Normal Form of a Line

If p is the length of 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 straight 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, so OA and OB becomes its X-intercept and Y-intercept as shown in the given diagram. Now using equation of 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 draw from origin O to the non-vertical straight line, then consider OCA is the right triangle as given in the diagram, so using 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, so \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 the 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}

Which is the equation of straight line in normal form.