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 given equation is $3x + 2y -12 = 0.$
It can be written as
$3x + 2y = 12$

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

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

$\frac{4x}{6} –\frac{ 3y}{6} = \frac{6}{6}$

$\frac{2x}{3} –\frac{ y}{2} = 1$

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

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