SHORTEST method for finding the third vertex of an equilateral triangle given two vertices? I know the usual method of calculating third vertex by using distance formula , forming quadratic and solving and stuff ,  but i was wondering if there was a shortcut method for finding it without much havoc ? 
Eg: Equilateral triangle ABC , A(3,2) and B(5,1) find third vertex C?
I tried by considering two circles centred at A,B respectively but ended up with another hopeless equation , how do i approach this kinds of questions analytically ?

Well given $A(x_1,y_1)$ and $B(x_2,y_2)$.
We know the midpoint is $M({x_1+x_2\over2},{y_1+y_2\over2})$.
Now we know the slope of $AB$ is $u=(x_2-x_1,y_2-y_1)$ so the perpendicular slope is $v=(y_1-y_2,x_2-x_1)$ and the unit perpendicular slope is $v_1=\sqrt{(y_1-y_2)^2+(x_2-x_1)^2}\cdot(y_1-y_2,x_2-x_1)$.
And hence the third vertex is $C=M+v_1\cdot \sqrt3$. (There are going to be two $C$s)
