ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line

Exercise 12.1

1. Find the slope of a line whose inclination is
(i) 45°
(ii) 30°
Solution:
The slope of a line having inclination:
(i) We have given that:
45°
Slope = tan 45°  = 1
(ii) We have given that:
30°
Slope = tan 30°  = 1/√3
 
2. Find the inclination of a line whose gradient is
(i) 1
(ii) √3
(iii) 1/√3
Solution:
We have given that:
(i) tan θ = 1
⇒ θ = 45°
(ii) tan θ = √3
⇒ θ = 60° 
(i) tan θ = 1/√3
⇒ θ = 30°
 
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-
 
5. Find the equation of a straight line parallel to y-axis and passing through the point ( – 3,5).
Solution:
The equation of the line parallel to y-axis passing through ( – 3,5) to x = -3
⇒ x + 3 = 0 
 
6. Find the equation of a line whose
(i) slope = 3,y-intercept = – 5
(ii) slope = -2/7,y-intercept = 3
(iii) gradient = √3,y-intercept = -4/3
(iv) inclination = 30°,y-intercept = 2
Solution:
We have given that:
Equation of a line whose slope and y-intercept is given by:
y = mx + c, where m is the slope and c is the y-intercept
(i) Given: slope = 3, y-intercept = – 5
⇒ y = 3x + (-5) 
Hence, the equation of line is y = 3x – 5.
(ii) Given: slope = -2/7, y-intercept = 3
⇒ y = (-2/7)x + 3
y = (-2x + 21)/7
7y = -2x + 21
Hence, the equation of line is 2x + 7y – 21= 0.
(iii) Given: gradient = √3, y-intercept = -4/3
⇒ y = √3x  + (-4/3)
y = (3√3x  – 4)/3
3y = 3√3x  – 4
Hence, the equation of line is 3√3x  – 3y – 4 = 0.
 
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-1
 
7. Find the slope and y-intercept of the following lines:
(i) x – 2y – 1 = 0
(ii) 4x – 5y – 9 = 0
(iii) 3x + 5y + 7 = 0
(iv) x/3 + y/4 = 1
(v) y – 3 = 0
(vi) x – 3 = 0
Solution:
We have given that:
We know that, equation of line whose slope and y-intercept is given by:
y = mx + c, where m is the slope and c is the y-intercept
Using the above and converting to this, we find
(i) x – 2y – 1 = 0
2y = x – 1
⇒ y = (½) x + (-½)
Hence, slope = ½ and y-intercept = – ½
(ii) 4x – 5y – 9 = 0
5y = 4x – 9 
⇒ y = (4/5) x + (-9/5) 
Hence, slope = 4/5 and y-intercept = -9/5
(iii) 3x + 5y + 7 = 0
5y = -3x – 7 
⇒ y = (-3/5) x + (-7/5) 
Hence, slope = -3/5 and y-intercept = -7/5
(iv) x/3 + y/4 = 1
(4x + 3y)/ 12 = 1 
4x + 3y = 12 
3y = -4x + 12  
⇒ y = (-4/3) x + 4 
Hence, slope = -4/3 and y-intercept = 4
(v) y – 3 = 0
y = 3 
⇒ y = (0) x + 3 
Hence, slope = 0 and y-intercept = 3
(vi) x – 3 = 0
Here, the slope cannot be defined as the line does not meet y-axis.
 
8. The equation of the line PQ is 3y – 3x + 7 = 0
(i) Write down the slope of the line PQ.
(ii) Calculate the angle that the line PQ makes with the positive direction of x-axis.
Solution:
Given, equation of line PQ is 3y – 3x + 7 = 0
Re-writing in form of y = mx + c, we have
3y = 3x – 7
⇒ y = x + (-7/3)
Here,
(i) Slope = 1
(ii) As tan θ = 1
θ = 45° 
Hence, the angle which PQ makes with the x-axis is Q.
 
9. The given figure represents the line y = x + 1 and y = √3x  – 1. Write down the angles which the lines make with the positive direction of the x-axis. Hence determine θ.
 
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-2
Solution:
We have given that:
Given line equations, y = x + 1 and y = √3x  – 1
On comparing with y = mx + c,
The slope of the line: y = x + 1 is 1 as m = 1
So, tan θ = 1 ⇒ θ = 45°
And,
The slope of the line: y = √3x  – 1 is √3 as m = √3
So, tan θ = √3  ⇒ θ = 60°
Now, in triangle formed by the given two lines and x-axis
Ext. angle = Sum of interior opposite angle
60°  = θ + 45°
θ = 60°  – 45°
Thus, θ = 15°
 
10. Find the value of p, given that the line y/2 = x – p passes through the point (– 4,4)
Solution:
We have given that:
Given, equation of line: y/2 = x – p
And, it passes through the point (-4,4)
Hence, it satisfies the line equation
So,
4/2 = (-4) – p
2 = -4 – p
p = -4 – 2
Thus, p = -6
 
11. Given that (a,2a) lies on the line y/2 = 3x – 6. Find the value of a.
Solution:
We have given that: equation of line: y/2= 3x – 6
And, it passes through the point (a,2a)
Hence, it satisfies the line equation
So,
2a/2 = 3(a) – 6 
a = 3a – 6
2a = 6
Thus, a = 3
 
12. The graph of the equation y = mx + c passes through the points (1,4) and (– 2,– 5). Determine the values of m and c.
Solution:
We have given that: equation of the line is y = mx + c
And, it passes through the points (1,4)
So, the point will satisfy the line equation
⇒ 4 = m x 1 + c 
4 = m + c 
m + c = 4 … (i)
Also, the line passes through another point (-2,-5)
So,
5 = m (-2) + c 
5 = -2 m + c 
⇒ 2m – c = 5 …(ii) 
Now, on adding (i) and (ii) we get
3m = 9
⇒ m = 3
Substituting the value of m in (i), we get
3 + c = 4
⇒ c = 4 – 3 = 1
Therefore, m = 3,c = 1
Hence, the given values m = 3,c = 1 are not roots of the equation.
 
13. Find the equation of the line passing through the point (2,– 5) and making an intercept of – 3 on the y-axis.
Solution:
We have given that: a line equation passes through point (2,-5) and makes a y-intercept of -3
We know that,
The equation of line is y = mx + c, where m is the slope and c is the y-intercept
So, we have
y = mx – 3 
Now, this line equation will satisfy the point (2,-5)
-5 = m (2) – 3
-5 = 2m – 3
2m = 3 – 5 = -2
⇒ m = -1
Hence, the equation of the line is y = -x + (-3) ⇒ x + y + 3 = 0
 
14. Find the equation of a straight line passing through (– 1,2) and whose slope is 2/5.
Solution:
We have given that: the equation of straight line passes through (-1,2) and having slope as 2/5
So, the equation of the line will be
y – y1 = m (x – x1
Here, (x1,y1) is (-1,2)
⇒ y – 2 = (2/5) [x – (-1)] 
5 (y – 2) = 2 (x + 1) 
5y – 10 = 2x + 2 
Thus, the line equation is 2x – 5y + 12 = 0.
 
15. Find the equation of a straight line whose inclination is 60° and which passes through the point (0,– 3).
Solution:
We have given that:
Inclination of a straight line is 60°
So, the slope = tan 60°  = √3  = m
And, the equation of line passes through the point (0,-3) = (x1,y1)
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-3
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-4
(iii) It’s seen that the co-ordinates of point of intersection of EF and the x-axis will be y = 0
So, substituting the value y = 0 in the above equation
x – y + 4 = 0
x – 0 + 4 = 0
x = -4
Hence, the given values x = -4 are not roots of the equation.
Hence, the co-ordinates are (-4,0).
 
18. Find the intercepts made by the line 2x – 3y + 12 = 0 on the co-ordinate axis.
Solution:
We have given that:
Given line equation is 2x – 3y + 12 = 0
On putting y = 0, we will get the intercept made on x-axis
2x – 3y + 12 = 0
2x – 3 × 0 + 12 = 0,
2x – 0 + 2 = 0,
2x = -12
⇒ x = -6
Now, on putting x = 0, we get the intercepts made on y-axis
2x – 3y + 12 = 0
2 × 0 – 3y + 12 = 0
-3y = -12
⇒ y = 4
Hence, the x-intercept and y-intercept of the given line is -6 and 4 respectively.
 
19. Find the equation of the line passing through the points P (5,1) and Q (1,– 1). Hence, show that the points P,Q and R (11,4) are collinear.
Solution:
We have given that:
Given, two points P (5,1) and G (1,-1)
Slope of the line (m) = y2  – y1/ x2  – x1
= -1 – 1/ 1 – 5
= -2/-4 = ½
So, the equation of the line is
y – y1  = m (x – x1)
y – 1 = ½ (x – 5)
2y – 2 = x – 5
x – 2y – 3 = 0
Now, if point R (11,4) is collinear to points P and Q then, R (11,4) should satisfy the line equation
On substituting, we have 
11 – 2(4) – 3 = 11 – 8 – 3 = 0 
As point R satisfies the line equation,
Hence, P, Q and R are collinear.
 
20. Find the value of ‘a’ for which the following points A (a,3),B (2,1) and C (5,a) are collinear. Hence find the equation of the line.
Solution:
We have given that:
Points A (a, 3), B (2,1) and C (5,a) are collinear.
So, slope of AB = slope of BC
 
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-5 
 
a (a – 4) + (a – 4) = 0
(a + 1) (a – 4) = 0
a = -1 or 4
As a = -1 doesn’t satisfy the equation
⇒ a = 4 
Now,
Slope of BC = (a – 1)/(5 – 2) = (4 – 1)/3 = 3/3 = 1 = m
So, the equation of BC is
(y – 1) = 1 (x – 2)
y – 1 = x – 2
x – y = -1 + 2
Thus, the equation of BC is x – y = 1.
 
21. Use a graph paper for this question. The graph of a linear equation in x and y, passes through A (– 1,– 1) and B (2,5). From your graph, find the values of h and k, if the line passes through (h,4) and (½,k).
Solution:
We have given that:
Points (h,4) and (1/2,k) lie on the line passing through A (-1,-1) and B (2,5)
 
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-6
From the graph, its clearly seen that
h = 3/2 and
k = 2 
 
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-7
7 = x + 4 and 4 = y + 7
x = 7 – 4 and y = 4 – 7
x = 3 and y = -3
Thus, the co-ordinates of A are (3,-3).
(ii) Equation of diagonal BD is given by
y – 8 = (-4 – 8)/(2 – 5) × (x – 5)
y – 8 = (-12/-3) × (x – 5)
y – 8 = 4 (x – 5)
y – 8 = 4x – 20
4x – y – 20 + 8 = 0
Hence, the equation of the diagonal is 4x – y – 12 = 0.
 
23. In ∆ABC,A (3,5),B (7,8) and C (1,– 10). Find the equation of the median through A.
Solution:
 
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-8
We have given that:
∆ABC and their vertices A (3,5),B (7,8) and C (1,– 10).
And, AD is median
So, D is mid-point of BC
Hence, the co-ordinates of D is ([7 + 1]/2,[8 – 10]/2) = (4,-1)
Now,
Slope of AD,m = y – y1/ x2  – x1
m = (5 + 1)/ (3 – 4) = 6/-1 = -6 
Thus, the equation of AD is given by
y – y1  = m (x – x1)
y + 1 = -6 (x – 4)
y + 1 = -6x + 24
⇒ 6x + y – 23 = 0 
 
24. Find the equation of a line passing through the point (– 2,3) and having x-intercept 4 units.
Solution:
We have given that: point (−2,3) and the x-intercept of the line passing through that point is 4 units.
So, the co-ordinates of the point where the line meets the x-axis is (4,0)
Now, slope of the line passing through the points (−2,3) and (4,0)
 
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-9
 
25. Find the equation of the line whose x-intercept is 6 and y-intercept is – 4.
Solution:
We have given that: x-intercept of a line is 6
So,
The line will pass through the point (6,0)
Also given, the y -intercept of the line is -4 ⇒ c = -4
So, the line will pass through the point (0,-4)
Now,
Slope, m = m = y2  – y1/ x2  – x1
m = (-4 – 0)/(0 – 6) = -4/-6 = 2/3 
Thus, the equation of the line is given by
y = mx + c
y = (2/3)x + (-4)
3y = 2x – 12
⇒ 2x – 3y – 12 = 0
 
 
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-10
 
Thus, the equation will be
y – y1  = m (x – x1)
y – 2 = 3/2 (x – 0)
2y – 4 = 3x
⇒ 3x – 2y + 4 = 0
 
27. Find the equation of the line passing through the point (1,4) and intersecting the line x – 2y – 11 = 0 on the y-axis.
Solution:
We have given that: line x – 2y – 11 = 0 passes through y-axis and point (1,4)
So, putting x = 0 in the line equation we get the y-intercept
0 – 2y – 11 = 0
y = -11/2
The co-ordinates are (0,-11/2)
Now, the slope of the line joining the points (1,4) and (0,-11/2) is given by
m = y2  – y/ x2  – x1
= (-11/2 – 4)/ (0 – 1)
= 19/2 
 
 
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-11
Thus, x = y
Now, the slope of the line (m) = y2  – y1/ x2  – x1
= 0 – y/ x – 0
= -x/x = -1
Hence, the equation of the line will be
y – y = m (x – x1)
y – 2 = -1 (x – 3)
y – 2 = -x + 3
⇒ x + y – 5 = 0
 
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-12
 
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-13
 
30. A and B are two points on the x-axis and y-axis respectively. P (2,– 3) is the mid point of AB. Find the
(i) the co-ordinates of A and B.
(ii) the slope of the line AB.
(iii) the equation of the line AB.
Solution:
We have given that:
 
 
 
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-14
Given, points A and B are on x-axis and y-axis respectively
Let co-ordinates of A be (x,0) and of B be (0,y)
And P (2,-3) is the midpoint of AB
So, we have
2 = (x + 0)/2 and -3 = (0 + y)/2
x = 4 and y = -6
(i) Hence, the co-ordinates of A are (4,0) and of B are (0,-6).
(ii) Slope of AB = y2  – y1/ x2  – x1
= (-6 – 0)/ (0 – 4)
= -6/-4 = 3/2 = m
(iii) Equation of AB will be
y – y1  = m (x – x1
y – (-3) = 3/2 (x – 2) [As P lies on it]
y + 3 = 3/2 (x – 2)
2y + 6 = 3x – 6
3x – 2y – 12 = 0
 
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-17 
 
 
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-15
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-18
 
33. Point A (3,– 2) on reflection in the x-axis is mapped as A’ and point B on reflection in the y-axis is mapped onto B’ ( – 4,3).
(i) Write down the co-ordinates of A’ and B.
(ii) Find the slope of the line A’B, hence find its inclination.
Solution:
We have given that:,
A’ is the image of A (3,-2) on reflection in the x-axis.
(i) The co-ordinates of A’ will be (3,2).
Again B’ (- 4,3) in the image of A’, when reflected in the y-axis
Hence, the co-ordinates of B will be (4,3)
(ii) Slope of the line joining, the points A’ (3,2) and B (4,3) will be
m = y2  – y1/ x2  – x1
= (2 – 3)/ (3 – 4)
= -1/-1 = 1
So, tan θ = 45°
Thus, the angle of inclination is 45°.  

Exercise 12.2

 

1. State which one of the following is true: The straight lines y = 3x – 5 and 2y = 4x + 7 are

(i) parallel

(ii) perpendicular

(iii) neither parallel nor perpendicular.

Solution:

We have given that:

Given straight lines: y = 3x – 5 and 2y = 4x + 7 ⇒ y = 2x + 7/2

And, their slopes are 3 and 2

The product of slopes is 3 x 2 = 6.

Hence, as the slopes of both the lines are neither equal nor their product is -1 the given pair of straight lines are neither parallel nor perpendicular.

 

2. If 6x + 5y – 7 = 0 and 2px + 5y + 1 = 0 are parallel lines, find the value of p.

Solution:

We have given that:

For two lines to be parallel, their slopes must be same.

Given line equations,

6x + 5y – 7 = 0 and 2px + 5y + 1 = 0

In equation 6x + 5y – 7 = 0,

5y = -6x + 7 

y = (-6/5) x + 7/5 

So, the slope of the line (m1) = -6/5

Again, in equation 2px + 5y + 1 = 0

5y = -2px – 1 

y = (-2p/5) x – 1/5 

So, the slope of the line (m2) = -2p/5

For these two lines to be parallel

m1  = m2

-6/5 = -2p/5

p = (-6/5) x (-5/2)

Thus, p = 3

Hence, the given values p = 3 are not roots of the equation.

 

3. Lines 2x – by + 5 = 0 and ax + 3y = 2 are parallel. Find the relation connecting a and b.

Solution:

We have given that:

Given lines are: 2x – by + 5 = 0 and ax + 3y = 2

If two lines to be parallel then their slopes must be equal.

In equation 2x – by + 5 = 0,

by = 2x + 5

y = (2/b) x + 5/b 

So, the slope of the line (m1) = 2/b

And in equation ax + 3y = 2, 

3y = -ax + 2

y = (-a/3) x + 2/3

So, the slope of the line (m2) = (-a/3)

As the lines are parallel

m1  = m2

2/b = -a/3

6 = -ab

Hence, the relation connecting a and b is ab + 6 = 0

 

4. Given that the line y/2 = x – p and the line ax + 5 = 3y are parallel, find the value of a.

Solution:

We have given that:

Line equation: y/2 = x – p

⇒ y = 2x – 2p 

Here, the slope of the line is 2.

And, another line equation: ax + 5 = 3y

⇒ 3y = ax + 5 

y = (a/3) x + 5/3 

Hence, the slope of the line is a/3

As the line are parallel, their slopes must be equal

⇒ 2 = a/3

a = 6

Hence, the given values a = 6 are not roots of the equation.

Thus, the value of a is 6.

 

5. If the lines y = 3x + 7 and 2y + px = 3 perpendicular to each other, find the value of p.

Solution:

We have given that:

If two lines are perpendicular, then the product of their slopes is -1

Now, slope of the line y = 3x + 7 is m = 3

And,

The slope of the line: 2y + px = 3

2y = -px + 3 

y = (-p/2) x + 3 

m = -p/2 

As the lines as perpendicular,

⇒ m1× m2= -1

3 × (-p/2) = -1

p = 2/3

Hence, the given valuesp = 2/3 are not roots of the equation.

Thus, the value of p is 2/3.

6. If the straight lines kx – 5y + 4 = 0 and 4x – 2y + 5 = 0 are perpendicular to each other. Find the value of k.
Solution:
We have given that:
In equation, kx – 5y + 4 = 0
⇒ 5y = kx + 4 
y = (k/5) x + 4/5 
So, the slope (m1) = k/5
And, in equation 4x – 2y + 5 = 0
⇒ 2y = 4x + 5 
y = 2x + 5/2 
So, the slope (m2) = 2
As the lines are perpendicular to each other
⇒ m1× m2= -1
k/5 × 2 = -1
k = (-1 × 5)/2
Hence, the value of k = -5/2
 
7. If the lines 3x + by + 5 = 0 and ax – 5y + 7 = 0 are perpendicular to each other, find the relation connecting a and b.
Solution:
We have given that: that the lines 3x + by + 5 = 0 and ax – 5y + 7 = 0 are perpendicular to each other
Then the product of their slopes must be -1.
Slope of line 3x + by + 5 = 0 is,
by = -3x – 5
y = (-3/b) – 5/b 
So, slope (m1) = -3/b
And,
The slope of line ax – 5y + 7 = 0 is
5y = ax + 7 
y = (a/5) x + 7/5 
So, slope (m2) = a/5
As the lines are perpendicular, we have
m1  × m2  = -1
-3/b x a/5 = -1
-3a/5b = -1
-3a = – 5b
3a = 5b
Hence, the relation connecting a and b is 3a = 5b.
 
8. Is the line through (– 2,3) and (4,1) perpendicular to the line 3x = y + 1?
Does the line 3x = y + 1 bisect the join of (– 2,3) and (4,1).
Solution:
We have given that:
Slope of the line passing through the points (-2,3) and (4,1) is given by
m1  = y2  – y1/ x2– x1
= (1 – 3)/ (4 + 2)
= -2/6
= -1/3
And, the slope of the line: 3x = y + 1
y = 3x -1 
Slope (m2) = 3
Now,
m1  × m2  = -1/3 × 3 = -1 
Thus, the lines are perpendicular to each other as the product of their slopes is -1.
Now,
Co-ordinates of the mid-point of the line joining the points (-2,3) and (4,1) is
([-2 + 4]/2,[3 + 1]/2) = (1,2) 
Now, if the line 3x = y + 1 passes through the mid-point then it will satisfy the equation
3(1) = (2) + 1
3 = 3
Hence, the line 3x = y + 1 bisects the line joining the points (– 2,3) and (4,1).
 
9. The line through A (– 2,3) and B (4,b) is perpendicular to the line 2x – 4y = 5. Find the value of b.
Solution:
We have given that:
The slope of the line passing through A (-2,3) and B (4,b) will be
m1  = (b – 3)/ (4 + 2) = (b – 3)/ 6 
Now, the gradient of the given line 2x – 4y = 5 is
4y = 2x + 5
y = (2/4) x + 5/4
y = ½ x + 5/4
So, m2  = ½
As the line are perpendicular to each other, we have
m1× m2= -1
(b – 3)/ 6 × ½ = -1
(b – 3)/ 12 = -1 
b – 3 = -12
b = -12 + 3 = -9
Hence, the value of b is -9.
 
10. If the lines 3x + y = 4,x – ay + 7 = 0 and bx + 2y + 5 = 0 form three consecutive sides of a rectangle, find the value of a and b.
Solution:
Given lines are:
3x + y = 4 … (i)
x – ay + 7 = 0 … (ii)
bx + 2y + 5 = 0 … (iii)
It’s said that these lines form three consecutive sides of a rectangle.
So,
Lines (i) and (ii) must be perpendicular
Also, lines (ii) and (iii) must be perpendicular
We know that, for two perpendicular lines the product of their slopes will be -1.
Now,
Slope of line (i) is
3x + y = 4 ⇒ y = -3x = 4 
Hence, slope (m1) = -3
And, slope of line (ii) is
x – ay + 7 = 0 ⇒ ay = x + 7
y = (1/a) x + 7/a
Hence, slope (m2) = 1/a
Finally, the slope of line (iii) is
bx + 2y + 5 = 0 ⇒ 2y = -bx – 5
y = (-b/2) x – 5/2
Hence, slope (m3) = -b/2
As lines (i), (ii) and (iii) are consecutive sides of rectangle, we have
m1  × m2 = -1 and m2× m3= -1 
(-3) x (1/a) = -1 and (1/a) x (-b/2) = -1
-3 = -a and -b/2a = -1 
a = 3 and b = 2a ⇒ b = 2(3) = 6 
Thus, the value of a is 3 and the value of b is 6.
 
11. Find the equation of a line, which has the y-intercept 4, and is parallel to the line 2x – 3y – 7 = 0. Find the coordinates of the point where it cuts the x-axis.
Solution:
We have given that:
Given line: 2x – 3y – 7 = 0 
Its slope is,
3y = 2x – 7
y = (2/3) x – 7/3
⇒ m = 2/3
So, the equation of the line parallel to the given line will be 2/3
Also given, the y-intercept is 4 = c
Hence, the equation of the line is given by
y = mx + c
y = (2/3) × + 4
3y = 2x + 12
2x – 3y + 12 = 0
Now, when this line intersects the x-axis the y co-ordinate becomes zero.
So, putting y = 0 in the line equation, we get
2x – 3(0) + 12 = 0
2x + 12 = 0
x = -12/2 = 6
Hence, the given values x=6 are not roots of the equation.
Hence, the co-ordinates of the point where it cuts the x-axis is (-6,0).
 
12. Find the equation of a straight line perpendicular to the line 2x + 5y + 7 = 0 and with y-intercept – 3 units.
Solution:
Given line: 2x + 5y + 7 = 0
So, its slope is given by
5y = -2x – 7
y = (-2/5) – 7/5
⇒ m = -2/5
Now, let the slope of the line perpendicular to this line be m’
Then,
m × m’ = -1
(-2/5) × m’ = -1
⇒ m’ = 5/2
Also given, the y-intercept (c) = -3
Hence, the equation of the line is given by
y = m’x + c 
y = (5/2) × + (-3)
2y = 5x – 6
5x – 2y – 6 = 0
 
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-19
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-20
y = mx + c
y = (3/2) x + 3
2y = 3x + 6
3x – 2y + 6 = 0
 
15. Find the equation of the line passing through (0,4) and parallel to the line 3x + 5y + 15 = 0.
Solution:
We have given that:
Given line: 3x + 5y + 15 = 0
5y = -3x – 15 
y = (-3/5) x – 3 
So, slope (m) = -3/5 
The slope of the line parallel to the given line will the same -3/5
And, the line passes through the point (0,4)
Hence, equation of the line will be
y – y = m (x – x1)
y – 4 = (-3/5) (x – 0)
5y – 20 = -3x
3x + 5y – 20 = 0
 
16. The equation of a line is y = 3x – 5. Write down the slope of this line and the intercept made by it on the y-axis. Hence or otherwise, write down the equation of a line which is parallel to the line and which passes through the point (0,5).
Solution:
We have given that:
Given line: y = 3x – 5
Here slope (m1) = 3
Substituting x = 0, we get y = – 5
Hence, the y-intercept = – 5
Now, the slope of the line parallel to the given line will be 3 and it passes through the point (0,5).
Thus, equation of the line will be
y – y1  = m (x – x1)
y – 5 = 3 (x – 0)
y = 3x + 5
 
17. Write down the equation of the line perpendicular to 3x + 8y = 12 and passing through the point (– 1,– 2).
Solution:
Given line: 3x + 8y = 12
8y = -3x + 12 
y = (-3/8) x + 12 
So, the slope (m1) = -3/8
Let’s consider the slope of the line perpendicular to the given line as m2
Then, m1  × m2  = -1
-3/8 × m2= -1 
m2  = 8/3 
 
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-21
x = -3
Hence, the line meets the x-axis at A (-3,0).
(ii) Now, the slope of the line is given by
4x – 3y + 12 = 0
3y = 4x + 12
y = (4/3) x + 4
⇒ m1  = 4/3
Let’s assume the slope of the line perpendicular to the given line be m2
Then, m1× m2= -1
4/3 × m2= -1 
m2= -3/4 
Thus, the equation of the line perpendicular to the given line passing through A will be
y – 0 = -3/4 (x + 3)
4y = -3(x + 3)
3x + 4y + 9 = 0
 
19. Find the equation of the line that is parallel to 2x + 5y – 7 = 0 and passes through the mid-point of the line segment joining the points (2,7) and (– 4,1).
Solution:
We have given that:
Given line: 2x + 5y – 7 = 0 
5y = -2x + 7 
y = (-2/5) x + 7/5 
So, the slope is -2/5
Hence, the slope of the line that is parallel to the given line will be the same, m = -2/5
Now, the mid-point of the line segment joining points (2,7) and (– 4,1) is
((2 – 4)/2,(7 + 1)/2) = (-1,4) 
Thus, the equation of the line will be
y –y1  = m (x – x1)
y – 4 = (-2/5) (x + 1)
5y – 20 = -2x -2
2x + 5y = 18
 
20. Find the equation of the line that is perpendicular to 3x + 2y – 8 = 0 and passes through the mid-point of the line segment joining the points (5,– 2) and (2,2).
Solution:
We have given that:
Given line: 3x + 2y – 8 = 0
2y = -3x + 8 
y = (-3/2) x + 4 
Here, slope (m1) = -3/2
Now, the co-ordinates of the mid-point of the line segment joining the points (5,-2) and (2,2) will be
((5 + 2)/7,(-2 + 2)/7) = (7/2,0) 
Let’s consider the slope of the line perpendicular to the given line be m2
Then,
m1× m2= -1
(-3/2) × m2= -1
m2= 2/3 
So, the equation of the line with slope m2 and passing through (7/2,0) will be
y – 0 = (2/3) (x – 7/2)
3y = 2x – 7
2x – 3y – 7 = 0
Thus, the required line equation is 2x – 3y – 7 = 0.
 
21. Find the equation of a straight line passing through the intersection of 2x + 5y – 4 = 0 with x-axis and parallel to the line 3x – 7y + 8 = 0.
Solution:
We have given that:
Let’s assume the point of intersection of the line 2x + 5y – 4 = 0 and x-axis be (x,0)
Now, substituting the value y = 0 in the line equation, we have
2x + 5(0) – 4 = 0 
2x – 4 = 0
x = 4/2 = 2
Hence, the co-ordinates of the point of intersection is (2,0)
Also given, line equation: 3x – 7y + 8 = 0
7y = 3x + 8 
y = (3/7) x + 8/7 
So, the slope (m) = 3/7
We know that the slope of any line parallel to the given line will be the same.
So, the equation of the line having slope 3/7 and passing through the point (2,0) will be
y – 0 = (3/7) (x – 2)
7y = 3x – 6
3x – 7y – 6 = 0
Thus, the required line equation is 3x – 7y – 6 = 0.
 
22. The equation of a line is 3x + 4y – 7 = 0. Find (i) the slope of the line. (ii) the equation of a line perpendicular to the given line and passing through the intersection of the lines x – y + 2 = 0 and 3x + y – 10 = 0.
Solution:
We have given that:
Given line equation: 3x + 4y – 7 = 0
(i) Slope of the line is given by,
4y = -3x + 7
y = (-3/4) x + 7
Hence, slope (m1) = -3/4
(ii) Let the slope of the perpendicular to the given line be m2
Then, m1× m2  = -1
(-3/4) × m= -1 
m= 4/3 
Now, to find the point of intersection of
x – y + 2 = 0 … (i)
3x + y – 10 = 0 … (ii)
On adding (i) and (ii), we get
4x – 8 = 0
4x = 8
x = 8/4 = 2
Putting x = 2 in (i), we get
2 – y + 2 = 0
y = 4
Hence, the point of intersection of the lines is (2,4) 
The equation of the line having slope m2 and passing through (2,4) will be
y – 4 = (4/3) (x – 2)
3y – 12 = 4x – 8
4x – 3y + 4 = 0
Thus, the required line equation is 4x – 3y + 4 = 0.
 
23. Find the equation of the line perpendicular from the point (1,– 2) on the line 4x – 3y – 5 = 0. Also find the co-ordinates of the foot of perpendicular.
Solution:
We have given that:
Given line equation: 4x – 3y – 5 = 0
3y = 4x – 5 
y = (4/3) x – 5 
Slope of the line (m1) = 4/3
Let the slope of the line perpendicular to the given line be m2
Then, m1× m2= -1
(4/3) × m= -1 
m= -3/4 
Now, the equation of the line having slope m2 and passing through the point (1,-2) will be
y + 2 = (-3/4) (x – 1)
4y + 8 = -3x + 3
3x + 4y + 5 = 0
Next, for finding the co-ordinates of the foot of the perpendicular which is the point of intersection of the lines
4x – 3y – 5 = 0 ….(1) and
3x + 4y + 5 = 0 …. (2)
On multiplying (1) by 4 and (2) by 3, we get
16x – 12y – 20 = 0
9x + 12y + 15 = 0
Adding we get,
25x – 5 = 0
x = 5/25
x = 1/5
Putting the value of x in (1), we have
4(1/5) – 3y – 5 = 0
4/5 – 3y – 5 = 0
3y = 4/5 – 5 = (4 – 25)/5
3y = -21/5
y = -7/5
Thus, the co-ordinates are (1/5,-7/5)
 
 
24. Prove that the line through (0,0) and (2,3) is parallel to the line through (2,– 2) and (6,4).
Solution:
We have given that:
Let the slope of the line through (0,0) and (2,3) be m1
So, m1= (y2  – y1)/ (x2  – x1)
= (3 – 0)/ (2 – 0)
= 3/2
And, let the slope of the line through (2,-2) and (6,4) be m2
So,   m= (y2  – y1)/ (x2  – x1)
= (4 + 2)/ (6 – 2) 
= 6/4 = 3/2 
It’s clearly seen that the slopes m1= m2
Thus, the lines are parallel to each other.
 
25. Prove that the line through (– 2,6) and (4,8) is perpendicular to the line through (8,12) and (4,24).
Solution:
We have given that:
Let the slope of the line through points (– 2,6) and (4,8) be m1
So, m1= (y2  – y1)/ (x2  – x1)
= (8 – 6)/ (4 + 2) 
= 2/6 
= 1/3 
And, let the slope of the line through (8,12) and (4,24) be m2
So,   m= (y2  – y1)/ (x2  – x1)
= (24 – 12)/ (4 – 8)
= 12/ (-4)
= -3
Now, product of slopes is
m1  ×  m2  = 1/3 x (-3) = -1 
Thus, the lines are perpendicular to each other.
 
26. Show that the triangle formed by the points A (1,3),B (3,– 1) and C (– 5,– 5) is a right-angled triangle by using slopes.
Solution:
We have given that:
Given, points A (1,3),B (3,– 1) and C (– 5,– 5) form a triangle
Now,
Slope of the line AB = m1  = (-1 – 3)/ (3 – 1) = -4/2 = -2
And,
Slope of the line BC = m2  = (-5 + 1)/ (-5 – 3) = -4/-8 = ½
Hence,
m1× m2= (-2) x (1/2) = -1 
So, the lines AB and BC are perpendicular to each other.
Therefore, ∆ABC is a right-angled triangle.
 
27. Find the equation of the line through the point (– 1,3) and parallel to the line joining the points (0,– 2) and (4,5).
Solution:
We have given that:
Slope of the line joining the points (0,-2) and (4,5) is
m = (5 + 2)/ (4 – 0)
= 7/4
Now, the slope of the line parallel to it and passing through (-1,3) will be also be 7/4
Hence, the equation of the line is
y – y1  = m (x – x1) ⇒ y – 3 = 7/4 (x + 1)
4y – 12 = 7x + 7
7x – 4y + 19 = 0
 
28. Are the vertices of a triangle. 
(i) Find the coordinates of the centroid G of the triangle. 
(ii) Find the equation of the line through G and parallel to AC.
Solution:
Given, A (– 1,3),B (4,2),C (3,– 2) 
(i) Co-ordinates of centroid G is
G (x,y) = ((x1  + x + x3)/2,(y1+ y2  + y3)/2)
= ((-1 + 4 + 3)/3,(3 + 2 – 2)/3)
= (6/3,3/3) = (2,1)
Hence, the co-ordinates of the centroid G of the triangle is (2,1)
 
(ii) Slope of AC = (y2– y1)/ (x2– x1) = (-2 – 3)/ (3 – (-1)) = -5/4
So, the slope of the line parallel to AC is also -5/4
Now, the equation of line through G is
y – 1 = (-5/4) (x – 2)
4y – 4 = -5x + 10
5x + 4y = 14
Thus, the required line equation is 5x + 4y = 14.
 
29. The line through P (5,3) intersects y-axis at Q. (i) Write the slope of the line. (ii) Write the equation of the line. (iii) Find the coordinates of Q.
 
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-22
Solution:
(i) Here, θ = 45°
So, the slope of the line = tan θ = tan 45°  = 1
(ii) Equation of the line through P and Q is
y – 3 = 1(x – 5)
x – y – 2 = 0
(iii) Let the co-ordinates of Q be (0,y)
Then, m = y2  – y1/ x2  – x1
1 = (3 – y)/ (5 – 0) 
5 = 3 – y 
y = 3 – 5 = -2 
Thus, co-ordinates of Q are (0,-2).
Hence, the given values y=-2 are not roots of the equation.
 
30. In the adjoining diagram, write down (i) the co-ordinates of the points A,B and C. (ii) the equation of the line through A parallel to BC.

ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line

Solution:

ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-
From the given figure, its clearly seen that
Co-ordinates of A are (2,3) and of B are (-1,2) and of C are (3,0).
Now,
Slope of BC = (0 – 2)/ (3 – (-1))
= -2/4 
= -1/2 
So, the slope of the line parallel to BC is also -1/2
And, the line passes through A (2,3)
Hence, the equation will be
y – 3 = (-1/2) (x – 2)
2y – 6 = -x + 2
x + 2y = 8
 
31. Find the equation of the line through (0,– 3) and perpendicular to the line joining the points (– 3,2) and (9,1).
Solution:
We have given that:
The slope of the line joining the points (-3,2) and (9,1) is
m2  = (1 – 2)/ (9 + 3) = -1/12 
Now, let the slope of the line perpendicular to the above line be m2
Then, m1× m2= -1
(-1/12) × m= -1 
m= 12 
So, the equation of the line passing through (0,-3) and having slope of mwill be
y – (-3) = 12 (x – 0)
y + 3 = 12x
12x – y = 3
Thus, the required line equation is 12x – y = 3.
 
32. The vertices of a triangle are A (10,4),B (4,– 9) and C (– 2,– 1). Find the equation of the altitude through A. The perpendicular drawn from a vertex of a triangle to the opposite side is called altitude.
Solution:
We have given that:
Given, vertices of a triangle are A (10,4),B (4,– 9) and C (– 2,– 1)
Now,
Slope of line BC (m_1) = (-1 + 9)/ (-2 – 4) = 8/ (-6) = -4/3
Let the slope of the altitude from A (10,4) to BC be m2
Then, m× m= -1
(-4/3) × m= -1 
m= ¾ 
So, the equation of the line will be
y – 4 = ¾ (x – 10)
4y – 16 = 3x – 30
3x – 4y – 14 = 0
 
33. A (2, – 4), B (3, 3) and C (– 1, 5) are the vertices of triangle ABC. Find the equation of: 
(i) the median of the triangle through A 
(ii) the altitude of the triangle through B.
Solution:

ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-1
Given, A (2,– 4),B (3,3) and C (– 1,5) are the vertices of triangle ABC
(i) D is the mid-point of BC
So, the co-ordinates of D will be
((3 – 1)/2,(3 + 5)/2) = (2/2,8/2) = (1,4) 
Now,
The slope of AC (m1) = (5 + 4)/ (-1 – 2) = 9/-3 = -3
Let the slope of BE be  m2
Then,  m× m2= -1
-3 ×  m2= -1 
 m= 1/3 
so, the equation of BE will be
y – 3 = 1/3 (x – 3)
3y – 9 = x – 3
x – 3y + 6 = 0
Thus, the required line equation is x – 3y + 6 = 0.
 
 
34. Find the equation of the right bisector of the line segment joining the points (1,2) and (5,– 6).
Solution:
We have given that:
The slope of the line joining the points (1,2) and (5,-6) is
m_1  = (-6 – 2)/ (5 – 1) = -8/4 = -2 
Now, if  mis the slope of the right bisector of the above line
Then,
 m× m2= -1
-2 ×  m= -1
 m= ½
The mid-point of the line segment joining (1,2) and (5,-6) will be
((1 + 5)/2,(2 – 6)/2) = (6/2,-4/2) = (3,-2) 
So, equation of the line is
y + 2 = ½ (x – 3)
2y + 4 = x – 3
x – 2y – 7 = 0
Thus, the equation of the required right bisector is x – 2y – 7 = 0.
 
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-2
y – 3 = ½ (x – 4)
2y – 6 = x – 4
x – 2y + 2 = 0
(iii) As point (-2,p) lies on the above line
The point will satisfy the line equation
-2 – 2p + 2 = 0
-2p = 0
p = 0
Hence, the given values p = 0 are not roots of the equation.
Thus, the value of p is 0.
 
36. The points B (1,3) and D (6,8) are two opposite vertices of a square ABCD. Find the equation of the diagonal AC.
Solution:
We have given that: points B (1,3) and D (6,8) are two opposite vertices of a square ABCD
Slope of BD is given by
m1  = (8 – 3)/ (6 – 1) = 5/5 = 1 
We know that, the diagonal AC is a perpendicular bisector of diagonal BD
So, the slope of AC (m2) will be
m× m= -1
1 × m= -1
m2 = -1
And, the co-ordinates of mid-point of BD and AC will be
((1 + 6)/2 ,(3 + 8)/2) = (7/2,11/2) 
So, the equation of AC is
y – 11/2 = -1 (x – 7/2)
2y – 11 = -2x – 7
2x + 2y – 7 – 11 = 0 ⇒ 2x + 2y – 18 = 0
Thus, the equation of diagonal AC is x + y – 9 = 0.
 
37. ABCD is a rhombus. The co-ordinates of A and C are (3,6) and ( – 1,2) respectively. Write down the equation of BD.
Solution:
We have given that: ABCD is a rhombus and co-ordinates of A are (3,6) and of C are (-1,2)
Slope of AC (m1) = (2 – 6)/ (-1 – 3) = -4/-4 = 1
We know that, the diagonals of a rhombus bisect each other at right angles.
So, the diagonal BD is perpendicular to diagonal AC
Let the slope of BD be m2
Then, m1× m= -1
m= -1/(m1)
= -1/ (1) = -1
Now, the co-ordinates of the mid-point of AC is given by
((3 – 1)/2,(6 + 2)/2) = (2/2,8/2) = (1,4) 
So, the equation of BD will be
y – 4 = -1 (x – 1)
y – 4 = -x + 1
x + y = 5
Thus, the equation of BD is x + y = 5.
 
38. Find the equation of the line passing through the intersection of the lines 4x + 3y = 1 and 5x + 4y = 2 and
(i) parallel to the line x + 2y – 5 = 0 
(ii) perpendicular to the x-axis.
Solution:
We have given that: line equations:
4x + 3y = 1 … (1)
5x + 4y = 2 … (2)
On solving the above equation to find the point of intersection, we have
Multiplying (1) by 4 and (2) by 3
16 x + 12y = 4
15x + 12y = 6
On subtracting, we get
x = -2 
Putting the value of x in (1), we have
4(-2) + 3y = 1
-8 + 3y = 1
3y = 1 + 8 = 9
y = 9/3 = 3
Hence, the point of intersection is (-2,3).
(i) Given line, x + 2y – 5 = 0
2y = -x + 5
y = -(1/2) x + 5/2
Slope (m) = -1/2
A line parallel to this line will have the same slope m = -1/2
So, the equation of line having slope m and passing through (-2,3) will be
y – 3 = (-1/2) (x + 2)
2y – 6 = -x – 2
x + 2y = 4
(ii) As any line perpendicular to x-axis will be parallel to y-axis.
So, the equation of line will be
x = -2 ⇒ x + 2 = 0
 
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-3
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-4
 
40. Find the image of the point (1,2) in the line x – 2y – 7 = 0.
Solution:
Given line equation: x – 2y – 7 = 0 … (i)
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-5
 
Draw a perpendicular from point P (1,2) on the line
Let P’ be the image of P and let its co-ordinates be (x,y)
The slope of the given line is given as,
2y = x – 7
y = (1/2) x – 7
Slope (m1) = ½
Let the slope of line segment PP’ be m2
As PP’ is perpendicular to the given line, product of slopes: m1× m= -1
So, ½ × m= -1
m= -2
So, the equation of the line perpendicular to the given line and passing through P (1,2) is
y – 2 = (-2) (x – 1) 
y – 2 = -2x + 2 
2x + y – 4 = 0 … (ii)
Let the intersection point of lines (i) and (ii) be taken as M.
Solving both the line equations, we have
Multiplying (ii) by 2 and adding with (i)
x – 2y – 7 = 0
4x + 2y – 8 = 0
———————
5x – 15 = 0
x = 15/5 = 3
Putting value of x in (i), we get
3 – 2y – 7 = 0
2y = -4
y = -4/2 = -2
So, the co-ordinates of M are (3,-2)
Hence, its seen that M should be the mid-point of the line segment PP’
(3,-2) = ((x + 1)/2,(y + 2)/2)
(x + 1)/2 = 3
x + 1 = 6
x = 6 – 1 = 5
And,
(y + 2)/2 = -2
y + 2 = -4
y = -4 – 2 = -6
Therefore, the co-ordinates of P’ are (5,-6).
 
 
41. If the line x – 4y – 6 = 0 is the perpendicular bisector of the line segment PQ and the co-ordinates of P are (1,3), find the co-ordinates of Q.
Solution:
We have given that: line equation: x – 4y – 6 = 0 … (i)
Co-ordinates of P are (1,3)
Let the co-ordinates of Q be (x ,y)
Now, the slope of the given line is
4y = x – 6
y = (1/4) x – 6/4
slope (m) = ¼
So, the slope of PQ will be (-1/m) [As the product of slopes of perpendicular lines is -1]
Slope of PQ = -1/ (1/4) = -4
Now, the equation of line PQ will be
y – 3 = (-4) (x – 1) 
y – 3 = -4x + 4 
4x + y = 7 … (ii)
On solving equations (i) and (ii), we get the coordinates of M
 
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-6
 
Multiplying (ii) by 4 and adding with (i), we get
x – 4y – 6 = 0
16x + 4y = 28
——————
17x = 34
x = 34/17 = 2
Putting the value of x in (i)
2 – 4y – 6 = 0
-4 – 4y = 0
4y = -4
y = -1
So, the co-ordinates of M are (2,-1)
But, M is the mid-point of line segment PQ
(2,-1) = (x + 1)/2 ,(y + 3)/2
(x + 1)/2 = 2
x + 1 = 4
x = 3
And,
(y + 3)/2 = -1
y + 3 = -2
y = -5
Thus, the co-ordinates of Q are (3,-5).
 
42. OABC is a square, O is the origin and the points A and B are (3,0) and (p,q). If OABC lies in the first quadrant, find the values of p and q. Also write down the equations of AB and BC.
Solution:
We have given that: OABC is a square
Co-ordinates of A and B are (3,0) and (p,q) respectively
By distance formula, we have
 
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-7
 
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-8
 
Substituting the value of p in (i) 
(3)2+q2-6(3)=0⇒9+q2-18=0
q2-9=0⇒q2-9⇒q-3
∴p=3,q=3
As AB parallel to y-axis
∴ Equation AB will be x=3
⇒x-3=0
And equation AB will be y=3
⇒y-3=0 (∵BC∥x-axis)
 
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line-9
 
 
Chapter Test
1. Find the equation of a line whose inclination is 60° and y-intercept is – 4.
Solution:
We have given that: inclination = 60° and y-intercept (c) = -4
So, slope (m) = tan 60°  = √3
Hence, the equation of the line is given by
y = mx + c
y = √3x – 4
 
2. Write down the gradient and the intercept on the y-axis of the line 3y + 2x = 12.
Solution:
We have given that:
Given line equation: 3y + 2x = 12
3y = -2x + 12
y = (-2/3) x + 12/3
y = (-2/3) x + 4
Hence, gradient = -2/3 and the intercept on the y-axis is 4.
 
3. If the equation of a line is y – √3x  + 1, find its inclination.
Solution:
We have given that:
Given line equation: y – √3x  + 1
y = √3x  – 1 
Here, slope = √3
⇒ tan θ = √3
θ = 60°
Hence, the inclination of the line is 60°.
 
ML Aggarwal Solutions Class 10 Maths Chapter 1 Goods and Service Tax (GST)
ML Aggarwal Solutions Class 10 Maths Chapter 2 Banking
ML Aggarwal Solutions Class 10 Maths Chapter 3 Shares and Dividends
ML Aggarwal Solutions Class 10 Maths Chapter 4 Linear Inequations
ML Aggarwal Solutions Class 10 Maths Chapter 5 Quadratic Equations in One Variable
ML Aggarwal Solutions Class 10 Maths Chapter 6 Factorization
ML Aggarwal Solutions Class 10 Maths Chapter 7 Ratio and Proportion
ML Aggarwal Solutions Class 10 Maths Chapter 7 Ratio and Proportion
ML Aggarwal Solutions Class 10 Maths Chapter 8 Matrices
ML Aggarwal Solutions Class 10 Maths Chapter 9 Arithmetic and Geometric Progression
ML Aggarwal Solutions Class 10 Maths Chapter 10 Reflection
ML Aggarwal Solutions Class 10 Maths Chapter 11 Section Formula
ML Aggarwal Solutions Class 10 Maths Chapter 12 Equation of Straight Line
ML Aggarwal Solutions Class 10 Maths Chapter 13 Similarity
ML Aggarwal Solutions Class 10 Maths Chapter 14 Locus
ML Aggarwal Solutions Class 10 Maths Chapter 15 Circles
ML Aggarwal Solutions Class 10 Maths Chapter 16 Constructions
ML Aggarwal Solutions Class 10 Maths Chapter 17 Mensuration
ML Aggarwal Solutions Class 10 Maths Chapter 18 Trigonometric Identities
ML Aggarwal Solutions Class 10 Maths Chapter 19 Trigonometric Tables
ML Aggarwal Solutions Class 10 Maths Chapter 20 Heights and Distances
ML Aggarwal Solutions Class 10 Maths Chapter 21 Measures Of Central Tendency
ML Aggarwal Solutions Class 10 Maths Chapter 22 Probability
NCERT Exemplar Solutions Class 10 Maths Areas related to Circles
NCERT Exemplar Solutions Class 10 Maths Arithmetic Progression
NCERT Exemplar Solutions Class 10 Maths Circles
NCERT Exemplar Solutions Class 10 Maths Construction
NCERT Exemplar Solutions Class 10 Maths Coordinate Geometry
NCERT Exemplar Solutions Class 10 Maths Linear Equations
NCERT Exemplar Solutions Class 10 Maths Polynomials
NCERT Exemplar Solutions Class 10 Maths Quadratic Equation
NCERT Exemplar Solutions Class 10 Maths Real Numbers
NCERT Exemplar Solutions Class 10 Maths Surface Area and Volume
NCERT Exemplar Solutions Class 10 Maths Triangles
NCERT Exemplar Solutions Class 10 Maths Trigonometry