Question

In $△ABC,A\left(z_1\right),B\left(z_2\right)$ and $C\left(z_3\right)$ are inscribed in the circle $|z|=5$. If $H\left(z_H\right)$ be the orthocentre of $△ABC$, then $z_H=$

• A. $\frac{z_1+z_2+z_3}{3}$
• B. $z_1+z_2+z_3$
• C. $\frac{z_1+z_2+z_3}{2}$
• D. $\frac{z_1+2z_2+3z_3}{3}$

Answer 1

The correct option is B $z_1+z_2+z_3$

Circumcentre of $△ABC$ is clearly origin. Let $G\left(z_G\right)$ be its centroid. Then,
$z_G=\frac{1}{3}(z_1+z_2+z_3)$
Now we know that,
$OG:GH=1:2\Rightarrow z_G=\frac{2\times 0+1\times z_H}{3}$
$\Rightarrow z_H=3z_G=z_1+z_2+z_3$