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.


point-slope-form

Now from the given diagram, consider the triangle \Delta QAP, by the definition of 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 definition we can use m instead of \tan \alpha , 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)}


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

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 subtraction equation (ii) form 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 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

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


Substitute the above values in the formula to get the equation of 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}


Which is the required equation of straight line.

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