Question

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

Answer 1

Let the required point be (0, a)

Given, distance of (0, a) from line $4x+3y-12=0$ is 4 units

$D=\left|\frac{a{x}_1+b{y}_{+}c}{\sqrt{a^2+b^2}}\right|$

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

$4=\left|\frac{3a-12}{5}\right|$

$\Rightarrow 4=\frac{-3a+12}{5}$

$\Rightarrow -3a=20-12$

$a=-\frac{8}{3}$

And $4=\frac{3a-12}{5}$

$3a=20+12$

$\Rightarrow a=\frac{32}{3}$

So, required points are

$(0,\frac{32}{3},(0,\frac{-8}{3}))$