Question

If the area of the triangle on the complex plane formed by the points z, z + i z and iz is 200, then the value of 3 |z| must be equal to

• A. 20
• B. 40
• C. 60
• D. 80

Answer 1

The correct option is C 60

Let $z=x+iy$, then

$z+iz=x+iy+i(x+iy)=(x-y)+i(x+y)$

and $iz=i(x+iy)=-y+ix$,

Then, the area of the triangle formed by these lines is

$\Delta =\frac{1}{2}\left|\begin{array}{ccc}x& y& 1\\ (x-y)& (x+y)& 1\\ -y& x& 1\end{array}\right|$

Applying $R_2\to R_2-(R_1+R_3)$,

$\Delta =\frac{1}{2}\left|\begin{array}{ccc}x& y& 1\\ 0& 0& -1\\ -y& x& 1\end{array}\right|=\frac{1}{2}(x^2+y^2)$

$\Rightarrow \frac{1}{2}|z{|}^2=200$ (given)

$|z{|}^2=400\Rightarrow |z|=20$

$\therefore 3|z|=3\times 20=60$

Hence, $3|z|=60$.