# 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.