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 – y1) (x1 – x2) = (x – x1) (y1 – y2)
(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 – y1) (x1 – x2) = (x – x1) (y1 – y2)
(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.
m1 = Slope of AB = $$\frac{y-7}{3-2}$$ = y – 7
m2 = Slope of PQ = $$\frac{4-6}{-1-0}=\frac{-2}{-1}=2$$ AB and PQ are parallel
m1 = m2 ⇒ 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 – y1 = m(x – x1)
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 – y1 = m(x – x1)
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 – y1) (x1 – x2) = (x – x1) (y1 – y2)
(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 