Let $A (z_{1}), B (z_{2}) and C (z_{3})$ represent three complex numbers on argnd plane, then:

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If $|\begin{array}{lll} z_{1} {\bar{z}}_{1} 1 \\ z_{2} {\bar{z}}_{2} 1 \\ z_{3} {\bar{z}}_{3} 1 \end{array}| = 0$ then A, B and C will become collinear

Equation of perpendicular bisectors of AB is $z ({\bar{z}}_{1} - {\bar{z}}_{2}) + \bar{z} (z_{1} - z_{2}) + {|z_{1}|}^{2} - {|z_{2}|}^{2} = 0$

Centroid of $∆$ ABC is given by, $z_{0} = \frac{1}{3} (z_{1} + z_{2} + z_{3})$

$z = z_{1} + t \{\frac{(z_{1} - z_{2})}{|z_{1} - z_{2}|} + \frac{(z_{1} - z_{3})}{|z_{1} - z_{3}|}\}, t \in R$ is a real parameter, represents equation of one of the angle bisector of angle BAC

answer is A, B, C, D.

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Detailed Solution

$z = z_{1} - t ({\bar{AC}}_{1} + {\bar{AB}}_{1}), t \in R$ represents angle bisector of $∠ BAC$
$∴ z = z_{1} + t \{\frac{z_{1} - z_{3}}{|z_{1} - z_{3}|} + \frac{(z_{1} - z_{2})}{|z_{1} - z_{2}|}\}$
represents angle bisector of $∠ BAC$
b) we know that $z = \frac{1}{3} (z_{1} + z_{2} + z_{3})$
c) $|z - z_{1}| = |z - z_{2}|$ represents perpendicular bisector of A(z₁) and B (z₂)
$\begin{array}{l} ∴ {|z - z_{1}|}^{2} = {|z - z_{2}|}^{2} \\ \Rightarrow (z - z_{1}) (\bar{z} - {\bar{z}}_{1}) = (z - z_{2}) (\bar{z} - {\bar{z}}_{2}) \\ \Rightarrow z ({\bar{z}}_{1} - {\bar{z}}_{2}) + \bar{z} (z_{1} - z_{2}) + {|z_{1}|}^{2} - {|z_{2}|}^{2} = 0 \end{array}$
represents perpendicular bisector of A(z₁) and B(z₂)