Question

If $z=x+iy$, where $x,y\in R$ and $z,-iz$ and $z+iz$ is represented on the argand plane, then the area of triangle formed (in sq.units) is

• A. $x^2+y^2$
• B. $2(x^2+y^2)$
• C. $\frac{x^2+y^2}{2}$
• D. $\frac{x^2+y^2}{4}$

Answer 1

The correct option is C $\frac{x^2+y^2}{2}$

Given : $z=x+iy$
Now,
$-iz=-ix+y=y-ixz+iz=(x-y)+i(x+y)$

Therefore, the vertices of triangle formed are
$(x,y),(y,-x),(x-y,x+y)$
Area of a triangle formed by coordinates $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by
$A=\frac{1}{2}\left|\left|\begin{array}{cccc}{x}_1& x_2& x_3& x_1\\ y_1& y_2& y_3& y_1\end{array}\right|\right|\Rightarrow A=\frac{1}{2}\left|\left|\begin{array}{cccc}x& y& x-y& x\\ y& -x& x+y& y\end{array}\right|\right|\Rightarrow A=\frac{1}{2}|-x^2-y^2+xy+y^2+x^2-xy+xy-y^2-x^2-xy|\Rightarrow A=\frac{1}{2}|-x^2-y^2|\therefore A=\frac{x^2+y^2}{2}\text{sq.units}$