For example, if a=100 , b =200, c=300, then what should be the 4th point equidistant to all the 3 points ie a,b,c. Similarly, is there is any solution for nth point equidistant from n-1 points. The values are 1 dimensional points.

  • $\begingroup$ No, unless you go to higher dimensional space, you cannot get the equidistant-point. $\endgroup$ – Ankit Seth Jul 18 '18 at 7:34
  • $\begingroup$ how can we go to higher dimensional space....Is there any mathematical formula or method to do ? $\endgroup$ – Sheetal Jul 18 '18 at 9:25
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    $\begingroup$ Your problem is purely mathematical, I think you would get more suggestions on mathoverflow.net $\endgroup$ – Ankit Seth Jul 18 '18 at 9:36

As Ankit said in comments, this is not possible in your concrete case, as you cannot just expand 1d point coordinates to 2d without inventing some information.

In 1d, where each point it represented by one single number, it would only be possible to find the 3rd point, equidistant to two original points. If you imagine the two points on a line, the third point would then just be the centre of those two points, which also happens to be the mean of those two points.

For three points, you would require a 2d space, i.e. each point must be a vector of length two: so basically an (x, y) coordinate pair.

If you did have (x,y) coordinates for three unique points, they would form a triangle, and the equidistant position (i.e. your fourth point) is called the circumcenter, and it found by finding the centre of each of the sides of the triangle, then drawing a line through each, which is perpendicular to its corresponding side of the triangle. These three lines will cross at the circumcenter - and it is not necessarily inside the triangle!


Equidistant point example for triangle


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