The point on Y-axis which is equidistant from (6, - 1) and (2, 3) is

Let A (6, - 1), B (2,3) and P (0, y) be the required point.
$\begin{array}{l}\mathrm{PA}=\mathrm{PB}\Rightarrow (6-0{)}^2+(-1-\mathrm{y}{)}^2=(2-0{)}^2+(3-\mathrm{y}{)}^2\Rightarrow {\mathrm{y}}^2+2\mathrm{y}+37={\mathrm{y}}^2-6\mathrm{y}+13\\ \Rightarrow 8\mathrm{y}=-24\Rightarrow \mathrm{y}=-3.\end{array}$