One vertex of the eqµilateral triangle with centroid at the origin and one side as $x + y - 2 = 0$ is 

Since the triangle is equilateral the centroid $G (0, 0)$ is also orthocentre or median $\mathrm{AD}$ is also its altitude and G divides it in the ratio $2: 1.$ The point $D(\alpha ,\beta )$ lies on $BC.$

 $\therefore \alpha +\beta -2=0 ...\left(1\right)$

If A be the point 

$(h,k),$ then 

$\begin{array}{l}\frac{2\alpha +h}{3}=0,\frac{2\beta +k}{3}=0\\ \therefore h=-2\alpha ,k=-2\beta ...\left(2\right)\end{array}$

AD is perpendicular to $BC$

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{h-\alpha }{k-\beta }\times (-1)=-1\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}h-\alpha =k-\beta \text{ or }-3\alpha =-3\beta \text{ by (2) }\end{array}$

or $\alpha =\beta =1\text{, by }\left(1\right)$ $\therefore h=-2,k=-2,\text{ by (2) }$

$\therefore A\text{ is }(-2,-2)\text{. }$