Question

If a triangle $ABC$ is formed by the points represented by $z,$ $iz$ and $z+iz$ then

• A. $ABC$ is an equilateral triangle
• B. $ABC$ is an isosceles right angled triangle
• C. Area of $△ABC$ is $\frac{1}{2}|z{|}^2$ sq. units
• D. Area of $△ABC$ is $\frac{\sqrt{3}}{4}|z{|}^2$ sq. units

Answer 1

Answer: (B), (C)

Area of $△ABC$ is $\frac{1}{2}|z{|}^2$ sq. units


Let $A\left(z\right),B\left(iz\right),C(z+iz)$ be vertices of $\Delta ABC$
Then $CA=|z+iz-z|=|iz|=|z|$
$CB=|z+iz-iz|=|z|$
$A{B}^2=|z-iz{|}^2=|z{|}^2+|iz{|}^2-2Re(z\cdot \overline{iz})$
$=|z{|}^2+|z{|}^2-2Re(-i|z{|}^2)$
$=2|z{|}^2-0$
$=2|z{|}^2=C{A}^2+C{B}^2$
$\therefore ABC$ is right angled isosceles triangle.
Area of $△ABC=\frac{1}{2}CA\times CB$
$=\frac{1}{2}|z{|}^2$