Question

If the square $ABCD$ where $A(0,0),B(2,0),C(2,2)$ and $D(0,2)$ undergoes the following transformations successively
$\left(i\right)$ image with respect to line $y=x$
$\left(ii\right)f_2(x,y)\to (x+3y,y)$
$\left(iii\right)f_3(x,y)\to (\frac{x-y}{2},\frac{x+y}{2}),$ then the final figure formed by the transformed vertices is

• A. square
• B. parallelogram
• C. rhombus
• D. None of these

Answer 1

The correct option is B parallelogram

$\because$ image of point $P(h,k)$ w.r.t line $y=x$ is $(k,h)$
$\therefore 1st$ transformation here is $f_1(x,y)\to (y,x)$
For $A(0,0)$
$f_1(0,0)\to (0,0)$
$f_2(0,0)\to (0+0,0)$
$f_3(0,0)\to (0,0)$
Final image of $A$ is $A^{\prime }\equiv (0,0)$

For $B(2,0)$
$f_1(2,0)\to (0,2)$
$f_2(0,2)\to (0+6,2)$
$f_3(6,2)\to (\frac{6-2}{2},\frac{6+2}{2})$
Final image of $B$ is $B^{\prime }=(2,4)$

For $C(2,2)$
$f_1(2,2)\to (2,2)$
$f_2(2,2)\to (2+3\times 2,2)$
$f_3(8,2)\to (\frac{8-2}{2},\frac{8+2}{2})$
Final image of $C$ is $C^{\prime }=(3,5)$

For $D(0,2)$
$f_1(0,2)\to (2,0)$
$f_2(2,0)\to (2+3\times 0,0)$
$f_3(2,0)\to (\frac{2-0}{2},\frac{2+0}{2})$
Final image of $D$ is $D^{\prime }=(1,1)$

$A^{\prime }=(0,0),B^{\prime }=(2,4),C^{\prime }=(3,5),D^{\prime }=(1,1)$
which represent a parallelogram as $A^{\prime }{D}^{\prime }=B^{\prime }{C}^{\prime },A^{\prime }{B}^{\prime }=C^{\prime }{D}^{\prime }$ and $A^{\prime }{C}^{\prime }\ne B^{\prime }{D}^{\prime }$