Question

The centroid of an equilateral triangle is origin. If the two vertices $A,B$ of the equilateral triangle lie on the line $x+y=2\sqrt{2}$ , then $AB$ equals $k\sqrt{3}$ , where $k$ equals:

Answer 1

$OD=2$
$D\left(\sqrt{2} , \, \sqrt{2}\right)$
$BD=AD=2\sqrt{3}$

$\therefore $ $A, \, B=\left(\sqrt{2} \pm \left(\frac{- 1}{\sqrt{2}}\right) 2 \sqrt{3} , \sqrt{2} \pm \left(\frac{1}{\sqrt{2}}\right) 2 \sqrt{3}\right)$
$=\left(\sqrt{2} - \sqrt{6} , \, \sqrt{2} + \sqrt{6}\right)$ and $\left(\sqrt{2} + \sqrt{6} , \, \sqrt{2} - \sqrt{6}\right)$
$\Rightarrow $ $AB=4\sqrt{3}$