# The Intercepts Form of a Line

If $a$ and $b$ are non-zero $X$ and $Y$ intercepts of a line $l$, then its equation is of the form
$\frac{x}{a} + \frac{y}{b} = 1$

Since $a$ is an $X$-intercept of the line $l$, and as we know that if any point lies on the $X$-axis its value of $Y$ is equal to zero, it passes through the point $A\left( {a,0} \right)$. Also if $b$ is the $Y$-intercept of the line $l$, and we know that any point that lies on the $Y$-axis has a value of $X$ equal to zero, it passes through the point $B\left( {0,b} \right)$ as shown in the given diagram. Now to prove the intercepts form of a line, use the formula for the two points form of a straight line as given by
$\frac{{y – {y_1}}}{{{y_2} – {y_1}}} = \frac{{x – {x_1}}}{{{x_2} – {x_1}}}$

Take $A\left( {a,0} \right) = \left( {{x_1},{y_1}} \right)$ and $B\left( {0,b} \right) = \left( {{x_2},{y_2}} \right)$, and put these values in the above formula:

$\begin{gathered} \Rightarrow \frac{{y – 0}}{{b – 0}} = \frac{{x – a}}{{0 – a}} \\ \Rightarrow \frac{y}{b} = \frac{x}{{ – a}} – \frac{a}{{ – a}} \\ \end{gathered}$

$\Rightarrow \boxed{\frac{x}{a} + \frac{y}{b} = 1}$

This is the required equation of a straight line in intercepts form.

Example: Find the equation of a straight line with $X$-intercept $A\left( {3,0} \right)$ and $Y$-intercept $B\left( {0,2} \right)$.

From the above information we know the $X$-intercept is $a = 3$ and the $Y$-intercept is $b = 2$. Now we put all these values in the formula of the intercepts form as given:
$\begin{gathered} \frac{x}{a} + \frac{y}{b} = 1 \\ \Rightarrow \frac{x}{3} + \frac{y}{2} = 1 \\ \Rightarrow 2x + 3y = 6 \\ \Rightarrow 2x + 3y – 6 = 0 \\ \end{gathered}$

This is the required equation of a straight line.