Question -
Answer -
(i) x – √3y + 8 = 0
Given:
The equation is x – √3y + 8 = 0
Equation of line in normal form is given by x cos θ + y sinθ = p where ‘θ’ is the angle between perpendicular and positive x axis and ‘p’is perpendicular distance from origin.
So now, x – √3y + 8 = 0
x – √3y = -8
Divide both the sides by √(12 + (√3)2)= √(1 + 3) = √4 = 2
x/2 – √3y/2 = -8/2
(-1/2)x + √3/2y = 4
This is in the form of: x cos 120o + y sin120o = 4
∴ The above equation is of the form x cos θ + y sin θ = p,where θ = 120° and p = 4.
Perpendicular distance of line from origin = 4
Angle between perpendicular and positive x – axis = 120°
(ii) y – 2 = 0
Given:
The equation is y – 2 = 0
Equation of line in normal form is given by x cos θ + y sinθ = p where ‘θ’ is the angle between perpendicular and positive x axis and ‘p’is perpendicular distance from origin.
So now, 0 × x + 1 × y = 2
Divide both sides by √(02 + 12) =√1 = 1
0 (x) + 1 (y) = 2
This is in the form of: x cos 90o + y sin 90o =2
∴ The above equation is of the form x cos θ + y sin θ = p,where θ = 90° and p = 2.
Perpendicular distance of line from origin = 2
Angle between perpendicular and positive x – axis = 90°
(iii) x – y = 4
Given:
The equation is x – y + 4 = 0
Equation of line in normal form is given by x cos θ + y sinθ = p where ‘θ’ is the angle between perpendicular and positive x axis and ‘p’is perpendicular distance from origin.
So now, x – y = 4
Divide both the sides by √(12 + 12)= √(1+1) = √2
x/√2 – y/√2 = 4/√2
(1/√2)x + (-1/√2)y = 2√2
This is in the form: x cos 315o + y sin 315o =2√2
∴ The above equation is of the form x cos θ + y sin θ = p,where θ = 315° and p = 2√2.
Perpendicular distance of line from origin = 2√2
Angle between perpendicular and positive x – axis = 315°