Find a point on the y-axis which is equidistant from A

Let the point on the y-axis be P(0, y)

Given: $P$ is equidistant from $A(-4,3)$ and $B(5,2)$.

i.e., PA = PB

$\Rightarrow \sqrt{(-4-0)^{2}+(3-y)^{2}}=\sqrt{(5-0)^{2}+(2-y)^{2}}$

Squaring both sides, we get

$\Rightarrow(-4-0)^{2}+(3-y)^{2}=(5-0)^{2}+(2-y)^{2}$

$\Rightarrow 16+9-6 y+y^{2}=25+4-4 y+y^{2}$

$\Rightarrow 25-6 y=29-4 y$

$\Rightarrow 2 y=-4$

$\Rightarrow y=-2$

Therefore, the required point on the y-axis is (0, -2).