Explanation

(d)
The point of intersection of $x + 3y = 2\ and\ \ x^{2} + xy - y^{2} = 1$ is given by
$(2 - 3y)^{2} + (2 - 3y)y - y^{2} = 1$
$\Rightarrow \ \ 4 + 9y^{2} - 12y + 2y - 3y^{2} - y^{2} = 1$
$\Rightarrow \ 5y^{2} - 10y + 3 = 0$
$\therefore\ \ y = \frac{10 \pm \sqrt{100 - 60}}{2 \times 5}$
$= \frac{10 \pm \sqrt{40}}{10}$
$\therefore\ x = 2 - 3\left( \frac{10 \pm \sqrt{40}}{10} \right)$
$= \frac{- 10 \mp \sqrt{40}}{10}$
$\therefore$ Points of intersection are
$A\left( - 1 - \frac{\sqrt{40}}{10},1 + \frac{\sqrt{40}}{10} \right)\$
and $B\left( - 1 + \frac{\sqrt{40}}{10},1 - \frac{\sqrt{40}}{10} \right)\$
$\therefore$ Mid point of$\ AB$ is $( - 1,1)$