Point P $(2, 4)$ is translated through a distance$3\sqrt{2}$ units measured parallel to the line$y - x -1 = 0$, in the direction of decreasing ordinates, to reach at$Q.$$If R$is the image ofQ unth respect to the line$y - x -1 = 0$, then coordinates of $R$ are

The equation of a line through P (2, 4) and parallel to y - x - 1 = 0 is

$\frac{x-2}{\cos \pi /4}=\frac{y-4}{\sin \pi /4}$

The coordinates of Q are given by

$\frac{x-2}{\cos \pi /4}=\frac{y-4}{\sin \pi /4}=-3\sqrt{2}\Rightarrow x=-1,y=1$

Thus, the coordinates of Q are $(-1, 1). R$s the image of Q with respect to the line$y - x -1 = 0.$

Therefore, coordinates of R are given by

$\begin{array}{r}\frac{x+1}{-1}=\frac{y-1}{1}=-2\frac{(1+1-1)}{(-1{)}^2+(1{)}^2}\\ \frac{x+1}{-1}=\frac{y-1}{1}=-1\Rightarrow x=0,y=0\end{array}$

Hence, the coordinates of R are (0, 0).