# 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 $Q$.
Form $P$ draw $PM$ perpendicular on X-axis and from point $Q$ draw $QL$ also perpendicular on X-axis. Also from $Q$ draw perpendicular $QA$ on $PM$.

Now from the given diagram, consider the triangle $\Delta QAP$, by the definition of slope we take

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

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

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

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

Now subtraction equation (ii) form equation (i), we get

Example: Find the equation of straight line having 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 slope point form equation straight line

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

Which is the required equation of straight line.

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