Given, square $A B C D$ when $A(0,0), B(2,0), C(2,2)$,$D(0,2)$ undergoes the following transformations.
(i) $f_1(x, y) \rightarrow(y, x)$
(ii) $f_2(x, y) \rightarrow(x+3 y, y)$
(iii) $f_3(x, y) \rightarrow\left(\frac{x-y}{2}, \frac{x+y}{2}\right)$
By the transformation $f_1(x, y) \rightarrow(y, x)$, we get
$\begin{aligned}
& A(0,0) \rightarrow A(0,0) \\
& B(2,0) \rightarrow B(0,2) \\
& C(2,2) \rightarrow C(2,2) \text { and } D(0,2) \rightarrow D(2,0)
\end{aligned}$
Now, perform $f_2(x, y) \rightarrow(x+3 y, y)$
So, $A(0,0) \rightarrow A(0,0), B(0,2) \rightarrow B(6,2)$
$\begin{aligned} & C(2,2) \rightarrow C(8,2), D(2,0) \rightarrow D(2,0) \\ & \text { Next perform } f_3(x, y) \rightarrow\left(\frac{x-y}{2}, \frac{x+y}{2}\right)\end{aligned}$
$\begin{aligned}
& \therefore A(0,0) \rightarrow A(0,0), B(6,2) \rightarrow B(2,4) \\
& C(8,2) \rightarrow C(3,5), D(2,0) \rightarrow D(1,1)
\end{aligned}$
Thus, in the final figure $A B C D$ with $A(0,0), B(2,4)$, $C(3,5), D(1,1)$
$\begin{aligned} & A B=\sqrt{(2-0)^2+(4-0)^2}=2 \sqrt{5} \\ & B C=\sqrt{1+1}=\sqrt{2}, A C=\sqrt{9+25}=\sqrt{34} \\ & C D=\sqrt{4+16}=2 \sqrt{5}, D A=\sqrt{2} \\ & B D=\sqrt{1+9}=\sqrt{10}\end{aligned}$
$\because A B=C D$ and $B C=D A$ and diagonals are not equal.
$\Rightarrow$ It is a parallelogram.