Question

If the area of the triangle on the complex plane formed by the points $z,iz,z+iz$ is $200$ sq. units then the value of $3\left|z\right|$ equals to:

• A. $20$
• B. $40$
• C. $60$
• D. $80$

Answer 1

The correct option is C

$60$

Step 1: Find the area of a triangle

Suppose there is a $3\times 3$ determinant given as $\left|\begin{array}{ccc}{a}_1& b_1& 1\\ a_2& b_2& 1\\ a_3& b_3& 1\end{array}\right|$, then the area of a triangle using this determinant is calculated as

$\mathit{A}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}{\mathbf{a}}_{\mathbf{1}}\left(b_2-b_3\right)\mathbf{-}{\mathbf{a}}_{\mathbf{2}}\left(b_1-b_3\right)\mathbf{+}{\mathbf{a}}_{\mathbf{3}}\left(b_1-b_2\right)\mathbf{)}$

Let, $z=x+iy,iz=-y+ix,z+iz=\left(x-y\right)+i\left(x+y\right)$

The area of a triangle using determinant is:

$\begin{array}{rcl}A& =& \left(\frac{1}{2}\right)\left|\begin{array}{ccc}x& y& 1\\ -y& x& 1\\ \left(x-y\right)& \left(x+y\right)& 1\end{array}\right|\\ & =& \left(\frac{1}{2}\right)\left[x\left(x-x-y\right)-y\left(-y-x+y\right)+\left(-yx-y^2-x^2+xy\right)\right]\\ & =& \left(\frac{1}{2}\right)\left|x^2+y^2\right|\\ & =& \frac{1}{2}|z{|}^2\end{array}$

Step 2: Find the value of $3\left|z\right|$:

Therefore, its area is given by:

$\begin{array}{rcl}\therefore A& =& \frac{1}{2}|z{|}^2\\ & \Rightarrow & 200=\frac{1}{2}|z{|}^2\\ & \Rightarrow & 400=|z{|}^2\\ & \Rightarrow & \pm 20=\left|z\right|\end{array}$

So, the value of $3\left|z\right|$:

$\begin{array}{rcl}3\left|z\right|& =& 3\left(20\right)\\ & =& 60\end{array}$

Hence, option (C) is correct.