# Thread: Point equidistant to 3 or more other points

1. ## Point equidistant to 3 or more other points

Hi!

Hope this is posted in the right section - if not please advise.

If there are 2 points, (x1 y1) and (x2 y2), the equidistant point is at the midpoint at location ((x1+x2)/2) ((y1+y2)/2)

eg if there are 2 points p and q at locations (1,2) and (3,0) the point equidistant is at (2,1) - the midpoint m

y
4 |
3 |
2 |p
1 | m
_ |__q_______ x
0 1 2 3 4

Suppose there are more than 2 points? Is there a formula for working out the point equidistant to the n points?

eg if there are 3 points p q r at locations (1,2) (3,0) and (3,2), the point equidistant is still at (2,1), but I can't see a way of computing this from the figures.

y
4 |
3 |
2 |p--r
1 | m
_ |__q_______ x
0 1 2 3 4 NB hyphens are spaces-board changes 2 spaces to 1

This is not a test question or homework crib - I am in my late fifties and just interested in mathematics!

The thought occured to me and I just cant see a solution for the general case Equidistant point of points (x1 y1) (x2 y2) (x3 y3) ... (xn yn)

2. Hello RSDonovan

Welcome to Math Help Forum!
Originally Posted by RSDonovan
Hi!

Hope this is posted in the right section - if not please advise.

If there are 2 points, (x1 y1) and (x2 y2), the equidistant point is at the midpoint at location ((x1+x2)/2) ((y1+y2)/2)

eg if there are 2 points p and q at locations (1,2) and (3,0) the point equidistant is at (2,1) - the midpoint m

y
4 |
3 |
2 |p
1 | m
_ |__q_______ x
0 1 2 3 4

Suppose there are more than 2 points? Is there a formula for working out the point equidistant to the n points?

eg if there are 3 points p q r at locations (1,2) (3,0) and (3,2), the point equidistant is still at (2,1), but I can't see a way of computing this from the figures.

y
4 |
3 |
2 |p--r
1 | m
_ |__q_______ x
0 1 2 3 4 NB hyphens are spaces-board changes 2 spaces to 1

This is not a test question or homework crib - I am in my late fifties and just interested in mathematics!

The thought occured to me and I just cant see a solution for the general case Equidistant point of points (x1 y1) (x2 y2) (x3 y3) ... (xn yn)

You're only partly correct in saying that the mid-point of the line joining two points is equidistant from the two points. This is true, but so are all the points on the perpendicular bisector of the line.

Therefore, to find a point that is equidistant from three points, draw the perpendicular bisectors of the lines joining two pairs of these points. Their point of intersection is equidistant from all three points (provided, of course, that they are not parallel).

This point is, of course, the centre of the circle passing through the three given points, and is always a unique point provided that the three points don't lie in a straight line. But in general, you won't be able to do this for four (or more) points, unless they happen to form a cyclic quadrilateral.

Perhaps I should have better said "point closest", rather than equidistant, then there's only one possible point, rather than the infinite number on the perpendicular, as you correctly state!

PS what part of the South Coast are you on? I'm in Hove.

4. Originally Posted by RSDonovan
...PS what part of the South Coast are you on? I'm in Hove.
Perhaps this will give you a clue!