Reduce the following equations into intercept form and find their intercepts on the axes.

Solution:

(i) The given equation is 3x + 2y – 12 = 0.

It can be written as

$3 x+2 y=12$

$\frac{3 x}{12}+\frac{2 y}{12}=1$

i.e., $\frac{x}{4}+\frac{y}{6}=1$ $\ldots(1)$

This equation is of the form $\frac{x}{a}+\frac{y}{b}=1$, where $a=4$ and $b=6$.

(ii) The given equation is 4x – 3y = 6.

It can be written as

$\frac{4 x}{6}-\frac{3 y}{6}=1$

$\frac{2 x}{3}-\frac{y}{2}=1$

i.e., $\frac{x}{\left(\frac{3}{2}\right)}+\frac{y}{(-2)}=1$ $\ldots(2)$

This equation is of the form $\frac{x}{a}+\frac{y}{b}=1$, where $a=\frac{3}{2}$ and $b=-2$.

Therefore, equation (2) is in the intercept form, where the intercepts on the $x$ and $y$ axes are $\frac{3}{2}$ and $-2$ respectively.

(iii) The given equation is $3 y+2=0$.

It can be written as

$3 y=-2$

i.e., $\frac{y}{\left(-\frac{2}{3}\right)}=1$ $\ldots(3)$

This equation is of the form $\frac{x}{a}+\frac{y}{b}=1$, where $a=0$ and $b=-\frac{2}{3}$.

Therefore, equation (3) is in the intercept form, where the intercept on the $y$-axis is $-\frac{2}{3}$ and it has no intercept on the $x$-axis.