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

The normal form of a linear equation of a straight line uses two parameters: p and α to describe the line. Here, p stands for the length of the perpendicular from the origin to the line. And α is the angle between this perpendicular and the x-axis. Let p be the length of the normal drawn from the origin to a line, which subtends an angle α with the +I ve direction of the x-axis as follows. Consider the graph below where p is the length of a perpendicular from origin to the non-vertical line l and α is the inclination of p, then show that the equation of the line is

$x\cos \alpha +y\sin \alpha =p $

To prove this equation of a straight line is in normal form, consider P(x,y) be any point on the straight line l. Since the line intersects the coordinate axes at points A and B, then OA and OB become its X-intercept and Y-intercept. Now using the equation of a straight line intercepts form, we have

$\frac{x}{OA}+\frac{y}{OB}=1$

If C is the foot of the perpendicular drawn from origin O to the non-vertical straight line, then consider OCA is the right triangle as given in the diagram. Use the trigonometric ratios to obtain:

$\frac{OC}{OA}=\cos \alpha $

$\frac{P}{OA}=\cos \alpha $

This means that OC =p

Since OCB is a right triangle, then OCOB=cos(90°−α)

$\frac{P}{OB}=\sin \alpha $

Now put the values of OA and OB in the equation above, we get:

$\frac{x/p}{cos\alpha }+\frac{y/p}{sin\alpha }=1$

$x\cos \alpha +y\sin \alpha =p$

This form of the equation is called the normal form.

To convert the general equation of a line into the normal form:

Begin with the general or linear form of the straight line equation:

ax+by+c=0

From the above equations, we obtain:

Cos α = p/m and Sin α = p/n

From Trigonometric identity, we have

Cos^{2}α + Sin^{2}α =1

$\frac{p^{2}}{m^{2}}+\frac{p^{2}}{n^{2}}=1$

Simplifying this equation gives:

$P=\frac{mn}{\sqrt{m^{2}+n^{2}}}$

Since P is the perpendicular distance from the origin to the line, we get:

$ax+by=0; \vert c\vert /\sqrt{a^{2}+b^{2}}$

Therefore, we convert the linear equation by diving it by the factor of $\sqrt{a^{2}+b^{2}}$

Hence, we obtain the following equation:

$

\frac{a}{\sqrt{a^{2}+b^{2}}}x+\frac{b}{\sqrt{a^{2}+b^{2}}}y+\frac{c}{\sqrt{a^{2}+b^{2}}}=0

$

Recall the trigonometric identity: Cos^{2}α + Sin^{2}α =1 and the trigonometric ratios:

$\frac{a}{\sqrt{a^{2}+b^{2}}}=\cos \alpha $

$\frac{b}{\sqrt{a^{2}+b^{2}}}y=\sin \alpha $

$\frac{c}{\sqrt{a^{2}+b^{2}}}=p$

Substituting the values, we obtain:

$x\cos \alpha +y\sin \alpha =p$

**Distance between a point and a line:**

Consider the graph below

The distance from point A to a line l is equal to the distance between the given line and the parallel line passing through point A. This means that the length p+d and coordinates of A are sufficient to obtain the line equation:

$x_{0}cos\phi + y_{0}sin\phi =p+d$

$d=x_{0}cos\phi + y_{0}sin\phi -p$