<s>[INST] 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.
 [/INST] 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 $y$.
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