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.
 A: 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.
