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

Given points are $(-5, -2)$ and $(3, 2)$.

To do:

We have to find the point(s) on y-axis which is equidistant from $(-5, -2)$ and $(3, 2)$.

Solution:

Let the coordinates of the two points be $A (-5, -2)$ and $B (3, 2)$.

We know that,

The x coordinate of a point on the y-axis is $0$.
Let the coordinates of the point which is equidistant from the points $A$ and $B$ be $C(0, y)$.

This implies,

$AC = CB$

The distance between two points \( \mathrm{A}\left(x_{1}, y_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}\right) \) is \( \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \).

Therefore,

\( AC=\sqrt{(0+5)^2+(y+2)^2} \)

\( =\sqrt{25+(y+2)^2} \)

\( CB=\sqrt{(0-3)^{2}+(y-2)^{2}} \)

\( =\sqrt{9+(y-2)^{2}} \)

\( \Rightarrow \sqrt{25+(y+2)^{2}}=\sqrt{9+(y-2)^{2}} \)

Squaring on both sides, we get,

\( 25+(y+2)^{2}=9+(y-2)^{2} \)

\( y^{2}+4 y+4+25=y^{2}-4 y+4+9 \)

\( 4 y+4 y=9-25 \)

\( 8y=-16 \)

\( \Rightarrow y=\frac{-16}{8} \)

\( y=-2 \)

Therefore, the required point is $(0, -2)$.