Explanation

(d)
Let $P\left( x_{1},\ y_{1} \right)$ be the image of point $Q(4,\ - 3)$
$Mid\ point\ of\ PQ\ is\ \left( \frac{x_{1} + 4}{2},\frac{y_{1} - 3}{2} \right).\ This\ point\ lies\ on\ y = x$
$\therefore\ \frac{x_{1} + 4}{2} = \frac{y_{1} - 3}{2} \Rightarrow x_{1} - y_{1} = - 7\ \ldots(i)$
$Slope\ of\ PQ = \frac{- 3 - y_{1}}{4 - x_{1}}\ and\ slope\ of\ y = x\ is\ 1$
$\because\$ $PQ$ is perpendicular to $y = x$
$\therefore\left( \frac{- 3 - y_{1}}{4 - x_{1}} \right)(1) = - 1$
$\Rightarrow y_{1} + x_{1} = 1$ …(ii)
On solving Eqs. (i) and (ii), we get $x_{1} = - 3,\ y_{1} = 4$