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 (0, t) be a point on the y-axis.

It is given that the perpendicular distance of the line $\frac{x}{3}+\frac{y}{4}=1$ from the point (0, t) is 4 units.

$$\begin{gathered} \therefore \left|\frac{{\displaystyle 0}+{\displaystyle \frac{t}{4}}-1}{\sqrt{{\displaystyle \frac{1}{3^2}}+{\displaystyle \frac{1}{4^2}}}}\right|=4 \\ \Rightarrow \left|t-4\right|=4\times 4\times \frac{5}{3\times 4} \\ \Rightarrow \left|t-4\right|=\frac{20}{3} \end{gathered}$$

$$\begin{gathered} \Rightarrow t-4=\pm \frac{20}{3} \\ \Rightarrow t=4\pm \frac{20}{3} \\ \Rightarrow t=4+\frac{20}{3},4-\frac{20}{3} \\ \Rightarrow t=\frac{32}{3},-\frac{8}{3} \end{gathered}$$

Hence, the required points on the y-axis are $\left(0,\frac{32}{3}\right)\mathrm{and}\left(0,-\frac{8}{3}\right)$