Question

$A,B,C$ are the points representing the complex numbers $z_1,z_2,z_3,$ respectively on the complex plane and the circumcenter of the triangle $ABC$ lies at the origin. If the altitude $AD$ of the triangle $ABC$ meets circumcircle again at $P$, then $P$ represents the complex number

• A. $-\frac{\overline{z_1}{z}_2}{z_3}$
• B. $-\frac{\overline{z_1}{z}_2}{\overline{z_3}}$
• C. $-\frac{\overline{z_1}{z}_3}{\overline{z_2}}$
• D. $-\frac{z_2{z}_3}{z_3}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $-\frac{\overline{z_1}{z}_2}{\overline{z_3}}$
C $-\frac{\overline{z_1}{z}_3}{\overline{z_2}}$
D $-\frac{z_2{z}_3}{z_3}$

We have, $\angle DAC=\frac{\pi }{2}-C$ and $OC=OD$

$\angle POC=2\angle DAC$ and $OA=OB$

$\therefore \frac{z}{z_3}=\cos (\pi -2C)+i\sin (\pi -2C)$

$\frac{z}{z_3}=-\cos 2C+i\sin 2C$ ...(1)

Again, $\angle AOB=2C$ and $OA=OB$

$\therefore \frac{z_1}{z_2}=\cos 2C+i\sin 2C$ ...(2)

Multiply (1) and (2), we get $\frac{{zz}_1}{z_2{z}_3}=-1$

$\Rightarrow z=\frac{-z_2{z}_3}{z_1}=\frac{-\overline{z_1}\overline{z_2}{z}_2{z}_3}{z_1\overline{z_1}\overline{z_2}}=\frac{-\overline{z_1}{z}_3}{\overline{z_2}}$

$(\because z_1\overline{z_1}=z_2\overline{z_2})$

$=\frac{-\overline{z_1}{z}_2}{\overline{z_3}}.$