Find the point on the y-axis that is equidistant from the points (5, –2) and (–3, 2).

We need to find a point on y-axis So, that means x coordinate of that point is zero.

Let us denote that point by A. So, we can write A(0,y) as it has y coordinate.

Also,given that A is equidistant from P(5,–2) and Q(–3,2).

∴ it means PA = PQ

⇒ PA² = PQ²

∴ By distance formula we get

PA² = (5 – 0)^{2 +} (– 2 – y)²

= 25 + 4 + y^{2 +} 4 y

= 29 + y^{2 +} 4y

QA² = (– 3 – 0)^{2 +} (2 – y)²

= 9 + 4 + y²–4y

= 13 + y²–4y

But PA² = PQ²

∴ 29 + y² + 4y = 13 + y²–4y

⇒ 8y = –16

⇒ y = –2

∴ point A(0,–2)

Hence A(0,–2) is the point equidistant from P(5,–2) and Q(–3,2).