<PRE><B>Question 168890</B>
find the point on the y-axis that is equidistant from (6,1) (-2,-3)
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The point must have the form (0,y)
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Equation:
distance^2 = distance^2
(6-0^2 + (y-6)^2 = (0--2)^2 + (y--30^2

36 + y^2 -12y + 12 = 4 + y^2 +6y + 9

48 + y^2 - 12y = 13 + y^2 + 6y

18y = 35
 y = 35/18
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Ans: (0, 35/18)
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Cheers,
Stan H.
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