A rectangle ABCD, where A(0,0),B(4,0),C(4,2),D(0,2), undergoes the following transformation successively : (i)f1(x,y)→(y,x) (ii)f2(x,y)→(x+3y,y) (iii)f3(x,y)→(x−y2,x+y2) Then the final figure is :
A
square
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B
parallelogram
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C
rhombus
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D
rectangle
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Solution
The correct option is B parallelogram Clearly, A will remain as (0,0);f1 will make B as (0,4),f2 will make it (12,4) and f3 will make it (5,9). Finally, f1 will make D as (2,0),f2 will make it (2,0) and f3 will make if (1,1). So finally we get A≡(0,0),B≡(4,8),C≡(5,9),D≡(1,1). Therefore, mAB=84,mBC=9−85−4=1 mCD=9−15−1=84 mAD=1,mAC=95,mBD=8−14−1=73 Hence, the final figure will be parallelogram.