Question

In an Argand plane $z_1,z_2$ and $z_3$ are, respectively, the vertices of an isosceles triangle $ABC$ with $AC=BC$ and $\angle CAB=\theta$. If $z_4$ is the centre of triangle, then the value of $AB\times \frac{AC}{(IA{)}^2}$ is

• A. $\frac{(z_2-z_1)(z_3-z_1)}{(z_4-z_1{)}^2}$
• B. $\frac{(z_2-z_1)(z_1-z_3)}{(z_4-z_1{)}^2}$
• C. $\frac{(z_4-z_1)}{(z_2-z_1)(z_3-z_1)}$
• D. none of these

Answer 1

The correct option is A $\frac{(z_2-z_1)(z_3-z_1)}{(z_4-z_1{)}^2}$

$\frac{AB\times AC}{(IA{)}^2}=\frac{AB}{IA}\times \frac{AC}{IA}$

$\angle IAB=\frac{\theta }{2}$, $\angle IAC=\frac{\theta }{2}$

$\frac{z_2-z_1}{z_4-z_1}=\frac{\mid z_2-z_1\mid }{\mid z_4-z_1\mid }{e}^{-\frac{i\theta }{2}}$

and

$\frac{z_3-z_1}{z_4-z_1}=\frac{\mid z_3-z_1\mid }{\mid z_4-z_1\mid }{e}^{\frac{i\theta }{2}}$

Multiplying,

$\frac{z_2-z_1}{z_4-z_1}\frac{z_3-z_1}{z_4-z_1}=\frac{z_2-z_1}{z_4-z_1}\frac{z_3-z_1}{z_4-z_1}$

$\Rightarrow$ $\frac{(z_2-z_1)(z_3-z_1)}{(z_4-z_1{)}^2}=\frac{AB\times AC}{I{A}^2}$