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.