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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
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Verified Answer
The correct answer is:
60
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)
$\begin{array}{l}
\Rightarrow \frac{1}{2}|z|^{2}=200 \\
\quad|z|^{2}=400 \Rightarrow|z|=20 \\
\therefore 3|z|=3 \times 20=60
\end{array}$
$z+i z=x+i y+i(x+i y)=(x-y)+i(x+y)$ 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)
$\begin{array}{l}
\Rightarrow \frac{1}{2}|z|^{2}=200 \\
\quad|z|^{2}=400 \Rightarrow|z|=20 \\
\therefore 3|z|=3 \times 20=60
\end{array}$
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