Construct an equilateral triangle given a line segment

Question

Given two vertices that make up a line segment (x1,y1) and (x2,y2), how can we find the third vertice that would make up an equilateral triangle? I'm looking to derive the third vertex algebraically, rather than drawing circles and whatnot.

Answer 1

First of all, you should know that there are two points that would make it an equilateral triangle. Also note that the third point is on a line perpendicular to the segment between the first two points. Call the first two points A and B. If you average their x and y values, you will get midpoint M, which is right between the first two points. Find the slope of the line segment AB, and use a negative reciprocal to find the perpendicular line that passes through point M. You should also know that the altitude of an equilateral triangle is $sqrt(3)$ times 1/2 its base. Then, find the two points C and D that are on the line containing M and perpendicular to AB that are $sqrt(3) * b/2$, where b is length AB.

Answer 2

By simple transformation, we can reasonably assume that the line lies on the x-axis with one end at (0, 0) and the other at (h, 0) where h is the length of the line segment.

The computation is included in the following diagram.

Edit: The image vertex is at (h/2, –√3h/6).