Question

A rectangle $ABCD$, where $A(0,0),B(4,0),C(4,2),D(0,2)$, undergoes the following transformation successively :
$\left(i\right)f_1(x,y)\to (y,x)$
$\left(ii\right)f_2(x,y)\to (x+3y,y)$
$\left(iii\right)f_3(x,y)\to (\frac{x-y}{2},\frac{x+y}{2})$
Then the final figure is :

• A. square
• B. parallelogram
• C. rhombus
• D. rectangle

Answer 1

The correct option is B parallelogram

Clearly, $A$ will remain as $(0,0);f_1$ will make $B$ as $(0,4),f_2$ will make it $(12,4)$ and $f_3$ will make it $(5,9)$. Finally, $f_1$ will make $D$ as $(2,0),f_2$ will make it $(2,0)$ and $f_3$ will make if $(1,1)$. So finally we get $A\equiv (0,0),B\equiv (4,8),C\equiv (5,9),D\equiv (1,1)$. Therefore,
$m_{AB}=\frac{8}{4},m_{BC}=\frac{9-8}{5-4}=1$
$m_{CD}=\frac{9-1}{5-1}=\frac{8}{4}$
$m_{AD}=1,m_{AC}=\frac{9}{5},m_{BD}=\frac{8-1}{4-1}=\frac{7}{3}$
Hence, the final figure will be parallelogram.