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Lets say you have 3 points in 2D, which are non-collinear: a,b and c.

How can you find a 4th point, when you have the distance from the 4th point to a, b and c ?

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The formula for distance is [(x-x1)^2+(y-y1)^2+(z-z1)^2]^(1/2). Just plug in your known points and you have three linear equations in three variables. – nw. Dec 22 '11 at 22:37
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Is the fourth point to be in the same plane as the other three? Then for almost all combinations $(d_1,d_2,d_3)$ of distances, there will not be such a fourth point. – André Nicolas Dec 22 '11 at 23:53

migrated from stackoverflow.com Dec 22 '11 at 23:30

1 Answer

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For each point a,b,c, draw a circle with center at that point, having a radius equal to its distance to 4th point. The 4th point is where all circles intersect.

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Thanks! this should work – user1112570 Dec 23 '11 at 0:00

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