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

Answer: 4

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

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