Question

Show that the points in the Argand plane represented by the complex number $-2+7i,-\frac{3}{2}+\frac{1}{2}i,4-3i$ and $\frac{7}{2}(1+i)$ are the vertices of a Rhombus.

Answer 1

Let the points be $A(-2,7),B(\frac{-3}{2},\frac{1}{2}),C(4,-3),D(\frac{7}{2},\frac{7}{2})$

$AB=\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(\frac{61}{2}\right)}^2}=\sqrt{\frac{170}{4}}$

$BC=\sqrt{{\left(\frac{51}{2}\right)}^2+{\left(\frac{31}{2}\right)}^2}=\sqrt{\frac{170}{4}}$

$CD=\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(\frac{61}{2}\right)}^2}=\sqrt{\frac{170}{4}}$

$DA=\sqrt{{\left(\frac{51}{2}\right)}^2+{\left(\frac{31}{2}\right)}^2}=\sqrt{\frac{170}{4}}$

as $AB=BC=CD=DA$ it is a rhombus.