# 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.

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 to$$3$$.

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.