Explanation

(a)
Vertices are interception points of line
$x + y = 2\sqrt{2}$ …(i)
with $y = x\tan{(105{^\circ})\ or\ y = x\tan(165{^\circ})}$
(lines through centroid )
$y = - x\tan{75{^\circ}}$ …(ii)
$y = - x\tan{15{^\circ}}\$…(iii)
For the interception point of Eqs. (i) and (ii)
$x - x\left( 2 + \sqrt{3} \right) = 2\sqrt{2}$
$\Rightarrow - x\left( 1 + \sqrt{3} \right) = 2\sqrt{2}$
$\Rightarrow x = - \frac{2\sqrt{2}(1 - \sqrt{3})}{\left( 1 + \sqrt{3} \right)(1 - \sqrt{3})}$
$\Rightarrow x = \sqrt{2} - \sqrt{6}$
$\therefore y = - \left( \sqrt{2} - \sqrt{6} \right)\left( 2 + \sqrt{3} \right)$
$= - (2\sqrt{2} + \sqrt{6} - 2\sqrt{6} - 3\sqrt{2}$
$= \sqrt{2} + \sqrt{6}$
and its image about $y = x\$ is $(\sqrt{2} + \sqrt{6},\ \sqrt{2} - \sqrt{6})$