$$\begin{gathered} \text{ The centroid of an equilateral triangle is }(0,0)\text{ . If two vertices of the triangle lie on }x+y=2\sqrt{2}\text{ ,} \\ \text{ then one of them will have its coordinates } \end{gathered}$$

Let the vertices B and C lie on the given line. Then, $OD=\frac{2\sqrt{2}}{\sqrt{2}}=2 (perpendicular dis\tan ce from origin=\frac{\left|c\right|}{\sqrt{a^2+b^2}})$

 $$\begin{gathered} Let equation od OD be y=mx, \sin ce slope is perpendicular to x+y=2\sqrt{2 } is 1, \therefore equation is x=y \\ solving the two equations we get D (\sqrt{2},\sqrt{2}) \end{gathered}$$=

$\text{ Also, }BD=OD\times \tan {60}^{\circ }=2\sqrt{3}\text{ for the coordinates of }B\text{ and }C\text{ . }$

Slope of the given line is -1, slope of OD is 1 ,hence angle of inclination is ${45}^0$

Using parametric equation of line, we get 

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{x-\sqrt{2}}{1/\sqrt{2}}=\frac{y-\sqrt{2}}{1/\sqrt{2}}=\pm 2\sqrt{3}\\ \text{ or }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}C\equiv (\sqrt{2}+\sqrt{6},\sqrt{2}-\sqrt{6})\\ \text{ and }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}B\equiv (\sqrt{2}-\sqrt{6},\sqrt{2}+\sqrt{6})\end{array}$