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 &amp; AC using distance formula where we get the equation in the form of x &amp; 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.
Distance between two given points will be the side of the triangle $a=\sqrt{(1+\sqrt{3})^2+(\sqrt{3}-1)^2}=2\sqrt{2}$. Now you just need to draw two circles of radius $a$ and with centers at the known vertices. The equations of the circles are: 
$(x-1)^2+(y+1)^2=8$
$(x+\sqrt{3})^2+(y+\sqrt{3})^2=8$
The circles will intersect at two points and to find the coordinates you need to solve this system:$$ \begin{cases}
(x-1)^2+(y+1)^2=(x+\sqrt{3})^2+(y+\sqrt{3})^2 \\[2ex]
(x-1)^2+(y+1)^2=8
\end{cases}$$
$$ \begin{cases}
x^2-2x+1+y^2+2y+1=x^2+2\sqrt{3}x+3+y^2+2\sqrt{3}y+3 \\[2ex]
(x-1)^2+(y+1)^2=8
\end{cases}$$
$$ \begin{cases}
x^2-2x+1+y^2+2y+1=x^2+2\sqrt{3}x+3+y^2+2\sqrt{3}y+3 \\[2ex]
(x-1)^2+(y+1)^2=8
\end{cases}$$

$$ \begin{cases}
y(2-2\sqrt{3})=x(2+2\sqrt{3})+4 \\[2ex]
(x-1)^2+(y+1)^2=8
\end{cases}$$

$$ \begin{cases}
y=\frac{x(1+\sqrt{3})+2}{1-\sqrt{3}} \\[2ex]
(x-1)^2+(y+1)^2=8
\end{cases}$$Finally, we get a quadratic equation for $x$: $$x^2-2x+1+\Big(\frac{x(1+\sqrt{3})+2}{1-\sqrt{3}}+1\Big)^2=8 $$ The solutions are: $x=-1, x=2-\sqrt{3}$. We already have $y$ expressed via $x$ so just plug the $x$ values in that equation and find $y$.
After writing the equations of the two circles to find the points of intersection, I found the resulting quadratic equations to be too tedious for my patience.  Instead, I determined the (parametric) vector from $(-\sqrt{3}, - \sqrt{3} )$ to $(1, -1)$ to be

$$A = \left(-\sqrt{3} + \left(1+\sqrt{3}\right) T, -\sqrt{3} + \left(-1+\sqrt{3}\right) T\right)$$ where $0 \le T \le 1$.

I then added $(\sqrt{3}, \sqrt{3})$ to this to translate it to the origin, then multiplied it 
by the rotation matrix $\begin{bmatrix} \cos &amp;  -\sin \\ \sin &amp; \cos \end{bmatrix}$, using $\theta = +60^{\circ}$ and $T=1$.  Lastly, 
I subtracted $(\sqrt{3}, \sqrt{3})$ to translate back.  After MANY sign corrections, I got 
one of the two possible points to be $(-1,1)$, which checks out using the distance formula.
No quadratics to solve, just vectors to add and rotate.

(Sorry about my lack of proper math notation formatting; if I find this site to be interesting, I will learn MathJax.)