AP EAPCET 23-Aug-2021 Shift 2 Paper

Given, square ABCD when A(0,0),B(2,0),C(2,2) , D(0,2) undergoes the following transformations.
(i) f1(x,y)→(y,x)
(ii) f2(x,y)→(x+3y,y)
(iii) f3(x,y)→(

x−y

2

,

x+y

2

)
By the transformation f1(x,y)→(y,x) , we get
A(0,0)→A(0,0)
B(2,0)→B(0,2)
C(2,2)→C(2,2) and D(0,2)→D(2,0)
Now, perform f2(x,y)→(x+3y,y)
So, A(0,0)→A(0,0) , B(0,2)→B(6,2)
C(2,2)→C(8,2),D(2,0)→D(2,0)
Next perform f3(x,y)→(

x−y

2

,

x+y

2

)
∴A(0,0)→A(0,0),B(6,2)→B(2,4)
C(8,2)→C(3,5),D(2,0)→D(1,1)
Thus, in the final figure ABCD with A(0,0),B(2,4) , C(3,5),D(1,1)
AB=√(2−0)2+(4−0)2=2√5
BC=√1+1=√2, AC=√9+25=√34
CD=√4+16=2√5,DA=√2
BD=√1+9=√10
∵AB=CD and BC=DA and diagonals are not equal.
⇒ It is a parallelogram.