Explanation

(d)
Let $\mathcal{Q}(x,\ y)$ be the image of the point $P(4,\ 1)$ to the line $y - x + 1 = 0$
Then, $P\mathcal{Q}$ is perpendicular to $y - x + 1 = 0$
$\therefore\ \frac{y + 1}{x - 4} \times 1 = - 1$
$\Rightarrow \ y + x = 4 + 1 = 5$ …(i)
$Also,\ mid\ point\ of\ P\mathcal{Q,\ }ie,\ \left( \frac{4 + x}{2},\frac{y + 1}{2} \right)\ lies\ on\ y - x + 1 = 0$
$\therefore\ \frac{y + 1}{2} - \frac{(4 + x)}{2} + 1 = 0$
$\Rightarrow \ y - x - 1 = 0\$ …(ii)
On solving Eqs. (i) and (ii), we get the required point (2, 3)