# The Normal Form of a Line

If $p$ is the length of a perpendicular from origin to the non-vertical line $l$ and $\alpha$ 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 is in normal form, let $P\left( {x,y} \right)$ 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 as shown in the given diagram. Now using the equation of a straight line intercepts form, we have
$\frac{x}{{OA}} + \frac{y}{{OB}} = 1\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

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. So we use the trigonometric ratio $\cos \alpha$ as follows:
$\begin{gathered} \frac{{OC}}{{OA}} = \cos \alpha \\ \Rightarrow \frac{p}{{OA}} = \cos \alpha \\ \Rightarrow OA = \frac{p}{{\cos \alpha }}\,\,\,\,\,\,\,\,\,\,\because OC = p \\ \end{gathered}$

Since $OCB$ is a right triangle, then $\frac{{OC}}{{OB}} = \cos \left( {{{90}^ \circ } – \alpha } \right)$
$\begin{gathered} \Rightarrow \frac{p}{{OB}} = \sin \alpha \\ \Rightarrow OB = \frac{p}{{\sin \alpha }} \\ \end{gathered}$

Now putting the values of $OA$ and $OB$ in equation (i), we get
$\begin{gathered} \frac{{\frac{x}{p}}}{{\cos \alpha }} + \frac{{\frac{y}{p}}}{{\sin \alpha }} = 1 \\ \Rightarrow \frac{{x\cos \alpha }}{p} + \frac{{y\sin \alpha }}{p} = 1 \\ \end{gathered}$

$\Rightarrow \boxed{x\cos \alpha + y\sin \alpha = p}$

This is the equation of a straight line in normal form.