Question

The complex numbers $z_1,z_2\mathrm{and}{z}_3$ satisfying $\frac{\left(z_1-z_3\right)}{\left(z_2-z_3\right)}=\frac{1-\sqrt{3}i}{2}$ 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

Explanation for the correct option:

Determine if the triangle is an equilateral triangle:

Given,

$\frac{\left(z_1-z_3\right)}{\left(z_2-z_3\right)}=\frac{1-\sqrt{3}i}{2}$

Multiply numerator and denominator with $(1+\sqrt{3}i)$

$$\begin{gathered} \frac{\left(z_1-z_3\right)}{\left(z_2-z_3\right)}=\frac{\left(1-\sqrt{3}i\right)\left(1+\sqrt{3}i\right)}{2\left(1+\sqrt{3}i\right)} \\ =\frac{1+1\left(3\right)}{2\left(1+\sqrt{3}i\right)} \\ =\frac{2}{\left(1+\sqrt{3}i\right)} \\ \mathrm{Taking}\mathrm{reciprocal} \\ \frac{\left(z_2-z_3\right)}{\left(z_1-z_3\right)}=\frac{\left(1+\sqrt{3}i\right)}{2} \\ =\cos \left(\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{3}\right)\left[\because \cos \left(\frac{\pi }{3}\right)=\frac{1}{2}\mathrm{and}\sin \left(\frac{\mathrm{\pi }}{3}\right)=\frac{\sqrt{3}}{2}\right] \\ \mathrm{Now}, \\ \left|\frac{\left({\mathrm{z}}_2-{\mathrm{z}}_3\right)}{\left({\mathrm{z}}_1-{\mathrm{z}}_3\right)}\right|=1 \\ \arg \left|\frac{\left({\mathrm{z}}_2-{\mathrm{z}}_3\right)}{\left({\mathrm{z}}_1-{\mathrm{z}}_3\right)}\right|=\frac{\mathrm{\pi }}{3} \end{gathered}$$

Therefore the triangle is equilateral.

Therefore the correct answer is option (C).