Question

Find the locus of a point equidistant from the point (2, 4) and the y-axis.

Answer 1

Let P(h, k) be the point which is equidistant from the point (2, 4) and the y-axis.
The distance of point P(h, k) from the y-axis is h.

$$\begin{gathered} \therefore h=\sqrt{{\left(h-2\right)}^2+{\left(k-4\right)}^2} \\ \Rightarrow h^2-4h+4+k^2-8k+16=h^2 \\ \Rightarrow k^2-4h-8k+20=0 \end{gathered}$$

Hence, the locus of (h, k) is $y^2-4x-8y+20=0$.