Question

The value of $(z_4-z_1{)}^2(\cos \theta +1)\sec \theta$ is

• A. $\frac{(z_2-z_1)(z_3-z_1)}{(z_4-z_1)}$
• B. $(z_2-z_1)(z_3-z_1)$
• C. $(z_2-z_1)(z_3-z_1{)}^2$
• D. $\frac{(z_2-z_1)(z_3-z_1)}{(z_4-z_1{)}^2}$

Answer 1

Answer: (B)

$(z_2-z_1)(z_3-z_1)$

$\frac{AB\times AC}{(IA{)}^2}=\frac{AB}{IA}\times \frac{AC}{IA}$
$\angle IAB=\frac{\theta }{2},\angle IAC=\frac{\theta }{2}$
$\frac{z_2-z_1}{z_4-z_1}=\frac{|z_2-z_1|}{|z_4-z_1|}{e}^{\frac{i\theta }{2}}$
and $\frac{z_3-z_1}{z_4-z_1}=\frac{|z_3-z_1|}{|z_4-z_1|}{e}^{\frac{i\theta }{2}}$
Multiplying,
$\frac{z_2-z_1}{z_4-z_1}\frac{z_3-z_1}{z_4-z_1}=\frac{|z_2-z_1|}{|z_4-z_1|}\frac{|z_3-z_1|}{|z_4-z_1|}$
$\Rightarrow \frac{(z_2-z_1)(z_3-z_1)}{(z_4-z_1{)}^2}=\frac{AB\times AC}{I{A}^4}$ (1)
From (1).
$\frac{(z_2-z_1)(z_3-z_1)}{(z_4-z_1{)}^2}=2{\left(\frac{AD}{IA}\right)}^2\left(\frac{AC}{AD}\right)(\because AB=2AD)$
$\Rightarrow (z_2-z_1)(z_3-z_1)=(z_4-z_1{)}^{2}2{\cos }^2\frac{\theta }{2}\sec \theta$
$=(z_4-z_1{)}^2(\cos \theta +1)\sec \theta$