# The Slope Intercept Form of the Equation of a Straight Line

Consider the straight line $l$. Let $P\left( {x,y} \right)$ be any point on the given line $l$. Suppose that $\alpha$ is the inclination of the line $l$ as shown in the given diagram, i.e. $\tan \alpha = m$

Take $c$ as a $Y$-intercept of the straight line because it cuts the $Y$-axis at the point $Q\left( {0,c} \right)$, i.e. $OQ = c = Y$-intercept.

From point $P$ draw $PM$ perpendicular to the $X$-axis, and from $Q$ draw $QR$ perpendicular to the $PM$.

Now from the given diagram, consider the triangle $\Delta PQR$, i.e. $m\angle RQP = \alpha$.

By the definition of slope we take
$\begin{gathered} \tan \alpha = \frac{{PR}}{{QR}} = \frac{{PM – RM}}{{OM}} \\ \Rightarrow \tan \alpha = \frac{{PM – OQ}}{{OM}} \\ \Rightarrow \tan \alpha = \frac{{y – c}}{x} \\ \end{gathered}$

Now by the definition we can use $m$ instead of $\tan \alpha$, and we get
$\begin{gathered} \Rightarrow m = \frac{{y – c}}{x} \\ \Rightarrow mx = y – c \\ \end{gathered}$
$\boxed{y = mx + c}$

This is the equation of a straight line having the slope $m$ and Y-intercept $c$.

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

Example: Find the equation of a straight line having the slope $3$ and $Y$-intercept equal to 8.
Here we have slope $m = 3$ and $Y$-intercept $c = 8$

Now using the formula of straight line having the slope and $Y$-intercept
$y = mx + c$

Substitute the above values in the formula to get the equation of a straight line
$y = 3x + 8$
$3x – y + 8 = 0$

This is the required equation of a straight line.