Question

In an argand plane $z_1,z_{2}and{z}_3$ are respectively the vertices of an isoceles $a△ABC$ with $AC=BC$ and $\angle CAB=\theta$. If $z_4$ is incentre of triangle. Then:
(Read the following carefully and answer the following questions) The value of $(z_4-z_1{)}^2(1+\cos \theta )\sec \theta$ is

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

Answer 1

The correct option is A $(z_2-z_1)(z_3-z_1)$

$AC=BC$ and $\angle CAB=\theta$

$z_4$ is the in center of the circle,

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

We have $arg\left(\frac{z_2-z_1}{z_4-z_1}\right)=-arg\left(\frac{z_3-z_1}{z_4-z_1}\right)$

By applying coni's method,

$\Rightarrow \frac{(z_2-z_1)}{|z_2-z_1|}.e^{iQ/2}=\frac{(z_4-z_2)}{|z_4-z_2|}$

$\frac{(z_1-z_2)}{|z_1-z_2|}.e^{iQ/2}=\frac{(z_1-z_4)}{|z_1-z_4|}$

By applying the law of since to $△ACI$ where $I\left(zu\right)$ gives

$\Rightarrow \frac{|z_4-z_1|}{\sin (\frac{\pi }{2}-\theta )}=\frac{|z_3-z_1|}{\sin (\frac{\pi }{2}+\frac{\theta }{2})}$

$\Rightarrow \frac{z_2-z_1}{z_4-z_1}\frac{z_3-z_1}{z_4-z_1}=\left|\frac{z_2-z_1}{z_4-z_1}\right|\left|\frac{z_3-z_1}{z_4-z_1}\right|$

$=\frac{2{\cos }^2\frac{\theta }{2}}{\cos \theta }=\frac{\cos \theta +1}{\cos \theta }=1+\sec \theta$

$\Rightarrow \frac{z_2-z_1}{|z_2-z_1|}.e^{\pm iQ/2}=\frac{z_4-z_1}{|z_4-z_1|}=\frac{z_3-z_1}{|z_3-z_1|}.e^{\pm iQ/2}$

$=\left({z_2-z}_1\right)\left({z_3-z}_1\right).$

Hence, the answer is $\left({z_2-z}_1\right)\left({z_3-z}_1\right).$