Question

Find the point (0, b) on the y-axis that is equidistant from the points (5, 5) and (3, -2)

Answer 1

I think this attachment will help you

Answer 2

Equidistant in this case means that the distance from (0, b) to (5, 5) is the same as the distance from (0, b) to (3, -2).

The distance from (0, b) to (5, 5) is

`√((0-5)^2+(b-5)^2) = √(25 + (b-5)^2)`.

The distance from (0, b) to (3, -2) is

`√((0-3)^2+(b+2)^2) = √(9 + (b+2)^2)`

These two distances are equal to each other, so

`√(25 + (b-5)^2) = √(9 + (b+2)^2) \\ 25 + (b-5)^2 = 9 + (b+2)^2 \\ 25 + b^2-10b+25 = 9 + b^2 + 4b + 4 \\ -14b = -37 \\ b = 37/14`