Question

If $3x-4y+12=0$ is a equation of line. This line cut $x-$axis at point $A$ and $y-$axis at point $B$. Find distance between $A$ and $B$.

Answer 1

$3x-4y+12=0$

$\Rightarrow 3x-4y=-12$

$\Rightarrow \frac{3x}{-12}-\frac{4y}{-12}=1$

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

so, intercepts on $x$-A=axis is $A\left(-4,0\right)$

and intercepts on $y$ -axis is $B\left(0,3\right)$

Therefore, $AB=\sqrt{{\left(0+4\right)}^2+{\left(3-0\right)}^2}$

$=\sqrt{16+9}$

$=\sqrt{25}$

$=5$ units.