# 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.