Question

What are the points on the $y-$axis whose distance from the line $\frac{x}{6}-\frac{y}{8}=1$ is $-3$ units.

Answer 1

The given line is $\frac{x}{6}-\frac{y}{8}=1$

$8x-6y-48=0$ ........$\left(1\right)$

Let $(0,k)$ be a point on the $y-$axis whose perpendicular distance from the line $\left(1\right)$ is $-3$ units, then

$\frac{|8\times 0-6\times k-48|}{\sqrt{8^2+6^2}}=-3$

$\Rightarrow \frac{|-6k-48|}{\sqrt{64+36}}=-3$

$\Rightarrow 6k+48=\pm (-30)$

$\Rightarrow 6k+48=\mp 30$

$\Rightarrow 2k+16=\mp 10$

$\Rightarrow 2k=-16\mp 10$

$\Rightarrow 2k=-26,-6$

$\therefore k=\frac{-26}{2},\frac{-6}{2}$

$\therefore k=\frac{-26}{2},-3$

Hence the required points are $(0,-13)$ and $(0,-3)$