Question

If $z_1,z_2,z_3$ are the vertices of a triangle in argand plane such that $|z_1-z_2|=|z_1-z_3|,$ then $\arg (\frac{2z_1-z_2-z_3}{z_3-z_2})$ is

• A. $\pm \frac{\pi }{3}$
• B. $0$
• C. $\pm \frac{\pi }{2}$
• D. $\pm \frac{\pi }{6}$

Answer 1

The correct option is C $\pm \frac{\pi }{2}$


$\frac{z_1-z_2}{z_3-z_2}=\frac{|z_1-z_2|}{|z_3-z_2|}{e}^{i\theta }$
$\Rightarrow \frac{z_1-z_3}{z_2-z_3}=\frac{|z_1-z_3|}{|z_2-z_3|}{e}^{-i\theta }$
$\Rightarrow \frac{z_1-z_2}{z_3-z_2}+\frac{z_1-z_3}{z_3-z_2}=$Purely imaginary number
So, $\arg (\frac{2z_1-z_2-z_3}{z_3-z_2})=\pm \frac{\pi }{2}$