The complex numbers $z_1, z_2,z_3$ are satisfying $\frac{z_1-z_3}{z_2-z_3}=\frac{1-i\sqrt{3}}{2}$..... Problem : 
The complex numbers $z_1, z_2,z_3$ are satisfying $\frac{z_1-z_3}{z_2-z_3}=\frac{1-i\sqrt{3}}{2}$ $z_1,z_2,z_3$ are vertices of a triangle , which type of triangle is it : 
My approach : 
$|\frac{z_1-z_3}{z_2-z_3}|=|\frac{1-i\sqrt{3}}{2}|  = 1 $
argument $\frac{z_1-z_3}{z_2-z_3} = \cos\frac{\pi}{3}-\sin\frac{\pi}{3}$ =$\frac{\pi}{3}$ 
So, I am unable to work out, whether it a isosceles or equilateral triangle. Please suggest will be of great help, thanks. 
 A: $$\left|\frac{z_1-z_3}{z_2-z_3}\right|= 1 \Rightarrow |z_1-z_3|=|z_2-z_3|$$
using proportion rule:
$$\frac{z_1-z_3}{z_2-z_3}=\frac{1-i\sqrt{3}}{2} \Rightarrow \frac{z_1-z_3-(z_2-z_3)}{z_2-z_3}=\frac{1-i\sqrt{3}-2}{2} \Rightarrow \left|\frac{z_1-z_2}{z_2-z_3}\right|=1 \Rightarrow |z_1-z_2|=|z_2-z_3|$$
So,
$$|z_1-z_3|=|z_2-z_3|=|z_1-z_2|$$
So it is equilateral.
A: Given $$\frac{z_1-z_3}{z_2-z_3} = \frac{1-i\sqrt{3}}{2} =\frac{1}{2}-i\frac{\sqrt{3}}{2} = \cos(-\frac{\pi}{3}) +i\sin (-\frac{\pi}{3})$$ $$\Rightarrow \frac{z_2-z_3}{z_1-z_3} =\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3})$$ $$\Rightarrow |\frac{z_2-z_3}{z_1-z_3}| = 1 \text{and arg}(\frac{z_2-z_3}{z_1-z_3}) =\frac{\pi}{3}$$ If $z_1, z_2, z_3$ represent the vertices $A, B, C$ of a triangle ABC, then, $$|\frac{z_2-z_3}{z_1-z_3}| =1 \Rightarrow |z_2-z_3| = |z_1-z_3|$$ This gives us $BC=AC$. Also, $\text{arg}(\frac{z_2-z_3}{z_1-z_3}) =\frac{\pi}{3}$ gives $\angle BCA =\frac{\pi}{3}$. So, ABC is equilateral.
A: Hint:
$\implies z_1=(\cdots)z_2-(\cdots)z_3$
$$\dfrac{z_1-z_2}{z_2-z_3}=?$$
