Question -
Answer -
(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).