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# Find a point on 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 on y-axis which is equidistant from $(5, -2)$ and $(-3, 2)$.

Solution:

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

We know that,

The x co-ordinate of a point on y-axis is $0$.

Let the co-ordinates 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 $(0, -2)$.

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