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

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

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

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

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

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

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

This is the required equation of a straight line.