RD Chapter 23 The Straight Lines Ex 23.13 Solutions
Question - 1 : - Find the angles between each of the following pairs of straight lines:
(i) 3x + y + 12 = 0 and x + 2y – 1 = 0
(ii) 3x – y + 5 = 0 and x – 3y + 1 = 0
Answer - 1 : -
(i) 3x + y + 12 = 0 and x+ 2y – 1 = 0
Given:
The equations of thelines are
3x + y + 12 = 0 … (1)
x + 2y − 1 =0 … (2)
Let m1 andm2 be the slopes of these lines.
m1 =-3, m2 = -1/2
Let θ be theangle between the lines.
Then, by using theformula
tan θ = [(m1 –m2) / (1 + m1m2)]
= [(-3 + 1/2) / (1 +3/2)]
= 1
So,
θ = π/4 or 45o
∴ The acute anglebetween the lines is 45°
(ii) 3x – y + 5 = 0 and x –3y + 1 = 0
Given:
The equations of thelines are
3x − y + 5 =0 … (1)
x − 3y + 1 =0 … (2)
Let m1 andm2 be the slopes of these lines.
m1 =3, m2 = 1/3
Let θ be theangle between the lines.
Then, by using theformula
tan θ = [(m1 –m2) / (1 + m1m2)]
= [(3 + 1/3) / (1 +1)]
= 4/3
So,
θ = tan-1 (4/3)
∴ The acute anglebetween the lines is tan-1 (4/3).
Question - 2 : - Find the acute angle between the lines 2x – y + 3 = 0 and x + y + 2 = 0.
Answer - 2 : -
Given:
The equations of thelines are
2x − y + 3 =0 … (1)
x + y + 2 = 0 … (2)
Let m1 andm2 be the slopes of these lines.
m1 =2, m2 = -1
Let θ be theangle between the lines.
Then, by using theformula
tan θ = [(m1 –m2) / (1 + m1m2)]
= [(2 + 1) / (1 + 2)]
= 3
So,
θ = tan-1 (3)
∴ The acute anglebetween the lines is tan-1 (3).
Question - 3 : - Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.
Answer - 3 : -
To prove:
The points (2, -1),(0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram
Let us assume thepoints, A (2, − 1), B (0, 2), C (2, 3) and D (4, 0) be the vertices.
Now, let us find theslopes
Slope of AB = [(2+1) /(0-2)]
= -3/2
Slope of BC = [(3-2) /(2-0)]
= ½
Slope of CD = [(0-3) /(4-2)]
= -3/2
Slope of DA = [(-1-0)/ (2-4)]
= ½
Thus, AB is parallelto CD and BC is parallel to DA.
Hence proved, thegiven points are the vertices of a parallelogram.
Now, let us find theangle between the diagonals AC and BD.
Let m1 andm2 be the slopes of AC and BD, respectively.
m1 =[(3+1) / (2-2)]
= ∞
m2 =[(0-2) / (4-0)]
= -1/2
Thus, the diagonal ACis parallel to the y-axis.
∠ODB = tan-1 (1/2)
In triangle MND,
∠DMN = π/2 – tan-1 (1/2)
∴ The angle between thediagonals is π/2 – tan-1 (1/2).
Question - 4 : - Find the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.
Answer - 4 : -
Given:
Points (2, 0), (0, 3)and the line x + y = 1.
Let us assume A (2,0), B (0, 3) be the given points.
Now, let us find theslopes
Slope of AB = m1
= [(3-0) / (0-2)]
= -3/2
Slope of the line x +y = 1 is -1
∴ m2 =-1
Let θ be theangle between the line joining the points (2, 0), (0, 3) and the line x + y =
tan θ = |[(m1 –m2) / (1 + m1m2)]|
= [(-3/2 + 1) / (1 +3/2)]
= 1/5
θ = tan-1 (1/5)
∴ The acute anglebetween the line joining the points (2, 0), (0, 3) and the line x + y = 1is tan-1 (1/5).
Question - 5 : - 
Answer - 5 : -
We need to prove:
Let us assume A (x1,y1) and B (x2, y2) be the given points and Obe the origin.
Slope of OA = m1 =y1×1
Slope of OB = m2 =y2×2
It is giventhat θ is the angle between lines OA and OB.
Hence proved.