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. $\frac{2ac+b(a-c)}{a+c+2b}$
• B. $\frac{2ac+b(a+c)}{a+c+2b}$
• C. $\frac{2ac+b(a+c)}{a+c-2b}$
• D. $\frac{2ac-b(a+c)}{a+c-2b}$

Answer 1

The correct option is D $\frac{2ac-b(a+c)}{a+c-2b}$

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Let complex number representing point ${}^{\prime }{D}^{\prime }$ is $d$ and $\angle DAB=\theta$ So, $\angle BCD=\pi -\theta$.
Now, applying rotation formula on $A$ and $C$ we get
$\frac{b-a}{d-a}=\frac{AB}{AD}{e}^{i\theta }\text{and}\frac{d-c}{b-c}=\frac{CD}{CB}{e}^{i(\pi -\theta )}$
Multiplying these two, we get
$\left(\frac{b-a}{d-a}\right)\left(\frac{d-c}{b-c}\right)=\frac{AB\times CD}{AD\times CB}{e}^{i\pi }$
$\Rightarrow \frac{d(b-a)-c(b-a)}{d(b-c)-a(b-c)}=-1(\because \frac{AD}{AB}=\frac{CD}{CB})$
$\Rightarrow d=\frac{2ac-b(a+c)}{a+c-2b}$