A rectangle $\mathrm{ABCD},$ where $\mathrm{A}\equiv \left(0,0\right),\mathrm{B}\equiv \left(4,0\right),\mathrm{C}\equiv \left(4,2\right),\mathrm{D}\equiv \left(0,2\right),$ undergoes the following transformation successively :

i. ${\mathrm{f}}_1\left(\mathrm{x},\mathrm{y}\right)\to \left(\mathrm{y},\mathrm{x}\right)$

ii. ${\mathrm{f}}_2\left(\mathrm{x},\mathrm{y}\right)\to \left(\mathrm{x}+3\mathrm{y},\mathrm{y}\right)$

iii. ${\mathrm{f}}_3\left(\mathrm{x},\mathrm{y}\right)\to \left(\left(\mathrm{x}-\mathrm{y}\right)/2,(\mathrm{x}+\mathrm{y}\right)/2)$

The final figure will be 

Clearly, A will remains as $\left(0,0\right);{\mathrm{f}}_1$will make B as(0, 4)

 ${\mathrm{f}}_2$ will make it $\left(12,4\right)$, and ${\mathrm{f}}_3$ will make it (4, 8)

 ${\mathrm{f}}_1$ will make C as $\left(2,4\right),{\mathrm{f}}_2$ will make it $\left(14,4\right),$

 and ${\mathrm{f}}_3$ will make it $\left(5,9\right)$. Finally, ${\mathrm{f}}_1$ will make D as $\left(2,0\right),{\mathrm{f}}_2$will make it $\left(2,0\right)$ 

and ${\mathrm{f}}_3$ will make it $\left(1,1\right).$ So, we finally get

 $\mathrm{A}\equiv \left(0,0\right)$, $\mathrm{B}\equiv \left(4,8\right)$, $\mathrm{C}\equiv \left(5,9\right)$ and $\mathrm{D}\equiv \left(1,1\right).$ Hence,

${\mathrm{m}}_{\mathrm{AB}}=\frac{8}{4},{\mathrm{m}}_{\mathrm{BC}}=\frac{9-8}{5-4}=1,{\mathrm{m}}_{\mathrm{CD}}=\frac{9-1}{5-1}=\frac{8}{4}$

${\mathrm{m}}_{\mathrm{AD}}=1,{\mathrm{m}}_{\mathrm{AC}}=\frac{9}{5},{\mathrm{m}}_{\mathrm{BD}}=\frac{8-1}{4-1}=\frac{7}{3}$

Hence, the final figure will be a parallelogram