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?

An easy way to solve it is as follows:
Because the side $x+y=1$ makes the 135° angle with the x-axis and the triangle is equilateral, the other two sides make the 135° - 60° = 75° and 135° - 120° = 15° angles with the x-axis, respectively. 
Therefore, their slopes are
$$ \tan(75^\circ) = 2+\sqrt{3}$$
$$ \tan(15^\circ) = 2-\sqrt{3}$$
and their equations passing the vertex C = (-1,-1) are
$$y+1=(2\pm \sqrt{3})(x+1)$$
Yours are the same after rationalizing the denominators.
