Equation of a Line with X-Intercept

Consider the straight line l and let \alpha be the inclination of the straight line as shown in the given diagram. Now the slope of the line is 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 pass through the point A\left( {a,0} \right).

Take a as the X-intercept of the straight line so the line must pass through the point A\left( {a,0} \right), i.e. OA = a = X-intercept. From point P draw PQ perpendicular to the X-axis.


line-with-x-intercept

Now from the given diagram, consider the triangle \Delta PAQ, i.e. m\angle PAQ = \alpha , and 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 the definition we can use m instead of \tan \alpha , and 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)}

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

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

Example: Find the equation of a straight line having the slope 8 and X-intercept equal to3.

Here we have slope m = 8 and X-intercept a = 3

Now using the formula of a straight line having a slope and X-intercept

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

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

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

This is the required equation of straight line.