Question

Reduce the equation $3x+4y-5=0$ to intercept form.Hence find the length of the segment intercepted between the axes.

Answer 1

The given equation is $3x+4y-5=0$

$\Rightarrow$3x+4y=5$

Dividing both sides by $5$, we get

$\Rightarrow \frac{3x}{5}+\frac{4y}{5}=1$

$\Rightarrow \frac{x}{\frac{5}{3}}+\frac{y}{\frac{5}{4}}=1$ which is the required intercept form.

Here $x-$intercept$=a=\frac{5}{3}$ and $x-$intercept$=b=\frac{5}{4}$

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

$=\sqrt{{\left(\frac{5}{3}\right)}^2+{\left(\frac{5}{4}\right)}^2}$

$=5\sqrt{\frac{1}{9}+\frac{1}{16}}$

$=5\sqrt{\frac{16+9}{9\times 16}}$

$=5\frac{\sqrt{25}}{12}$

$=\frac{25}{12}$ units