# Slope of a Line Through Two Points

Let $P\left( {{x_1},{y_1}} \right)$ and $Q\left( {{x_2},{y_2}} \right)$ be any two points on the given line $l$. Also let $\alpha$ be the inclination of the line $l$ as shown in the given diagram. From point $P$ draw $PM$ perpendicular to the X-axis, and from $Q$ draw $QN$ perpendicular to the X-axis.

Now from the given diagram, consider the triangle $\Delta PQR$. From the definition of a slope we take
$\begin{gathered} \tan \alpha = \frac{{QR}}{{PR}} = \frac{{QN – RN}}{{MN}} \\ \Rightarrow \tan \alpha = \frac{{QN – RN}}{{ON – OM}} \\ \Rightarrow \tan \alpha = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}} \\ \end{gathered}$

Now by the definition we can use $m$ instead of $\tan \alpha$, and we get the slope of a line through two points:
$\boxed{m = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}}}$

Example: Find the slope of a straight line passing through the pair of points $\left( {3,4} \right)$ and $\left( {7,9} \right)$.

Here we have two points. Suppose that $P\left( {3,4} \right) = \left( {{x_1},{y_1}} \right)$ and $Q\left( {7,9} \right) = \left( {{x_2},{y_2}} \right)$. Now using the formula of a slope passing through two given points of the straight line:
$m = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}}$

By substituting the above points in the formula we get the slope of the line as
$m = \frac{{9 – 4}}{{7 – 3}} = \frac{5}{4}$

Here $m = \frac{5}{4}$ is the required slope of the line.