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 point on y - axis be (0, y).

The given equation of line is $\frac{x}{3}+\frac{y}{4}=1$

$\Rightarrow 4x+3y=12\Rightarrow 4x+3y-12=0$

Now perpendicular distance from point (0, y) to line 4x + 3y - 12 = 0 is

$\left|\frac{4\times 0+3y-12}{\sqrt{(4{)}^2+(3{)}^2}}\right|=\left|\frac{3y-12}{\sqrt{25}}\right|=\left|\frac{3y-12}{5}\right|$

It is given that

$\left|\frac{3y-12}{5}\right|=4\Rightarrow \frac{3y-12}{5}=\pm 4$

When $\frac{3y-12}{5}=-4\Rightarrow 3y-12=-20$

$\Rightarrow y=\frac{-8}{3}$

Thus required points are $(0,\frac{32}{3})$ and $(0,\frac{-8}{3})$