Find the coordinate of third point of equilateral triangle. I have two points A and B whose coordinates are $(3,4)$ and $(-2,3)$ The third point is C. We need to calculate its coordinates.
I think there will be two possible answers, as the point C could be on the either side of line joining A and B.
Now I put AB = AC = BC.
We calculate AB by distance formula : $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
= $ \sqrt{5^2 + 1^2} = \sqrt{26}$
I plug this value into the distance of AC and BC and finally equate to get :
$5x + y = 6$
Now what can I do? There are two variables, I am getting equation of a line! How can I solve this?

Calculate slope of AB (m), distance AB = s
Point A is at (h,k)
tan-1 m = angle N
s x sin (60 + angle N) = y
s x cos (60 + angle N) = x
Third point is at x+h, y+k
Use (180 + 60 + angle N) for second answer
and h,k is now at point B
