For complex numbers z1, z2 and z3 satisfying z1−z2z2−z3=1−i 32 are the vertices of a triangle which is
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a
of area 3
b
right-angled and isosceles
c
equilateral
d
obtuse-angled and isosceles
answer is C.
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Detailed Solution
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.