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Question

Let A,B,C,D be four concyclic points in order in which AD:AB=CD:CB. If A,B,C are represented by complex numbers a,b,c, then vertex D can be represented as

A
2ac+b(ac)a+c+2b
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B
2ac+b(a+c)a+c+2b
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C
2ac+b(a+c)a+c2b
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D
2acb(a+c)a+c2b
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Solution

The correct option is D 2ac−b(a+c)a+c−2b​​ Let complex number representing point ′D′ is d and ∠DAB=θ So, ∠BCD=π−θ. Now, applying rotation formula on A and C we get b−ad−a=ABADeiθ and d−cb−c=CDCBei(π−θ) Multiplying these two, we get (b−ad−a)(d−cb−c)=AB×CDAD×CBeiπ ⇒d(b−a)−c(b−a)d(b−c)−a(b−c)=−1 (∵ADAB=CDCB) ⇒d=2ac−b(a+c)a+c−2b

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