Question

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

Answer 1

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
$\sqrt{(0-5)^2+(b-5)^2} = \sqrt{25 + (b-5)^2}$.

The distance from (0, b) to (3, -2) is
$\sqrt{(0-3)^2+(b+2)^2} = \sqrt{9 + (b+2)^2}$

These two distances are equal to each other, so
$\sqrt{25 + (b-5)^2} = \sqrt{9 + (b+2)^2} \\ 25 + (b-5)^2 = 9 + (b+2)^2 \\ 25 + b^2-10b+25 = 9 + b^2 + 4b + 4 \\ -14b = -37 \\ b = 37/14$

Answer 2

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