Find a point so that the triangle is equilateral We have O(0,0), A(3,4) and B(x,y). Find $x,y\in{R}$ so that the OAB triangle is equilateral.
I tried using the fact that the median is also the altitude(height) of the equilateral triangle. I calculated that distance from O to A(it's 5), meaning all the sides of the triangle have to be 5. Then, I computed the line equation for OA. I set M as the middle point of line OA. I then computed the equation for the line BB' which would be perpendicular on line OA(I knew the slope, since I knew the slope of line OA and I knew that M was a point on the line) and pass through point M(the median/height of the triangle).
After all this I could say that:
$$d(O,B)=\sqrt{x^2+y^2}=5$$
and $$d(A,B)=\sqrt{(3-x)^2+(4-y)^2}=5$$
The solutions I get from this equation system should be in the form of x,y, where x and y are the coordinates of point B. The system should return me more than one solution, but in order for it to be right I must also check that the points I get are also on line BB'.
Anyway, that was my thought process. The problem is that I'm getting into really 'icky' equations that I deem hard to solve, giving me really odd solutions for my coordinates. 
Isn't there a better way to solve this problem? I've been busting my brain on this one for about 3 hours now.

Using rotation transformation rotate the vector on x-axis,  $ 5\,i, $  by angle
$$ \pi/3 + tan^{-1} \frac 43 $$ 
for vector tip to get to required position $(x,y).$ It simplifies easily.
