Question

What are the points on the $y-$axis whose distance from the line $\frac{x}{6}-\frac{y}{8}=1$ is $5$ 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 $5$ units, then

$\frac{|8\times 0-6\times k-48|}{\sqrt{8^2+6^2}}=5$

$\Rightarrow \frac{|-6k-48|}{\sqrt{64+36}}=5$

$\Rightarrow 6k+48=\pm 50$

$\Rightarrow 3k+24=\pm 25$

$\Rightarrow 3k=-24\pm 25$

$\Rightarrow 3k=-49,1$

$\therefore k=\frac{-49}{3},\frac{1}{3}$

Hence the required points are $(0,\frac{-49}{3})$ and $(0,\frac{1}{3})$