The point on the y-axis which is equidistant from A(–5, –2) and B(3, 2) is

Concept

A point is equidistant from two fixed points when its distances to both are equal. By the distance formula, the distance between (x1, y1) and (x2, y2) is √[(x2−x1)2 + (y2−y1)2]. Any point on the y-axis has the form (0, y), so x is fixed at 0 and only y is unknown.

Application

1. Let the required point be P(0, y), since it lies on the y-axis.

2. Set the two distances equal, PA = PB: √[(–5−0)2 + (–2−y)2] = √[(3−0)2 + (2−y)2].

3. Square both sides to remove the roots: 25 + (y+2)2 = 9 + (y−2)2.

4. Expand both squares: 25 + y2 + 4y + 4 = 9 + y2 − 4y + 4.

5. Cancel y2 and the +4 common to both sides: 29 + 4y = 13 − 4y.

6. Collect the y-terms: 8y = −16, so y = −2.

7. Therefore the required point is (0, −2).

Cross-check

Substitute back: PA = √[(0+5)2 + (–2+2)2] = 5 and PB = √[(0−3)2 + (–2−2)2] = √(9+16) = 5. Both distances equal 5, confirming the point is equidistant.