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Question 584185: Find the point on the x-axis which is equidistant from (1,2) and (-5, -15)
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! Since the point is on the x-axis, the y-value is zero, so we can write is as (x,0)
The squares of the distances from this point to the other points are
(x-1)^2 + 2^2
(x+5)^2 + 15^2
Set them equal and solve for x:
(x-1)^2 + 2^2 = (x+5)^2 + 15^2
x^2 - 2x + 5 = x^2 + 10x + 25 + 225
12x = -245
x = -245/12
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