Question

Reduce the following equations into intercept form and find their intercepts on
the axes.
(i)$3x+2y-12=0$
(ii)$4x-3y=6$
(iii)$3y+2=0$

Answer 1

(i)The equation is $3x+2y-12=0$
It can be written as
$3x+2y=12$
$\frac{3x}{12}+\frac{2y}{12}=1$
i.e;$\frac{x}{4}+\frac{y}{6}=1$
This equation is of the form $\frac{x}{a}+\frac{y}{b}=1$,where $a=4$ and $b=6$.
Therefore equation (1) is in the intercept form,where the intercepts on the $x$ and $y$ axes are 4 and 6 respectively.
(ii)The given equation is $4x-3y=6$
It can be written as
$\frac{4x}{6}-\frac{3y}{6}=1$
$\frac{2x}{3}-\frac{y}{2}=1$
i.e;$\frac{x}{\left\{\frac{3}{2}\right\}}+\frac{y}{(-2)}=1$
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 $3y+2=0$
It can be written as
$3y=-2$
i.e;$\frac{y}{\{-\frac{2}{3}\}}=1$
This equation is of the form$\frac{x}{a}+\frac{y}{b}=1$, where $a=\infty$ 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.