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.
 A: How about using complex numbers?
Let $x+iy$ be the point we want where $x,y\in\mathbb R$.
We have
$$x+iy=(3+4i)\left(\cos\left(\pm\frac{\pi}{3}\right)+i\sin\left(\pm\frac{\pi}{3}\right)\right),$$
i.e.
$$x+iy=3\cos\left(\pm\frac{\pi}{3}\right)-4\sin\left(\pm\frac{\pi}{3}\right)+i\left(3\sin\left(\pm\frac{\pi}{3}\right)+4\cos\left(\pm\frac{\pi}{3}\right)\right)$$
Hence, we have
$$(x,y)=\left(3\cos\left(+\frac{\pi}{3}\right)-4\sin\left(+\frac{\pi}{3}\right),3\sin\left(+\frac{\pi}{3}\right)+4\cos\left(+\frac{\pi}{3}\right)\right),\left(3\cos\left(-\frac{\pi}{3}\right)-4\sin\left(-\frac{\pi}{3}\right),3\sin\left(-\frac{\pi}{3}\right)+4\cos\left(-\frac{\pi}{3}\right)\right)$$
A: If complex are forbidden (are not in your program), you could consider solving trinomials.
You know that $x^2 + y^2 = 25$ and $(3-x)^2 + (4-y)^2 = 25$. Develop the second equality into $25 + x^2 + y^2 - 6x - 8y = 25$. Since $x^2 + y^2 = 25$, it turns into $6x + 8y = 25$, hence $x = \frac{25}{6} - \frac{4}{3}y$.
Now replace $x$ with this expression in the first equation and you will get a classic trinomial in $y$. Use the technique seen in class to solve it. You should get two solutions. You easily derive the value of $x$ then.
Your solutions should be something like $y = 2 \pm \frac{3\sqrt{3}}{2}$ once simplified.
A: 1351957
Given:
$P_0(x_0\mid y_0)$
$P_1(x_1\mid y_1)$ .
Object:
Find $P_2(x_2\mid y_2)$, such that $P_0P_1P_2$ mark an equilateral triangle.    
Solution:
$x_2=\frac{x_1+x_0\mp(y_1-y_0)\sqrt3}2$
$y_2=\frac{y_1+y_0\pm(x_1-x_0)\sqrt3}2$    
Choose the upper sign for anticlockwise,
lower sign for clockwise.
A: 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.
