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.