Question

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

Answer 1

Let $(0,b)$ be the point on the $y$-axis whose distance from line $\frac{x}{3}+\frac{y}{4}=1$ is $4$ units.
The given line can be written as $4x+3y-12=0$....(1)
Thus using distance formula,
$4=\frac{\left|4\right(0)+3(b)-12|}{\sqrt{4^2+3^2}}$
$\Rightarrow 4=\frac{|3b-12|}{5}$
$\Rightarrow 20=|3b-12|$
$\Rightarrow 20=\pm (3b-12)$
$\Rightarrow 20=(3b-12)$ or $20=-\{3b-12\}$
$\Rightarrow 3b=20+12$ or $3b=-20+12$
$\Rightarrow b=\frac{32}{3}$ or $b=-\frac{8}{3}$
Thus, the required points are $(0,\frac{32}{3})\text{and}(0,-\frac{8}{3})$.