Question

A, B, C are the points representing the complex numbers $z_1,z_2,z_3$ respectively on the complex plane and the circumcentre of the triangle ABC lies at the origin. If the altitude of the triangle through the vertex A meets the circumcircle again at P, then show that P represents the complex number $-\frac{z_2{z}_3}{z_1}$.

Answer 1

$|z_1|=|z_2|=|z_3|=|z|$

$z_1\overline{z_1}=z_2\overline{z_2}=z_3\overline{z_3}=z\overline{z}$..........(1)

$AP\perp BC$

$\frac{z-z_1}{\overline{z}-\overline{z_1}}+\frac{z_2-z_3}{\overline{z_2}-\overline{z_3}}=0$

$\Rightarrow \frac{z-z_1}{\frac{z_1\overline{z_1}}{z}-\overline{z_1}}+\frac{z_2-z_3}{\frac{z_3\overline{z_3}}{z_2}-\overline{z_3}}=0$ $\text{[From (1)]}$

$\Rightarrow \frac{z(z-z_1)}{z_1\overline{z_1}-z\overline{z_1}}+\frac{z_2(z_2-z_3)}{z_3\overline{z_3}-z_2\overline{z_3}}=0$

$\Rightarrow \frac{z(z-z_1)}{\overline{z_1}(z_1-z)}+\frac{z_2(z_2-z_3)}{\overline{z_3}(z_3-z_2)}=0$

$\Rightarrow \frac{-z(z_1-z)}{\overline{z_1}(z_1-z)}-\frac{z_2(z_3-z_2)}{\overline{z_3}(z_3-z_2)}=0$

$\Rightarrow \frac{-z}{\overline{z_1}}-\frac{z_2}{\overline{z_3}}=0$

$\Rightarrow \frac{-z}{\overline{z_1}}=\frac{z_2}{\overline{z_3}}$

$\Rightarrow z=-z_2\left(\frac{\overline{z_1}}{\overline{z_3}}\right)$..........(2)

$\because z_1\overline{z_1}=z_3\overline{z_3}$

$\therefore \frac{\overline{z_1}}{\overline{z_3}}=\frac{z_3}{z_1}$

Putting in (2), we get

$z=-z_2\left(\frac{z_3}{z_1}\right)$ or $\frac{-z_2{z}_3}{z_1}$