SOLUTION: find the locus of all points in a plane the are equidistant from points (-3,0) and (0,3)

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Question 318204: find the locus of all points in a plane the are equidistant from points (-3,0) and (0,3)
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Use the distance formula,
D%5E2=%28x1-x2%29%5E2%2B%28y1-y2%29%5E2
Distance from (-3,0),
D1%5E2=%28x-%28-3%29%29%5E2%2B%28y-0%29%5E2
D1%5E2=%28x%2B3%29%5E2%2By%5E2
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Distance from (0,3),
D2%5E2=%28x-0%29%5E2%2B%28y-3%29%5E2=0
D2%5E2=x%5E2%2B%28y-3%29%5E2=0
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D1=D2
%28x%2B3%29%5E2%2By%5E2=x%5E2%2B%28y-3%29%5E2
x%5E2%2B3x%2B9%2By%5E2=x%5E2%2By%5E2-3y%2B9
cross%28x%5E2%29%2B3x%2B9%2Bcross%28y%5E2%29=cross%28x%5E2%29%2Bcross%28y%5E2%29-3y%2B9
3x%2B9=-3y%2B9
highlight%28y=-x%29
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All of the points on the line y=-x are equidistant from the two points.
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