(- 4, 5)

-4 + 5 = a ⇒ a = 1

Equation of the required line is x + y = 1 or x + y – 1 = 0

Question 9.

Find the equation of the straight line passing through (-2, 4) and making non¬zero intercepts whose sum is zero.

Solution:

Equation of the line in the intercept form is

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

Given a + b = 0 ⇒ b = -a

Equation of the line is \(\frac{x}{a}-\frac{y}{b}\)

⇒ x – y = a

This line passes through P(-2,4)

∴ -2 – 4 = a ⇒ a = -6

Equation of the required line is x – y = -6

⇒ x – y + 6 = 0

III.

Question 1.

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.

Solution:

Equation of the line in the intercept form is x y

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

Given \(\frac{a}{b}\) = \(\frac{2}{3}\) ⇒ b = \(\frac{3a}{2}\)

Equation of the line is \(\frac{x}{a}+\frac{2 y}{3 a}=1\)

⇒ 3x + 2y = 3a

This line passes through P(3, – 4)

9 – 8 = 3a ⇒ 3a = 1

Equation of the required line is 3x + 2y = 1

⇒ 3x + 2y – 1 = 0

Question 2.

Find the equation of the straight line passing through the point (4, -3) and perpendicular to the line passing through the points (1, 1) and (2, 3).

Solution:

A(1, 1), B(2, 3) are the given points.

m = Slope of AB = \(\frac{1-3}{1-2}=\frac{-2}{-1}=2\)

PQ is perpendicular to AB

Slope of PQ = –\(\frac{1}{m}\) = – \(\frac{1}{2}\)

PQ passes through P(4, -3)

Equation of PQ is y – y_{1} = m(x – x_{1})

y + 3 = –\(\frac{1}{2}\)(x – 4)

2y + 6 = -x + 4 ⇒ x + 2y + 2 = 0

Question 3.

Show that the following sets of points are collinear and find the equation of the line L containing them.

i) (-5, 1), (5, 5), (10, 7)

ii) (1, 3), (-2, – 6), (2, 6)

iii) (a, b + c), (b, c + a), (c, a + b)

Solution:

i) A(-5, 1), B(5, 5), C(10, 7) are the given points.

Equation of AB is

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

(y- 1) (-5 – 5) = (x + 5) (1 – 5)

– 10y + 10 = -4x – 20

4x – 10y + 30 = 0

or 2x – 5y + 15 = 0

C(10, 7)

2x – 5y + 15 = 2.10 – 5.7 + 15

= 20 – 35 + 15 = 0

A, B, C are collinear.

Equation of the line containing them is 2x – 5y + 15 = 0

ii) A(1, 3), B(-2, -6), C(2, 6)

Equation of AB is

(y – 3) (1 + 2) = (x – 1) (3 + 6)

3(y – 3) = 9(x – 1)

y – 3 = 3x – 3

3x – y = 0

C(2, 6)

3x – y = 3.2 – 6 = 6 – 6 = 0

∴ The given points A, B, C are collinear.

Equation of the line containing A,B,C is 3x – y = 0

iii) A(a, b f c), B(b, c + a), C(c, a + b)

Equation of AB is

(y – (b + c)) (a-b) = (x – a)(b + c – c – a)

(y – b – c) (a – b) = -(a – b) (x – a)

y – b – c = -x + a

or x + y – (a + b + c) = 0

C (c, a + b)

c + a + b – a – b – c = 0

C lies on AB

A, B, C are collinear.

Equation of the line containing them is x + y = a + b + c

Question 4.

A(10, 4), B(-4, 9) and C(-2, -1) are the vertices of a triangle. Find the equations of

i) \(\stackrel{\leftrightarrow}{A B}\)

ii) the median through A

iii) the altitude through B

iv) the perpendicular bisector the side of \(\stackrel{\leftrightarrow}{A B}\)

Solution:

i) A(10,4), B(-4, 9) are the given points.

Equation of AB is

(y – 4) (10 + 4) = (x – 10) (4 – 9)

14y – 56 = -5x + 50

5x + 14y – 106 = 0

ii) D is the mid-point of BC

A (10,4) is the other vertex Equation of AD is

(y – 4) (10 + 3) = (x + 3) (4 – 4)

13(y – 4) = 0 ⇒ y – 4 = 0 (or) y = 4

iii)

Slope of AC = \(\frac{4+1}{10+2}=\frac{5}{12}\)

BE is perpendicular to AC

Slope of BE = \(\frac{-1}{m}=\frac{-12}{5}\)

BE passes through B(-4, 9)

Equation of the altitude BE is

y – 9 = \(\frac{-12}{5}\)(x + 4)

5y – 45 = -12x – 48

12x + 5y + 3 = 0

iv) O is the mid-point of AB