Distance Formula Problem If two vertices of an equilateral triangle are $(1, -1)$ and $(-\sqrt{3}, - \sqrt{3})$, find the coordinates of the third vertex.
Step by Step procedure to get the answer.
Take $A=(1, -1)$, $B=(-\sqrt{3}, - \sqrt{3})$.
Let the third vertex be $C=(x, y)$.
The distance d between the two points $A=(x_1, y_1)$ and $B=(x_2, y_2)$ is given by the formula 
$$
d= \sqrt{ (x_2-x_1)^2 + (y_2 - y_1)^2 }
$$
So we get the distance $AB = 2 \sqrt{2}$
. Similarly we have to find the distance between BC & AC using distance formula where we get the equation in the form of x & y even though the distance is $2\sqrt{2}$ because it is equilateral triangle. But the correct solution is not able to get. Please help.

Equilateral triangle
$A=(1,-1)$
$B=(-\sqrt{3},-\sqrt{3})$
$C=(x,y)$
$|AB|=2\sqrt{2}$
height of an equilateral triangle:
$h=\frac{1}{2}\sqrt{3}a$
$a=|AB|$
$h=\frac{1}{2}\sqrt{3}.2\sqrt{2}=\sqrt{6}$
noting that
$|BO|=\sqrt{(0-(-\sqrt{3}))^2+(0-(-\sqrt{3}))^2}=\sqrt{6}$
O = (0,0) origin Cartesian Cordinate System
BO is foot of the altitude
O is midpoint de CA
$\frac{xC+1}{2}=0⇒xC=-1$
$\frac{yC+(-1)}{2}=0⇒yC=1$

C=(-1,1)
