Question

If the complex numbers $z_1,z_2$ and $z_3$ satisying $\frac{z_1-z_3}{z_2-z_3}=\frac{1-i\sqrt{3}}{2}$ are the vertices of a triangle, then the triangle is/has

• A. area $=0$
• B. a right-angled isosceles triangle
• C. an equilateral triangle
• D. an obtuse-angled isosceles triangle

Answer 1

The correct option is C an equilateral triangle

Let $z_1,z_2$ and $z_3$ represent the vertices of a triangle $ABC.$
Given,

$\frac{z_1-z_3}{z_2-z_3}=\frac{1-i\sqrt{3}}{2}$
$\Rightarrow (z_1-z_3)=e^{-i\frac{\pi }{3}}(z_2-z_3)$ ...(1)

$\overline{CA}$ is obtained by rotating $\overline{CB}$ through $\frac{\pi }{3}$ in clockwise sense.
$\Rightarrow$ Mod on both side in equation (1) is equal.

So, $CA=CB$.
$\Rightarrow CA=CB$ and $\angle ACB={60}^0$

Hence, $△ABC$ is an equilateral triangle.

Hence, option C.