Question

What are the points on the $y-$axis whose distance from the line $\frac{x}{3}-\frac{y}{4}=1$ is $2$ 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 $2$ units, then

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

$\Rightarrow \frac{|-3k-12|}{\sqrt{16+9}}=2$

$\Rightarrow 3k+12=\pm 10$

$\Rightarrow 3k=-12\pm 10$

$\Rightarrow 3k=-2,-22$

$\therefore k=\frac{-2}{3},\frac{-22}{3}$

Hence the required points are $(0,\frac{-22}{3})$ and $(0,\frac{-2}{3})$