Of the three lines $x+\sqrt3y=0,x+y=1$ and $x-\sqrt3y=0$,two are equations of two altitudes of an equilateral triangle Of the three lines $x+\sqrt3y=0,x+y=1$ and $x-\sqrt3y=0$,two are equations of two altitudes of an equilateral triangle.The centroid of the equilateral triangle is
$(A)(0,0)\hspace{1cm}(B)\left(\frac{\sqrt3}{\sqrt3-1},\frac{-1}{\sqrt3-1}\right)\hspace{1cm}(C)\left(\frac{\sqrt3}{\sqrt3+1},\frac{1}{\sqrt3+1}\right)\hspace{1cm}(D)$none of these
All options (A),(B),(C) appears to be the answers.But correct answer is given (A).It is not given which two are altitude equations and which are not.Please guide me.

Notice, we have the following equations of the lines $$x+\sqrt 3y=0\iff y=\frac{-1}{\sqrt 3}x\tag 1$$
$$x+y=1\iff y=-x+1\tag 2$$
$$x-\sqrt 3y=0\iff y=\frac{1}{\sqrt 3}x\tag 3$$
Let the slopes of the above lines be denoted by $m_1=-\frac{1}{\sqrt 3}$, $m_2=-1$ & $m_3=\frac{1}{\sqrt 3}$ then the angles between them are calculated as follows 


*

*The angle between lines (1) & (2)  $$\theta_{12}=\tan^{-1}\left|\frac{m_1-m_2}{1+m_1m_2}\right|$$ 
$$=\tan^{-1}\left|\frac{-\frac{1}{\sqrt 3}-(-1)}{1-\frac{1}{\sqrt 3}(-1)}\right|=\tan^{-1}\left|2-\sqrt 3\right|=15^\circ$$

*The angle between lines (2) & (3)  $$\theta_{23}=\tan^{-1}\left|\frac{-1-\frac{1}{\sqrt 3}}{1+(-1)\frac{1}{\sqrt 3}}\right|=\tan^{-1}\left|2+\sqrt 3\right|=75^\circ$$

*The angle between lines (1) & (3)  $$\theta_{13}=\tan^{-1}\left|\frac{-\frac{1}{\sqrt 3}-\frac{1}{\sqrt 3}}{1-\frac{1}{\sqrt 3}\frac{1}{\sqrt 3}}\right|=\tan^{-1}\left|\sqrt 3\right|=60^\circ$$
We know that the angle between any two altitudes of an equilateral triangle is $60^\circ$ or $120^\circ$ Hence, the lines (1) & (3) are the altitudes of an equilateral triangle hence the centroid of the triangle is the intersection point of (1) & (3) i.e. origin $(0, 0)$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{Centriod of equilateral triangle}\equiv\color{blue}{(0, 0)}}}$$
Hence, option (A) $(0, 0)$ is correct.
