For complex numbers z1, z2 and z3 satisfying z1−z2z2−z3=1−i 32 are the vertices of a triangle which is

We have |z1−z3z2−z3|=|1−i32|=14(1+3)=1 ⇒ |z1−z3|=|z2−z3| Also, z1−z3z2−z3−1=1−i32−1⇒z1−z2z2−z3=−1−i32 ⇒ |z1−z2||z2−z3|=14(1+3)=1 ⇒ |z1−z2|=|z2−z3| Thus, |z1−z3|=|z2−z3|=|z2−z1| Hence, z1, z2 and z3 are the vertices of an equilateral triangle.

For complex numbers z1, z2 and z3 satisfying z1−z2z2−z3=1−i 32 are the vertices of a triangle which is

We have |z1−z3z2−z3|=|1−i32|=14(1+3)=1 ⇒ |z1−z3|=|z2−z3| Also, z1−z3z2−z3−1=1−i32−1⇒z1−z2z2−z3=−1−i32 ⇒ |z1−z2||z2−z3|=14(1+3)=1 ⇒ |z1−z2|=|z2−z3| Thus, |z1−z3|=|z2−z3|=|z2−z1| Hence, z1, z2 and z3 are the vertices of an equilateral triangle.

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

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of area $\sqrt{3}$

right-angled and isosceles

equilateral

obtuse-angled and isosceles

answer is C.

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Detailed Solution

We have $| \frac{z_{1} - z_{3}}{z_{2} - z_{3}} | = | \frac{1 - i \sqrt{3}}{2} | = \sqrt{\frac{1}{4} (1 + 3)} = 1$

$\Rightarrow$ $| z_{1} - z_{3} | = | z_{2} - z_{3} |$

Also, $\frac{z_{1} - z_{3}}{z_{2} - z_{3}} - 1 = \frac{1 - i \sqrt{3}}{2} - 1 \Rightarrow \frac{z_{1} - z_{2}}{z_{2} - z_{3}} = \frac{- 1 - i \sqrt{3}}{2}$