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Question 168890: find the point on the y-axis that is equidistant from (6,1) (-2,-3)
Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! find the point on the y-axis that is equidistant from (6,1) (-2,-3)
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Interesting problem.
The point will be (0,y). x is zero since it's on the y-axis
Use Pythagoras:
d1^2 = (6-0)^2 + (1-y)^2
d2^2 = (-2-0)^2 + (-3 -y)^2
The two expressions are equal, so
36 + (1-y)^2 = 4 + (-3-y)^2
36 + 1 -2y +y^2 = 4 + 9 + 6y + y^2
24 -2y = 6y
24 = 8y
y = 3
The point is (0,3)
Check:
(6-0)^2 + (1-3)^2 =? (-2-0)^2 + (-3-3)^2
36 + 4 =? 4 + 36
40 = 40, so it's equidistant.
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! 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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