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 ?

midpoint of $AB$ = $(4, 1.5)$
slope of $AB = -\frac{1}{2} $
right bisector of $AB$ ... $(y-1.5)=2(x-4)$
parametrize bisector ... $$\vec \ell(t)= (4, 1.5) + \frac{t}{\sqrt 5}(1,2)  $$ where I have put in the factor of $\sqrt 5$ so that the distance from (4, 1.5) is given by $|t|$
now the altitude of an equilateral triangle is $\frac{\sqrt 3}{2}$ times the length of each side ( in this case $\sqrt 5$)
so the co-ordinates of the point $C$ will be given by $\vec\ell( \frac{\pm\sqrt{15}}{2})$
