Question

What are the points on the $y-$axis whose distance from the line $\frac{x}{3}-\frac{y}{4}=1$ is $5$ units.

Answer 1

The given line is $\frac{x}{3}-\frac{y}{4}=1$

$4x-3y-12=0$ ........$\left(1\right)$

Let $(0,k)$ be a point on the $y-$axis whose perpendicular distance from the line $\left(1\right)$ is $5$ units, then

$\frac{|4\times 0-3\times k-12|}{\sqrt{4^2+3^2}}=5$

$\Rightarrow \frac{|-3k-12|}{\sqrt{16+9}}=5$

$\Rightarrow 3k+12=\pm 25$

$\Rightarrow 3k=-12\pm 20$

$\Rightarrow 3k=8,-32$

$\therefore k=\frac{8}{3},\frac{-32}{3}$

Hence the required points are $(0,\frac{-32}{3})$ and $(0,\frac{8}{3})$