# Symmetric Form of 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 straight line is

Now to prove this formula of 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.
Form $P$ draw a line $PM$ which is perpendicular on X-axis and from point $Q$ draw another line $QL$ which is also perpendicular on X-axis. Also from $Q$ draw a line $QA$ perpendicular on $PM$.

Now from the given diagram, consider the triangle $\Delta QAP$, by the definition of slope we take

Now by using trigonometric ratio formula as $\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }}$, the above form can be written as

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

Example: Find the equation of 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 $\left( {{x_1},{y_1}} \right) = \left( {2,\sqrt 2 } \right)$
Equation of line in symmetric form is

Substitute the above values in the formula to get the equation of straight line

This is the required equation of straight line.