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.


symmetric-line-form

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.