# Normal Form of a Line Equation

Here you will learn normal form of a line equation with proof and examples.

Let’s begin –

## Normal Form of a Line (Perpendicular form of line)

The equation of the straight line upon which the length of of the perpendicular from the origin is p and this perpendicular makes an angle $$\alpha$$ with x-axis is

$$xcos\alpha$$ + $$ysin\alpha$$ = p.

Proof

Let the line AB be such that the length of the perpendicular OQ from the origin O to the line be p and $$\angle$$ XOQ = $$\alpha$$

Let P(x, y) be any point on the line. Draw PL $$\perp$$ OX, LM $$\perp$$ OQ and PN $$\perp$$ LM. Then,

OL = x and LP = y

In triangle OLM, we have

$$cos\alpha$$ = $$OM\over OL$$

$$\implies$$ OM = OL $$cos\alpha$$ = x $$cos\alpha$$.

In triangle PNL, we have

$$sin\alpha$$ = $$PN\over PL$$

$$\implies$$ PN = PL $$sin\alpha$$ = y $$sin\alpha$$

$$\implies$$ MQ = PN = y $$sin\alpha$$

Now, p = OQ = OM + MQ = $$xcos\alpha$$ + $$ysin\alpha$$

Hence, the equation of the required line is

$$xcos\alpha$$ + $$ysin\alpha$$ = p.

Example : Find the equation of the line which is at a distance 3 from the origin and the perpendicular from the origin to the line makes an angle of 30 with the positive direction of the x-axis.

Solution : Here, p = 3, $$\alpha$$ = 30

Equation of the line in the normal form is

x cos 30 + y sin 30 = 3

$$\implies$$ x $$\sqrt{3}\over 2$$ + $$y\over 2$$ = 3

$$\implies$$ $$\sqrt{3}$$ x + y = 6.

which is the required equation line.