Question

The complex numbers $z_1,z_2$ and $z_3$ satisfying $\frac{z_1-z_3}{z_2-z_3}=\frac{1-i\sqrt{3}}{2}$ are the vertices of a triangle which is

• A. of area zero
• B. right-angled isosceles
• C. equilateral
• D. obtuse-angled isosceles

Answer 1

The correct option is C equilateral

Taking mod of both sides of given relation $\left|\frac{z_1-z_3}{z_2-z_3}\right|=|\frac{1}{2}-i\frac{\sqrt{3}}{2}|=\sqrt{\frac{1}{4}+\frac{3}{4}}=1$
So, $|z_1-z_3|=|z_2-z_3|$. Also, amp $\left(\frac{z_1-z_3}{z_2-z_3}\right)=ta{n}^{-1}(-\sqrt{3})=-\frac{\pi }{3}$ or amp $\left(\frac{z_2-z_3}{z_1-z_3}\right)=\frac{\pi }{3}$ or $\angle z_2{z}_3{z}_1={60}^{\circ }$
$\therefore$ The triangle has two sides equal and the angle between the equal sides $={60}^{\circ }$. So, it is an equilateral triangle.