# 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.