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Question

# If a point in argand plane A(3,2) rotated through B(1,1) about π4 to obtain C, then area of △ABC is

A
54 sq. unit
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B
52 sq. unit
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C
502 sq. unit
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D
504 sq. unit
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Solution

## The correct option is D √504 sq. unitLet zA=3+2i,zB=1+i and after rotating zA the complex number be zC then |zA−zB|=|zC−zB| Using rotation property we have zC−zBzA−zB=|zC−zB||zA−zB|eiπ/4⇒zC−zB2+i=1√2+i√2⇒zC=(1+i)+(1√2+3i√2)∴zC=1+√2√2+(3+√2)i√2 Now AC=|zA−zC|=1√2√(2√2−1)2+(√2−3)2 =√10−5√2 unit Now BD=∣∣∣zB−(zA+zC2)∣∣∣ [∵AB=CB] BD=√10+5√22 unit Hence required area =AC×BD2 =√504 sq. unit (Here we have considers anticlockwise direction however clock wise direction will also give same answer) Alternte Solution: Using sine rule required area will be =12(AB)(BC)sin(π/4) =12√2(AB)2=12√2[22+12] =√504 sq. unit

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