How many lines can be drawn in a plane such that they are equidistant from 3 non-collinear points?

@John Bentin has shown below that there are at least 3. Why are there no more than 3?

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    $\begingroup$ You can't have a line equidistant from a 3 points in a plane in general except in a special case. Can you see why? $\endgroup$ – user21436 Feb 2 '12 at 8:16
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    $\begingroup$ No, especially considering John's answer. $\endgroup$ – user57159 Feb 2 '12 at 8:40
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    $\begingroup$ If the three points are not collinear, then a candidate equidistant line must separate the points into two sets: either (1) three points vs no points, or (2) two points vs one point. Case (1) requires the points be collinear; case (2) --which can be accomplished in one of exactly three ways (because three non-collinear points must be distinct)-- is addressed by @John. $\endgroup$ – Blue Feb 2 '12 at 9:46
  • $\begingroup$ @DayLateDon Thanks Don. I see that by a simple counting argument, there are 3 ways to split the points into two sets. That combined with Ben's lemma shows that there are exactly 3 lines. $\endgroup$ – Joe Feb 2 '12 at 20:10

Assuming that the line is in the plane of the points, there are three. Consider a triangle ABC. Draw a line parallel to BC so that it is half way between A and BC. The other two are constructed similarly.

(Added for completeness) Consider any line in the plane equidistant from the points. Suppose for the present that it is not parallel to any side. Then it intersects all three sides of the triangle. The three points of intersection cannot all be internal: say the point D on the line lies on BC produced. But then C would be nearer the line than B, contradicting the given conditions. Hence the supposition can be ruled out. Therefore the line is parallel to a side of the triangle, and so it must be one of the three lines mentioned above.

  • $\begingroup$ This gives the construction to find the 3 lines. To prove that there are exactly 3 lines (i.e. there aren't any left out by this construction) consider the comment of Day Late Don on my question, and the answer written by Ben Bogosel and my comment on his answer. $\endgroup$ – Joe Feb 5 '12 at 1:51

You can use the following lemma:

If $\ell$ is equidistant from $A$ and $B$ then $\ell$ passes through the midpoint of $AB$ or $\ell$ is parallel to $AB$.

As a consequence, if $\ell$ is equidistant from $A,B$ and $\ell$ separates $A,B$ then $\ell$ passes through the midpoint of $AB$.

  • $\begingroup$ That's a helpful lemma. But why is it true? If $\ell$ is not parallel to $AB$, then it intersects the line through $AB$ at some point. The intersection point must either be through the midpoint, or closer to $A$, or $B$. Which, since it's a line, implies that the line passes closer to one than the other. $\endgroup$ – Joe Feb 2 '12 at 20:15

This had me stumped for a while, because I didn't realize that in geometry, a line is equidistant from a point not if the distance from every point on the line to the point stays the same, but if the line's perpendicular distance from the point stays the same. Here's an attempt at an explanation:

enter image description here

The three non-colinear points A, B, and C form a triangle. There is only one point that is equidistant from all three, marked as D... but there are three lines that are. To construct one of them:

  • Draw line BC
  • Draw a line from A that intersects BC perpendicularly at point A'.
  • Find the midpoint E of the line AA'.

And now the red line parallel to BC and going through E is equidistant to point A and line BC. You can also think of this line being equidistant to points A and A', or to line BC and its parallel (the red dotted line) passing through point A.

Repeat for lines AB, AC to get the other two lines.


A and b and c are not in the same line ,L parallel to $AB$ and $CP$ perpendicular on $AB$

If we want $A$ , $B$ and $C$ to be equidistant $L$ must be parallel to $AB$ and pass through midpoint of $CP$
Making line $BC$ we want $B$ and $C$ and $A$ to be equidistant to some line we make the same process draw line perpendicular on $BC$ and Line parallel to $BC$ and pass through midpoint of the line perpendicular on $BC$
so we will have three lines such that they are equidistant from these three points
why not more than three?
because we can't draw more than line that is parallel to $AB$ or $BC$ or $AC$
and pass through midpoint of the line perpendicular on the line


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