Equations of the sides an equilateral triangle with centroid at the origin and one side is $x+y=1$ 
An equilateral triangle has its centroid at the origin and one side is $x+y=1$. Find the equations of the other sides.

My Attempt

$$
OD=\frac{1}{\sqrt{2}}\implies OC=\sqrt{2}\implies C=(-1,-1)\\
m_{AB}=m_1=-1\implies\tan60=\sqrt{3}=|\frac{m+1}{1-m}|\\
\sqrt{3}-m\sqrt{3}=m+1\quad\text{or}\quad m\sqrt{3}-\sqrt{3}=m+1\\
m(1+\sqrt{3})=\sqrt{3}-1\quad\text{or}\quad m(\sqrt{3}-1)=\sqrt{3}+1\\
m=\frac{\sqrt{3}-1}{\sqrt{3}+1}\quad\text{or}\quad m=\frac{\sqrt{3}+1}{\sqrt{3}-1}\\
y+1=\frac{\sqrt{3}-1}{\sqrt{3}+1}(x+1)\quad\text{or}\quad y+1=\frac{\sqrt{3}+1}{\sqrt{3}-1}(x+1)
$$
But, my reference gives the solutions 
$$y+1=\frac{\sqrt{3}-1}{\sqrt{3}+1}(x+1)$$ 
and 
$$y\pm1=3+\sqrt{3}(x-1)$$
So, what are the actual solution to the problem and the easiest way to solve it?

The coordinates of the vertices of the triangle are found  to be $$A=(\frac {1+\sqrt 3}{2}, \frac {1-\sqrt 3}{2})$$
$$B=(\frac {1-\sqrt 3}{2}, \frac {1+\sqrt 3}{2})$$
$$C= (-1,-1)$$
The equations of lines passing through these points are 
$$ CA: y+1=\frac {\sqrt 3 -1}{\sqrt 3+1}(x+1)$$
$$ CB:   y+1=\frac {\sqrt 3 +1}{\sqrt 3-1}(x+1)$$
$$ AB: x+y=1$$
Thus your calculations are correct.
