Question

The area of the triangle with vertices $z,iz,z+iz$ is $50$, then $|z|=$

• A. $0$
• B. $5$
• C. $10$
• D. $15$

Answer 1

The correct option is C $10$

Let $A\equiv z,B\equiv iz$ & $C\equiv z+iz$
$\therefore$ BC is represented by $(z+iz)-iz=z$
$\therefore$ $\left|BC\right|=\left|z\right|$
The side AC is represented by $(z+iz)-z=iz$
$\therefore \left|AC\right|=\left|iz\right|=\left|z\right|$
Now the complex numbers $z$ & $iz$ are at ${90}^0$
$\therefore$ Area $\Delta ABC=\frac{1}{2}BC\times AC$
$=\frac{1}{2}\left|z\right|\times \left|z\right|$
$50=\frac{{\left|z\right|}^2}{2}$
${\left|z\right|}^2=100$
$\left|z\right|=10$