Mathematics - Complex Number Question with Solution | TestHub

Vertices of triangle in complex form is $z,iz,z+iz$ In cartesian form vertices are $(x,y),(-y,x)$ and $(x-y,x+y)$ $$\begin{array}{rl}& \therefore \text{ Area of triangle }=\frac{1}{2}\left|\begin{array}{ccc}x& y& 1\\ -y& x& 1\\ x-y& x+y& 1\end{array}\right|\\ & =\frac{1}{2}[x(x-x-y)-y(-y-x+y)+1\\ & \left(-yx-y^2-x^2+xy\right)]\end{array}$$ $$\begin{array}{rl}& =\frac{1}{2}\left[-xy+xy-y^2-x^2\right]=\frac{1}{2}\left(x^2+y^2\right)\\ & (\because \text{ Area can not be negative })\\ & =\frac{1}{2}|z{|}^2\left(\because z=x+iy,|z{|}^2=x^2+y^2\right)\end{array}$$