Finding the third vertex of an equilateral triangle If two vertices of an equilateral triangle are (0,0) and (3,√3) then find the third vertex.
 The first thing I did was calculated the distance of the given points and tried to make an equation including the third vertex but instead of being simplified, the term went on being complex.
 A: Use complex numbers. Multiply $3+i\sqrt{3}$ by $e^{i \pi /3}$ to get the third vertex. This is because the origin is already one of the vertices.
The other possible vertex you can get by rotating counterclockwise, i.e. multilying by $e^{-i\pi /3}$
A: Using matrices, going anticlockwise the other vertex is $$\left(\begin {matrix}\cos 60 &-\sin 60 \\ \sin 60 & \cos 60\end{matrix}\right)\left (\begin{matrix}3\\ \sqrt{3}\end{matrix}\right)$$
For the clockwise possibility, multiply by the inverse matrix instead i.e. replace $60$ with $-60$ in the above.
A: If you notice that the angle from the $x$-axis to $(3,\sqrt3)$ is $30°$, then you will see that the third vertex must be either on the $y$-axis or the reflection of $(3,\sqrt3)$ over the $x$-axis. (Note that $(3,\sqrt3)$ is a scalar multiple of $(\frac{\sqrt3}2,\frac12)$.)
If you didn't see that, then here's a general geometric solution. Let $A=(0,0)$ and $B=(3,\sqrt3)$. We can find the third vertex by looking at the perpendicular bisector of $AB$ and locating the points on it at the right distance from $A$. The midpoint $M$ of $AB$ is $(\frac32,\frac{\sqrt3}2)$. If $P$ is the third vertex, then the distances $\overline{MP}$ and $\overline{AM}$ are in a ratio of $\sqrt3:1$. Therefore, since $\overline{AM} = \sqrt3$, we have $\overline{MP}=\sqrt3\cdot\sqrt3=3$.
The unit vector perpendicular to $AB$ is
$\frac1{2\sqrt3}(-\sqrt3,3) = \left(-\frac12,\frac{\sqrt3}2\right)$. Therefore, the two choices for $P$ are:
$$\left(\frac32,\frac{\sqrt3}2\right) \pm 3\left(-\frac12,\frac{\sqrt3}2\right)$$
giving $(0,2\sqrt3)$ and $(3,-\sqrt3)$.
