Question

Reduce the equation $x+4y-6=0$ to intercept form.Hence find the length of the segment intercepted between the axes.

Answer 1

The given equation is $x+4y-6=0$

$\Rightarrow x-4y=6$

Dividing both sides by $6$, we get

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

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

$\Rightarrow \frac{x}{6}+\frac{y}{\frac{-3}{2}}=1$ which is the required intercept form.

Here $x-$intercept$=a=6$ and $x-$intercept$=b=\frac{-3}{2}$

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

$=\sqrt{{\left(6\right)}^2+{\left(\frac{-3}{2}\right)}^2}$

$=\sqrt{36+\frac{9}{4}}$

$=\sqrt{\frac{144+9}{4}}$

$=\frac{\sqrt{153}}{2}$ units