Q. If complex numbers $\mathcal{z}_{1},\ \mathcal{z}_{2}$ and $\mathcal{z}_{3}$ represent the vertices $A,\ B\ and\ C$ respectively of an isosceles triangle $ABC\$of which $\angle C$ is right angle, then correct statement is
Explanation
(d)
Since, $ABC$ is a right angled isosceles triangle
$\therefore\ \ BC = AC\ and\ \angle\ C = \frac{\pi}{2}$
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By rotation about $C$ in anti-clockwise sense
$\ CB = CA\ e^{i\pi/2}$
$\Rightarrow \ \ \left( \mathcal{z}_{2} - \mathcal{z}_{3} \right) = (\mathcal{z}_{1} - \mathcal{z}_{3})e^{i\pi/2}$
$= i\ \left( \mathcal{z}_{1} - \mathcal{z}_{3} \right)\ \ \ \ \ (\because e^{i\pi/2} = i)$
On squaring both sides, we get
$\left( \mathcal{z}_{2} - \mathcal{z}_{3} \right)^{2} = - \left( \mathcal{z}_{1} - \mathcal{z}_{3} \right)^{2}$
$\Rightarrow \ \ \mathcal{z}_{2}^{2} + \mathcal{z}_{3}^{2} - 2\mathcal{z}_{2}\mathcal{z}_{3} = - \mathcal{z}_{1}^{2} - \mathcal{z}_{3}^{2} + 2\mathcal{z}_{1}\mathcal{z}_{3}$
$\Rightarrow \ \ \mathcal{z}_{1}^{2} + \mathcal{z}_{2}^{2} - 2\mathcal{z}_{1}\mathcal{z}_{2} = 2\mathcal{z}_{1}\mathcal{z}_{3} + 2\mathcal{z}_{2}\mathcal{z}_{3} - 2\mathcal{z}_{3}^{2} - 2\mathcal{z}_{1}\mathcal{z}_{2}$
$\Rightarrow \ \ \left( \mathcal{z}_{1} - \mathcal{z}_{2} \right)^{2} = 2\left\lbrack \left( \mathcal{z}_{1}\mathcal{z}_{3} - \mathcal{z}_{3}^{2} \right) - \left( \mathcal{z}_{1}\mathcal{z}_{2} - \mathcal{z}_{2}\mathcal{z}_{3} \right) \right\rbrack$
$\Rightarrow \ \ \left( \mathcal{z}_{1} - \mathcal{z}_{2} \right)^{2} = 2\left( \mathcal{z}_{1} - \mathcal{z}_{3} \right)(\mathcal{z}_{3} - \mathcal{z}_{2})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.