Question

Reduce the equation $3x-2y+4=0$ to intercept form.Hence find the length of the segment intercepted between the axes.

Answer 1

The given equation is $3x-2y+4=0$

$\Rightarrow 3x-2y=-4$

Dividing both sides by $-4$, we get

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

$\Rightarrow \frac{x}{\frac{-4}{3}}+\frac{y}{2}=1$ which is the required intercept form.

Here $x-$intercept$=a=\frac{-4}{3}$ and $x-$intercept$=b=2$

$\therefore$The length of the segment intercepted between the axes$=\sqrt{a^2+b^2}$

$=\sqrt{{\left(\frac{-4}{3}\right)}^2+{\left(2\right)}^2}$

$=\sqrt{\frac{16}{9}+4}$

$=\sqrt{\frac{16+36}{9}}$

$=\frac{\sqrt{52}}{3}$

$=\frac{\sqrt{4\times 13}}{3}$

$=\frac{2}{3}\sqrt{13}$ units.