Question

What are the points on the $y-$axis whose distance from the line $\frac{x}{3}+\frac{y}{4}=1$ is $8$ 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 $8$ units, then

$\frac{|4\times 0+3\times k-12|}{\sqrt{4^2+3^2}}=8$

$\Rightarrow \frac{|3k-12|}{\sqrt{16+9}}=8$

$\Rightarrow 3k-12=\pm 8$

$\Rightarrow 3k=12\pm 40$

$\Rightarrow 3k=52,-28$

$\therefore k=\frac{52}{3},\frac{-28}{3}$

$\therefore k=14,\frac{-28}{3}$

Hence the required points are $(0,14)$ and $(0,\frac{-28}{3})$