# The Point Slope Equation of a Line

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

From $$P$$ draw $$PM$$ perpendicular to the X-axis and from point $$Q$$ draw $$QL$$ also perpendicular to the X-axis. Also from $$Q$$ draw perpendicular to $$QA$$ on $$PM$$.

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 the definition we can use $$m$$ instead of $$\tan \alpha $$, and we get

\[\begin{gathered} \Rightarrow m = \frac{{y – {y_1}}}{{x – {x_1}}} \\ \Rightarrow m\left( {x – {x_1}} \right) = y – {y_1} \\ \end{gathered} \]

\[\boxed{y – {y_1} = m\left( {x – {x_1}} \right)}\]

This is the equation of a straight line having slope $$m$$ and passing through the point $$\left( {{x_1},{y_1}} \right)$$.

__NOTE__**:** There is an alternate way to prove the slope point form of the equation of a line. Let $$P\left( {x,y} \right)$$ be any point on the line.

Considering the slope intercept form of the equation of a line, we have

\[y = mx + c\,\,\,{\text{ – – – }}\left( {\text{i}} \right)\]

Since the line passing through the point $$\left( {{x_1},{y_1}} \right)$$ above equation (i) becomes

\[{y_1} = m{x_1} + c\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right)\]

Now subtracting equation (ii) from equation (i), we get

\[y – {y_1} = mx – m{x_1}\]

\[\boxed{y – {y_1} = m\left( {x – {x_1}} \right)}\]

__Example__**:** Find the equation of a straight line having the slope $$ – 4$$ and passing through the point $$\left( {1, – 2} \right)$$

Here we have slope $$m = 8$$ and point $$\left( {{x_1},{y_1}} \right) = \left( {1, – 2} \right)$$

Now the slope point form of the equation of a straight line is

\[y – {y_1} = m\left( {x – {x_1}} \right)\]

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

\[\begin{gathered} y – \left( { – 2} \right) = – 4\left( {x – 1} \right) \\ \Rightarrow y + 2 = – 4x + 4 \\ \Rightarrow 4x + y – 2 = 0 \\ \end{gathered} \]

This is the required equation of a straight line.