Answer :

We have A → (–5, –2)

B → (0, y) equidistant from A and C

(x = 0 since B lies on Y-axis)

C → (3, 2)

Using the distance formula,

Since B is equidistant from A and C

∴ AB = BC

Squaring both sides

8y + 16 = 0

y = –2

The required point is B (0, –2).

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