<s>[INST] 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?
 [/INST] Rotating the coordinate system by $\pm \frac{2 \pi}{3}$ gives the new coordinates $(x', y')$ defined by
$$(x', y') = \left(\frac{-x \mp \sqrt{3} y}{2} , \frac{\pm \sqrt{3} x - y}{2}\right) .$$
So, rotating the line $x + y = 1$ gives the equation $x' + y' = 1$, and substituting the above transformation equation to write this in terms of $x, y$ gives:
$$1 = \left(\frac{-x \mp \sqrt{3} y}{2}\right) + \left( \frac{\pm \sqrt{3} x - y}{2}\right) = \frac{1}{2}\left[\left(-1 \pm \sqrt 3\right) x + \left(-1 \mp \sqrt 3\right) y\right] .$$
But rearranging shows that this equation is equivalent to your solution.
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