Question

The complex numbers $z_1,z_2$ and $z_3$ satisfying $\left[\right(z_1-z_3\left)/\right(z_2-z_3\left)\right]=\left[\right(1-i\sqrt{3}\left)/2\right]$ are the vertices of a triangle which is

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

Answer 1

Answer: (C)

equilateral

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

Apply argument on both sided

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

$\Rightarrow Arg\frac{(z_1-z_3)}{(z_2-z_3)}=\frac{\pi }{3}$

Angle between $z_{i_3^2}\&z_2-z_3$ is $60°$

By similarly substract $1$ on both sides of $\left(1\right)$

$\Rightarrow \frac{z_1-z_3}{z_2-z_3}-1=\frac{-1}{2}-\frac{i\sqrt{3}}{2}$

$\Rightarrow \frac{z_1-z_3-z_2{+z}_3}{z_2-z_3}-1=\frac{-1}{2}-\frac{i\sqrt{3}}{2}$

Apply argument on both sides we get

$\Rightarrow Arg\frac{(z_1-z_3)}{(z_2-z_3)}=\frac{\pi }{3}$

Angle between $z_1{z}_2\&z_3{+z}_2$ is $60°$ Obviously other angle is $60°$ equilateral triangle.

Hence, the answer is equilateral triangle.