A) \[2|z{{|}^{2}}\]
B) \[\frac{1}{2}|z{{|}^{2}}\]
C) \[|z{{|}^{2}}\]
D) \[\frac{3}{2}|z{{|}^{2}}\]
Correct Answer: B
Solution :
Since, \[z=x+iy\] \[\therefore \] \[iz=-y+xi\] and \[z+iz=x+iy+i(x+iy)\] \[=(x-y)+i(x+y)\] Since, \[z+iz\] and \[z+iz\] are the vertices of a triangle. \[\therefore \] Area of triangle\[=\frac{1}{2}\left| \left| \begin{matrix} x & y & 1 \\ -y & x & 1 \\ x-t & x+y & 1 \\ \end{matrix} \right| \right|\] \[=\frac{1}{2}|[x(x-x-y)-y(-y-x+y)\] \[+1(-xy-{{y}^{2}}-{{x}^{2}}+xy)]|\] \[=\frac{1}{2}|[-xy+xy-{{y}^{2}}-{{x}^{2}}]|\] \[=\frac{1}{2}[{{x}^{2}}+{{y}^{2}}]\] \[=\frac{|z{{|}^{2}}}{2}\]
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