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.