Question

Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.

(i) X - √3y + 8=0

(ii) y – 2 = 0

(iii) x – y = 4

Answer 1

(i) The given equation is, x -√3y + 8=0

It can be reduced as:

x- √3y = -8

⇒ -x + √3y = 8

Equation (1) is in the normal form.
On comparing equation (1) with the normal form of equation of line x cos ω + y sin ω = p, we obtain ω = 120° and p = 4.
Thus, the perpendicular distance of the line from the origin is 4, while the angle between the perpendicular and the positive x-axis is 120°.

(ii) The given equation is y – 2 = 0.
It can be reduced as 0.x + 1.y = 2

On dividing both sides by √0² + 1² = 1 , we obtain 0.x + 1.y = 2

⇒ x cos 90° + y sin 90° = 2 … (1)
Equation (1) is in the normal form.
On comparing equation (1) with the normal form of equation of line x cos ω + y sin ω = p, we obtain ω = 90° and p = 2.
Thus, the perpendicular distance of the line from the origin is 2, while the angle between the perpendicular and the positive x-axis is 90°.
(iii) The given equation is x – y = 4.
It can be reduced as 1.x + (–1) y = 4
Equation (1) is in the normal form.
On dividing both sides by √((1² +(-1)²)=√2 ,we obtain