A) \[2|z{{|}^{2}}\]
B) \[1/2|z{{|}^{2}}\]
C) \[4|z{{|}^{2}}\]
D) \[|z{{|}^{2}}\]
Correct Answer: B
Solution :
Vertices of triangle in complex form is z, iz, z + iz In cartesian form vertices are (x,y),(-y,x) and (x-y,x+y) So,\[y=-\frac{1}{2}x\pm \sqrt{2}\] \[\therefore \]Area of triangle \[=\frac{1}{2}\left| \begin{matrix} x & y & 1 \\ -y & x & 1 \\ x-y & x+y & 1 \\ \end{matrix} \right|\] \[=\frac{1}{2}[x(x-x-y)-y(-y-x+y)+1\] \[(-yx-{{y}^{2}}-{{x}^{2}}+xy)]\] \[=\frac{1}{2}[-xy+xy-{{y}^{2}}-{{x}^{2}}]=\frac{1}{2}({{x}^{2}}+{{y}^{2}})\] (\[\because \] Area can not be negative) \[=\frac{1}{2}{{\left| z \right|}^{2}}\left( \because z=x+iy,{{\left| z \right|}^{2}}={{x}^{2}}+{{y}^{2}} \right)\]
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