Question

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

(i) 3x + 2y - 1 2 = 0, (ii) 4x- 3y = 6, (iii) 3y + 2 = 0

Answer 1

(i) Here 3x + 2y - 12 = 0

$\Rightarrow$ 3x + 2y = 12

$\Rightarrow \frac{3x}{12}+\frac{2y}{12}=1\Rightarrow \frac{x}{4}+\frac{y}{6}=1$

which is required intercept form.

Comparing it with $\frac{x}{a}+\frac{y}{b}=1$, we have

a = 4 and b=6

(ii) Here 4x - 3y = 6

$\Rightarrow \frac{4x}{6}-\frac{3y}{6}=1\Rightarrow \frac{2x}{3}-\frac{y}{2}=1$

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

which is required intercept form.

Comparing it with $\frac{x}{a}+\frac{y}{b}=1$, we have

$a=\frac{3}{2}$ and b =-2

(iii) Here 3y + 2 = 0

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

$\Rightarrow \frac{0x}{-2}+\frac{3y}{-2}=1\Rightarrow \frac{0x}{-2}+\frac{y}{-\frac{2}{3}}=1$

which is required intercept form.

Comparing it with $\frac{x}{a}+\frac{y}{b}=1$, we have

a = 0 and $b=-\frac{2}{3}$.