Richard Donovan wrote: > Suppose there are more than 2 points? Is there a formula for working out > the point equidistant to the n points?
The points equidistant from P and Q lie on the perpendicular bisector of the line segment PQ. The points equidistant from P, Q and R lie on the intersection of the perpendicular bisectors of PQ, QR and PR. This intersection may not exist (e.g. P=(0,0), Q=(0,1), R=(0,2)). Alternatively, a point equidistant from P, Q, R is the center of the circle through P, Q, R (if it exists). This is easily generalizable to n points. Best wishes, John ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
