Question 727251
The point (3, y) is equidistant from the point (10, 3) and (-5, -1). Find the value of y. 
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From (3,y) to (10,3):
{{{d^2 = (y-3)^2 + (3-10)^2) = y^2 - 6y + 58}}}
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From (3,y) to (-5,-1):
{{{d^2 = (y+1)^2 + (3+5)^2) = y^2 + 2y + 65}}}
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From (3,y) to (10,3):
{{{y^2 + 2y + 65 = y^2 - 6y + 58}}}
8y = -7
y = -7/8
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Another approach:
Find the perpedicular bisector of the line between the 2 points, then find y when x = 3.