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.