# 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.