Reduce the following equations into normal form. Find their perpendicular distances from the origin

(i) x – \( \sqrt{3}y\) + 8 = 0

Given:

The equation is x – \( \sqrt{3}y\) + 8 = 0

Equation of line in normal form is given by x cos θ + y sin θ = p where ‘θ’ is the angle between perpendicular and positive x axis and ‘p’ is perpendicular distance from origin.

So now, x – \( \sqrt{3}y\) + 8 = 0

x – \( \sqrt{3}y\) = -8

Divide both the sides by 

\( \sqrt{(-1)^2 + (\sqrt{3})^2)}\)

 = \( \sqrt{(1 + 3)} = \sqrt{4} = 2\)

\( \dfrac{x}{\sqrt{2}}-\dfrac{y}{\sqrt{2}}=\dfrac{4}{\sqrt{2}}\)

(\(- \dfrac{1}{2}\))x + \( \dfrac{\sqrt{3}}{2}y\) = 4

This is in the form of: x cos 120° + y sin 120° = 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 sin θ = 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 \( \sqrt{(0^2 + 1^2)}\) = 

\( \sqrt{1}=1\) 

0 (x) + 1 (y) = 2

This is in the form of: x cos 90°+ y sin 90° = 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 sin θ = 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 \( \sqrt{(1^2 + 1^2)}\) 

= \( \sqrt{(1+ 1)}\)= \( 2 \sqrt{2}\)

\( \dfrac{x}{\sqrt{2}}-\dfrac{y}{\sqrt{2}}=\dfrac{4}{\sqrt{2}}\)

\( \dfrac{1}{\sqrt{2}x}+(-\dfrac{1}{\sqrt{2}y})= 2 \sqrt{2}\)

This is in the form: x cos 315° + y sin 315° = \( 2 \sqrt{2}\)

The above equation is of the form x cos θ + y sin θ = p, where θ = 315° and p = \(2 \sqrt{2}\).

Perpendicular distance of line from origin = \( 2 \sqrt{2}\)

Angle between perpendicular and positive x – axis = 315°