Question

Let z, $\omega z$ and $z+\omega z$ represent three vertices of $\Delta ABC$, where $\omega$ is cube root unity, then

• A. centroid of $\Delta ABC$ is $\frac{2}{3}(z+\omega z)$
• B. orthocenter of $\Delta ABC$ is $\frac{2}{3}(z+\omega z)$
• C. ABC is an obtuse angled triangle
• D. ABC is an acute angled triangle

Answer 1

Answer: (A), (C)

The correct options are
A centroid of $\Delta ABC$ is $\frac{2}{3}(z+\omega z)$
C ABC is an obtuse angled triangle

Centroid will be
$\frac{z_1+z_2+z_3}{3}$
$=\frac{z+wz+z+wz}{3}$
$=\frac{2(z+wz)}{3}$
Considering the third vertex, we get
$z(1+w)$
$=-z{w}^2$
Hence obtuse angled triangle.
Had it been $z,zw,z{w}^2$ the triangle would have been an equilateral triangle.