Question

The area of the triangle with vertices $P\left(z\right)$, $Q\left(iz\right)$ and $R(z+iz)$ is:

• A. $\left(\frac{1}{2}\right){\left|z+iz\right|}^2$
• B. $1$
• C. $\frac{1}{2}$
• D. $\left(\frac{1}{2}\right){\left|z\right|}^2$

Answer 1

The correct option is D

$\left(\frac{1}{2}\right){\left|z\right|}^2$

Explanation for the correct option:

Step 1: Illustrating the figure using the given information

Firstly, draw the diagram.

Here,

$$\begin{gathered} \left|OP\right|=\left|z\right|(\text{length between points)} \\ \left|OQ\right|=\left|iz\right|=|z| \\ \left|PR\right|=|z+iz-z|=\left|iz\right|=\left|z\right| \\ \left|RQ\right|=\left|z+iz-iz\right|=\left|z\right| \end{gathered}$$

All sides of the quadrilateral $OQRP$ is equal.

And the angle between $zandiz$is $90°$.

Therefore, quadrilateral $OQRP$ forms a square.

Step 2: Finding the area

So, the area of triangle $QRP$

$$\begin{gathered} \text{Area}=\frac{1}{2}\text{Area of square} \\ =\frac{1}{2}{\left|z\right|}^2 \end{gathered}$$

Therefore, option (D) is the correct answer.