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.


point-slope-form

Now from the given diagram, consider the triangle \Delta QAP. By the definition of a slope we take

\begin{gathered} \tan \alpha = \frac{{AP}}{{QA}} = \frac{{MP - MA}}{{LM}} \\ \Rightarrow \tan \alpha = \frac{{MP - MA}}{{OM - OL}} \\ \end{gathered}

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

\begin{gathered} \Rightarrow m = \frac{{y - {y_1}}}{{x - {x_1}}} \\ \Rightarrow m\left( {x - {x_1}} \right) = y - {y_1} \\ \end{gathered}


\boxed{y - {y_1} = m\left( {x - {x_1}} \right)}

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

y = mx + c\,\,\,{\text{ - - - }}\left( {\text{i}} \right)

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

{y_1} = m{x_1} + c\,\,\,{\text{ - - - }}\left( {{\text{ii}}} \right)

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

y - {y_1} = mx - m{x_1}


\boxed{y - {y_1} = m\left( {x - {x_1}} \right)}

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

y - {y_1} = m\left( {x - {x_1}} \right)

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

\begin{gathered} y - \left( { - 2} \right) = - 4\left( {x - 1} \right) \\ \Rightarrow y + 2 = - 4x + 4 \\ \Rightarrow 4x + y - 2 = 0 \\ \end{gathered}

This is the required equation of a straight line.