Question

Prove that the area of the triangle whose vertices are the points represented by the complex numbers $z_1,z_2,z_3$ on the Argand diagram is $\sum \left[\left(z_2{-z}_3\right){\left|z_1\right|}^2╱4i{z}_1\right]$

Answer 1

Let $z_1=x_1+i{y}_1=(x_1,y_1)$.
$z_2=x_2+i{y}_2=(x_2,y_2)$,
$z_3=x_3+i{y}_3=(x_3,y_3)$.
$\therefore$ The required area
$=\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|=\frac{1}{2i}\left|\begin{array}{ccc}{x}_1& {iy}_1& 1\\ x_2& {iy}_2& 1\\ x_3& {iy}_3& 1\end{array}\right|$
$=\frac{1}{2i}\left|\begin{array}{ccc}{x}_1& {x_1+iy}_1& 1\\ x_2& {x_2+iy}_2& 1\\ x_3& {x_3+iy}_3& 1\end{array}\right|$ by $C_2+C_1$
$=\frac{1}{2i}\left|\begin{array}{ccc}{x}_1& z_1& 1\\ x_2& z_2& 1\\ x_3& z_3& 1\end{array}\right|=\frac{1}{4i}\left|\begin{array}{ccc}{2x}_1& z_1& 1\\ {2x}_2& z_2& 1\\ {2x}_3& z_3& 1\end{array}\right|$
$=\frac{1}{4i}\left|\begin{array}{cc}{z}_1+{\overline{z}}_1& z_{1}1\\ z_2+{\overline{z}}_2& z_{2}1\\ z_3{+\overline{z}}_3& z_{3}1\end{array}\right|=\frac{1}{4i}\left|\begin{array}{ccc}{\overline{z}}_1& z_1& 1\\ {\overline{z}}_2& z_2& 1\\ {\overline{z}}_3& z_3& 1\end{array}\right|$,
by $C_1-C_2$
$=\frac{1}{4i}\Sigma {\overline{z}}_1(z_2-z_3)=\frac{1}{4i}\frac{\Sigma {\overline{z}}_1{z}_1(z_2-z_3)}{z_1}$
$=\frac{1}{4i}\Sigma \frac{{\left|z_1\right|}^2}{z_1}(z_2-z_3)=\Sigma \frac{{\left|z_1\right|}^2(z_2-z_3)}{4i{z}_1}$