If in an argand plane points z1, z2, z3 are the vertices of an isosceles triangle right angled at z2, then

Let BA = BC⇒z1−z22=z3−z22⇒ z1−z2z1−z2=z3−z2z3−z2 (1)Again,∴ ∠ABC=90∘∴ arg⁡BABC=90∘⇒ arg⁡z1−z2z3−z2=90∘⇒ real part of z1−z2z3−z2=0⇒ 12z1−z2z3−z2+z1−z2z3−z2=0⇒z1−z2z3−z2=−z1−z2z3−z2⇒z1−z2z1−z2=z2−z3z3−z2 (2)(1)×(2)⇒z1−z22=−z2−z32⇒ z12+z22+z32=2z2z1+z3

If in an argand plane points z1, z2, z3 are the vertices of an isosceles triangle right angled at z2, then

Let BA = BC⇒z1−z22=z3−z22⇒ z1−z2z1−z2=z3−z2z3−z2 (1)Again,∴ ∠ABC=90∘∴ arg⁡BABC=90∘⇒ arg⁡z1−z2z3−z2=90∘⇒ real part of z1−z2z3−z2=0⇒ 12z1−z2z3−z2+z1−z2z3−z2=0⇒z1−z2z3−z2=−z1−z2z3−z2⇒z1−z2z1−z2=z2−z3z3−z2 (2)(1)×(2)⇒z1−z22=−z2−z32⇒ z12+z22+z32=2z2z1+z3

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If in an argand plane points z₁, z₂, z₃ are the vertices of an isosceles triangle right angled at z₂, then

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$z_{1}^{2} + z_{2}^{2} + 2 z_{3}^{2} = 2 z_{2} (z_{1} + z_{3})$

$z_{1}^{2} + 2 z_{2}^{2} + z_{3}^{2} = 2 z_{2} (z_{1} + z_{3})$

$z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 2 z_{2} (z_{1} + z_{3})$

$2 z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 2 z_{2} (z_{1} + z_{3})$

answer is B.

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Detailed Solution

Let BA = BC

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

$\Rightarrow (z_{1} - z_{2}) (z_{1} - z_{2}) = (z_{3} - z_{2}) (z_{3} - z_{2})$ (1)

Again,

$\begin{array}{l} ∴ ∠ ABC = 90^{\circ} \\ ∴ \arg \frac{BA}{BC} = 90^{\circ} \\ \Rightarrow \arg \frac{z_{1} - z_{2}}{z_{3} - z_{2}} = 90^{\circ} \Rightarrow real part of \frac{z_{1} - z_{2}}{z_{3} - z_{2}} = 0 \\ \Rightarrow \frac{1}{2} [\frac{z_{1} - z_{2}}{z_{3} - z_{2}} + \frac{z_{1} - z_{2}}{z_{3} - z_{2}}] = 0 \end{array}$

$\Rightarrow \frac{z_{1} - z_{2}}{z_{3} - z_{2}} = - \frac{z_{1} - z_{2}}{z_{3} - z_{2}} \Rightarrow \frac{z_{1} - z_{2}}{z_{1} - z_{2}} = \frac{z_{2} - z_{3}}{z_{3} - z_{2}}$ (2)

$\begin{array}{l} (1) \times (2) \Rightarrow {(z_{1} - z_{2})}^{2} = - {(z_{2} - z_{3})}^{2} \\ \Rightarrow z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 2 z_{2} (z_{1} + z_{3}) \end{array}$