I have the coordinates (x,y) of 2 points. I want to build the third point so that these 3 points make an equilateral triangle.
How can I calculate the third point?
After reading the posts (specially vkit's) I produced this simple piece of code which will do the trick for one direction (remember that there are two points). The modification for the other case shold be trivial.
Let's call your two points A and B. Bisect AB, call this point C. Find the slope of AB (YA-YB / XA-XB), call it m. Find the perpendicular to that (-1/m) and call it m2. Then compute a segment CD whose length is sin(60) * length(AB), at the slope m2 (there will be two such points, one to each side of AB). ABD is then your equilateral triangle.
That, obviously, is a "constructive" method. You should also be able to do it by solving a set of linear equations. I haven't tried to figure out the right system of equations for this case, but this approach tends to be somewhat more stable numerically, and has fewer special cases (e.g., with the constructive version, a slope of 0 has to be dealt with specially).
For BlueRaja's challenge go to end of post:
Answer using translation and rotation:
Says points are P(x1,y1) and Q(x2,y2).
Since it is graphics, you can use tranforms to get the point.
First translate axes so P is the origin. Next rotate Q around P by 60 degrees (or -60 to get the other possible point).
This gives you the coordinates of the third point say R, when P is the origin.
Translate back and there you have it.
You can use standard graphics API which take care of precision etc issues for you. No headaches.
Of course you could do the math and actually come up with a formula and use that and that might be faster, but then the question might get closed as off-topic ;-)
To take up BlueRaja's challenge: Here is a method which does not use trigonometry.
Given points P(x1,y1) and Q(x2,y2) Say the point we need (R) to find is (x3,y3).
Let T be midpoint of PQ.
We know the area of triangle PQR (as it is equilateral and we know the side)
and we know the area of triangle PRT (1/2 the earlier area).
Now area of a triangle can be written as a determinant having the co-ordinates as entries:
We have two such equations (which are linear), solve for x3 and y3.
You could rotate the second point 60° around first to find the location of the third point.
Something like this:
//find offset from point 1 to 2 dX = x2 - x1; dY = y2 - y1; //rotate and add to point 1 to find point 3 x3 = (cos(60°) * dX - sin(60°) * dY) + x1; y3 = (sin(60°) * dX + cos(60°) * dY) + y1;