Q72 / 120 · VITEEE 2014 · Mathematics

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
  • Correct answer
    C
    60
  • D
    80

Solution

Let $z=x+i y$, then $z+i z=x+i y+i(x+i y)=(x-y)+i(x+y)$ and $i z=i(x+i y)=-y+i x$, Then, the area of the triangle formed by these lines is $\Delta=\frac{1}{2}\left\|\begin{array}{ccc} \mathrm{x} & \mathrm{y} & 1 \\ (\mathrm{x}-\mathrm{y}) & (\mathrm{x}+\mathrm{y}) & 1 \\ -\mathrm{y} & \mathrm{x} & 1 \end{array}\right\|$ Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\left(\mathrm{R}_{1}+\mathrm{R}_{3}\right)$ $\Delta=\frac{1}{2}\left\|\begin{array}{ccc} \mathrm{x} & \mathrm{y} & 1 \\ 0 & 0 & -1 \\ -\mathrm{y} & \mathrm{x} & 1 \end{array}\right\|=\frac{1}{2}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)$ (given) $\begin{array}{l} \Rightarrow \frac{1}{2}|z|^{2}=200 \\ \quad|z|^{2}=400 \Rightarrow|z|=20 \\ \therefore 3|z|=3 \times 20=60 \end{array}$

Hence, the correct answer is option: 'C'.

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