Discussion Forum

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

Answer:

Explanation:

 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 

   $\triangle$=$\frac{1}{2} \begin{bmatrix}x &y&1 \\(x-y) & (x+y)&1\\-y&x&1 \end{bmatrix}$

 Applying  $R_{2} \rightarrow  R_{2}-(R_{1}+R_{3})$

 $\triangle$=$\frac{1}{2} \begin{bmatrix}x &y&1 \\0& 0&-1\\-y&x&1 \end{bmatrix}$=$\frac{1}{2}(x^{2}+y^{2})$

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

 $|z|^{2}=400 \Rightarrow |z|=20$

 $\therefore$  $3|z|=3 \times 20=60$