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

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

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

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  • $\begingroup$ 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. $\endgroup$ – nw. Dec 22 '11 at 22:37
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    $\begingroup$ 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. $\endgroup$ – André Nicolas Dec 22 '11 at 23:53
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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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