# Symmetric Form of a Straight Line

If $\alpha$ is the incrimination of a straight line $l$ passing through the point $Q\left( {{x_1},{y_1}} \right)$ then its equation of a straight line is $\frac{{x – {x_1}}}{{\cos \alpha }} = \frac{{y – {y_1}}}{{\sin \alpha }}$

Now to prove this formula of a straight line, let $P\left( {x,y} \right)$ be any point on the given line $l$. Consider another point $Q\left( {{x_1},{y_1}} \right)$, since the line passes through the point $Q$.

Let $\alpha$ be the inclination of the straight line $l$ as shown in the given diagram.

From $P$ draw a line $PM$ which is perpendicular to the X-axis and from point $Q$ draw another line $QL$ which is also perpendicular to the X-axis. Also from $Q$ draw a line $QA$ perpendicular to $PM$.

Now from the given diagram, consider the triangle $\Delta QAP$. By the definition of a slope we take
$\begin{gathered} \tan \alpha = \frac{{AP}}{{QA}} = \frac{{MP – MA}}{{LM}} \\ \Rightarrow \tan \alpha = \frac{{MP – MA}}{{OM – OL}} \\ \end{gathered}$

Now by using the trigonometric ratio formula as $\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }}$, the above form can be written as
$\begin{gathered} \Rightarrow \frac{{\sin \alpha }}{{\cos \alpha }} = \frac{{y – {y_1}}}{{x – {x_1}}} \\ \Rightarrow \frac{{x – {x_1}}}{{\cos \alpha }} = \frac{{y – {y_1}}}{{\sin \alpha }} \\ \end{gathered}$
$\boxed{\frac{{x – {x_1}}}{{\cos \alpha }} = \frac{{y – {y_1}}}{{\sin \alpha }}}$

This is called the symmetric form of an equation of a straight line having point $\left( {{x_1},{y_1}} \right)$ and inclination $\alpha$.

Example: Find the equation of a straight line with inclination ${45^ \circ }$ and passing through the point $\left( {2,\sqrt 2 } \right)$
Here we have inclination $\alpha = {45^ \circ }$ and point $\left( {{x_1},{y_1}} \right) = \left( {2,\sqrt 2 } \right)$

The equation of line in its symmetric form is
$\frac{{x – {x_1}}}{{\cos \alpha }} = \frac{{y – {y_1}}}{{\sin \alpha }}$

Substitute the above values in the formula to get the equation of a straight line
$\begin{gathered} \frac{{x – 2}}{{\cos {{45}^ \circ }}} = \frac{{y – \sqrt 2 }}{{\sin {{45}^ \circ }}} \\ \Rightarrow \sin {45^ \circ }\left( {x – 2} \right) = \cos {45^ \circ }\left( {y – \sqrt 2 } \right) \\ \Rightarrow \frac{1}{{\sqrt 2 }}\left( {x – 2} \right) = \frac{1}{{\sqrt 2 }}\left( {y – \sqrt 2 } \right) \\ \Rightarrow x – y – 2 + \sqrt 2 = 0 \\ \end{gathered}$

This is the required equation of a straight line.