Question

$A\left(z_1\right),B\left(z_2\right)$ and $C\left(z_3\right)$ are the vertices of an isosceles triangle in anticlockwise direction with origin as in-centre. If $AB=AC$, then $z_2,z_1\text{and}k{z}_3\text{will form}(\text{where}k=\frac{|z_1{|}^2}{|z_2||z_3|})$

• A. $G.P.$
• B. $A.G.P.$
• C. None of these
• D. $A.P.$

Answer 1

The correct option is A $G.P.$


Let $\angle BAC=2\theta$
$\because AB=AC$
$\therefore \angle B=\angle C=\frac{1}{2}(\pi -2\theta )=\frac{\pi }{2}-\theta$
$\Rightarrow \angle AIC=\pi -(\theta +\frac{\angle C}{2})=(\frac{3\pi }{4}-\frac{\theta }{2}),\angle BIA=(\frac{3\pi }{4}-\frac{\theta }{2})$
$\Rightarrow \frac{z_3}{z_1}=\left|\frac{z_3}{z_1}\right|e^{(-i(\frac{3\pi }{4}-\frac{\theta }{2}))}...\left(1\right)$
$\text{and}\frac{z_2}{z_1}=\left|\frac{z_2}{z_1}\right|e^{i(\frac{3\pi }{4}-\frac{\theta }{2})}...\left(2\right)$

On multiplying equation $\left(1\right)$ and $\left(2\right)$, we get
$\frac{z_2{z}_3}{z_1^2}=\frac{|z_2||z_3|}{|z_1{|}^2}$
$\Rightarrow z_1^2=k{z}_2{z}_3$
Hence, $z_2,z_1\text{and}k{z}_3$ are in G.P.