Find the point on the y-axis which is equidistant from the points $(5,-2)$ and $(-3,2)$

Let the given point be P(5,-2) and Q(-3,2)

And the point required be R(0,a)

As per question,

Point R is equidistant from P & Q

Hence, $PR=QR$

Finding PR

$x_1=5,\ y_1=-2$

$x_2=0,\ y_2=a$

$PR=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$=\sqrt{(0-5)^2+(a+2)^2}$

$=\sqrt{25+a^2+4a+4}$

$=\sqrt{a^2+4a+29}$

Finding QR

$x_1=-3,\ y_1=2$

$x_2=0,\ y_2=a$

$PR=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$=\sqrt{(0+3)^2+(a-2)^2}$

$=\sqrt{9+a^2+4-4a}$

$=\sqrt{a^2-4a+13}$

As per question

$PR=QR$

$\sqrt{a^2+4a+29}=\sqrt{a^2-4a+13}$

Squaring both sides

$a^2+4a+29=a^2-4a+13$

$4a+4a=13-29$

$8a=-16$

$a=\dfrac{-16}{8}$

Hence the required point is $R(0,-2)$