Equations of the sides an equilateral triangle with centroid at the origin and one side is $x+y=1$

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.