The point $(4,1)$ undergoes the following three transformations successively.

i) Reflection about the line $x=y$

ii) Translation through a distance 2 units along the positive direction of X - axis.

iii) Rotation through an angle $\frac{\pi }{4}$ about the origin in the anti clockwise direction.

The final position of the point is given by the coordinates

Given, let $P=(4,1)$

i) Reflection (or image) of $P(4,1)$ w.r.t to the line $x+(-1).y+0=0$ is $Q$(say).

$\therefore$ By image formula we have$=(h,k)$

$\frac{h-4}{1}=\frac{k-1}{1}=\frac{-2(4-1+0)}{{\left(1\right)}^2+{(-1)}^2}=-3$

$\Rightarrow h=-3+4=1;k=3+1=4$

$\Rightarrow Q=(h,k)=(1,4)$

ii) Given, a point $Q(1,4)$ translated through a distance 2 units along the positive direction of x-axis.

Now, let new position of Q be R

Given, R lies on a line parallel to positive x-axis and the distance between Q, R is 2 units.

$\Rightarrow QR=2\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}}R=(x,4)\Rightarrow {(x-1)}^2+{\left(0\right)}^2=4,R\in Q,$

$\Rightarrow x-1=2\Rightarrow x=3$

$\therefore R=(3,4)$

iii) Point $R=(3,4)$(In old system) ;  $\alpha =\frac{\pi }{4}=$angle of rotation

$O=(0,0)=$origin

$OR=\sqrt{9+16}=5$

$\therefore OR=5\text{units}$

Given $\angle ROS=45°=\alpha =$angle of rotation

Let $\angle XOR=\theta$. Here X is a point on positive x-axis

$OR=OS=\sqrt{9+16}=5$

Let $\angle XOR=\theta$

Given $\angle ROS=45°=\alpha \left(\text{say}\right)$

$\angle XOS=\angle XOR+\angle ROS$

$=\theta +45°$

$\therefore \cos (\theta +45°)=\cos \theta \cos 45°-\sin \theta \sin 45°$

$=\left(\frac{3}{5}\right)\left(\frac{1}{\sqrt{2}}\right)-\left(\frac{4}{5}\right)\left(\frac{1}{\sqrt{2}}\right)$

$=\frac{-4+3}{5\sqrt{2}}$

$=\frac{-1}{5\sqrt{2}}$

$\sin (\theta +45°)=\sin \theta \cos 45°+\cos \theta \sin 45°$

$=\frac{4}{5}\left(\frac{1}{\sqrt{2}}\right)+\frac{3}{5}\left(\frac{1}{\sqrt{2}}\right)=\frac{7}{5\sqrt{2}}$

Point $S$ lies on the line $\overleftrightarrow{OS}$ such that $OS=r=5$

$\therefore$Parametric equation of line $\overleftrightarrow{OS}$ are

$x=x_1\pm r\cos \theta ;y=y_1\pm r\sin \theta$

$\Rightarrow x=0\pm 5\left(\frac{-1}{5\sqrt{2}}\right);y=0\pm 5\left(\frac{7}{5\sqrt{2}}\right)$

$\Rightarrow x=\pm \frac{(-1)}{\sqrt{2}};y=\pm \frac{7}{\sqrt{2}}$

$\therefore S=(-\frac{1}{\sqrt{2}},\frac{7}{\sqrt{2}})\left(\text{or}\right)S=(\frac{1}{\sqrt{2}},\frac{-7}{\sqrt{2}})$

Since, $\cos (\theta +45°)<0,\sin (\theta +45°)>0$, we have $(\theta +45°)\in Q_2$

$\Rightarrow S$lies in 2^nd quadrant $\left(Q_2\right)$

$S=(-\frac{1}{\sqrt{2}},\frac{7}{\sqrt{2}})$ is the required point