Equation of a Line with X Intercept

Consider the straight line l and \alpha  be the inclination of the straight line as shown in the given diagram now the slope of the represented by \tan  \alpha  = m. Let P\left( {x,y} \right) be any point on the given line l. Let a be the X-intercept of the straight line, so the line must passes through the point A\left(  {a,0} \right).
Take a as X-intercept of the straight line so the line must passes through the point A\left(  {a,0} \right), i.e. OA = a = X-intercept. From point P draw PQ perpendicular on X-axis.


line-with-x-intercept

Now from the given diagram, consider the triangle \Delta PAQ, i.e. m\angle PAQ = \alpha , by the definition of slope we take

\begin{gathered} \tan \alpha   = \frac{{PQ}}{{AQ}} = \frac{{PQ}}{{OQ - OA}} \\ \Rightarrow \,\tan \alpha  = \frac{y}{{x - a}} \\ \end{gathered}


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

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


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


Which is the equation of straight line having slope m and X-intercept a.

NOTE: It may be noted that if the line passes through the origin \left( {0,0} \right), then take X-intercept is equal to zero i.e. a = 0, so the equation of straight line becomes y = mx.

Example: Find the equation of straight line having slope 8 and X-intercept is equal to3.
Here we have slope m =  8 and X-intercept a = 3
Now using the formula of straight line having slope and X-intercept

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


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

\begin{gathered} y = 8\left( {x - 3} \right) \\ \Rightarrow y = 8x - 24 \\ \Rightarrow 8x - y - 24 = 0 \\ \end{gathered}


Which is the required equation of straight line.

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