# Equation of a Line – Normal Form

**Normal:
**

A normal to a line is a line segment drawn from a point perpendicular to the given line.

Let ** p** be the length of the normal drawn from the origin to a line, which subtends an angle ø with the positive direction of x-axis as follows.

Then, we have Cos ø = p/m à m = p/Cos ø

And Sin ø = p/n à n = p/Sin ø

The equation of line in intercept form is,

x/m + y/n =1

x/(p/Cos ø) + y/(p/Sin ø) = 1

x Cos ø/p + y Sin ø/p =1

*x Cos ø + y Sin ø = p.*

This is called the normal form of equation of the given line making the angle ø with the positive direction of x-axis and whose perpendicular distance from the origin is ** p**.

**Converting the general equation of a line into normal form:**

The equation of a straight line in general form is,

ax+by+c=0

From the above equations, we have

Cos ø = p/m and Sin ø = p/n

From Trigonometric identity, we have

Cos^{2 }ø + Sin^{2}ø=1

p^{2}/m^{2} + p^{2}/n^{2}=1

Solving this equation gives us,

P = mn/√(m^{2}+n^{2}), which is the perpendicular distance from the origin to the line x/m + y/n =1 (i.e., nx + my – mn=0)

Thus, we have the distance from origin to the line ax+by+c=0, = |c|/√(a^{2}+b^{2})

Thus, for converting the given line into normal form, divide the equation ax+by+c=0 by √(a^{2}+b^{2}).

Thus,

*[a/ √(a ^{2}+b^{2})] x + [b/√(a^{2}+b^{2})] + c/√(a^{2}+b^{2})=0*

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