# Converting Linear Equations in Standard Form to the Intercepts Form

The general equation or standard equation of a straight line is:
$ax + by + c = 0$

In which, $a$ and $b$ are constants and either $a \ne 0$ or $b \ne 0$.

Now to convert this linear equation in standard form to the intercepts form, i.e. $X$-intercept and $Y$-intercept, by definition the intercepts form is written as $\frac{x}{a} + \frac{y}{b} = 1$

To convert an equation from standard form to intercepts form, take the constant value $c$ and move it to the left hand side. Then divide both sides of the equation by $c$ and $1$ on the right hand side as follows:
$\begin{gathered} ax + by + c = 0 \\ \Rightarrow ax + by = – c \\ \end{gathered}$

Divide both sides of the above equation by  $– c$:
$\begin{gathered} \Rightarrow \frac{{ax}}{{ – c}} + \frac{{by}}{{ – c}} = 1 \\ \Rightarrow \frac{x}{{ – \frac{c}{a}}} + \frac{y}{{ – \frac{c}{b}}} = 1 \\ \end{gathered}$

This is the equation of a line intercepts form with the $X$-intercept $– \frac{c}{a}$ and the $Y$-intercept $– \frac{c}{b}$.

Example: Convert the equation $2x + 5y – 6 = 0$ into the intercepts form.

We have the equation of a line in standard form as $2x + 5y – 6 = 0$
$\begin{gathered} \Rightarrow 2x + 5y = 6 \\ \Rightarrow \frac{{2x}}{6} + \frac{{5y}}{6} = 1 \\ \Rightarrow \frac{x}{3} + \frac{{5y}}{{\frac{6}{5}}} = 1 \\ \end{gathered}$

Compare with the intercepts form $\frac{x}{a} + \frac{y}{b} = 1$, where the $X$-intercept is $3$ and the $Y$-intercept is $\frac{6}{5}$.