If the vertices $A, B$ and $C$ of an isosceles $\triangle A B C$ are respectively $z_1, z_2$ and $z_3$ and…

If the vertices $A, B$ and $C$ of an isosceles $\triangle A B C$ are respectively $z_1, z_2$ and $z_3$ and if $\angle C=90^{\circ}$, then
  1. $\left(z_1-z_2\right)=\left(z_1-z_3\right)\left(z_3-z_2\right)$
  2. $\left(z_1-z_2\right)^2=\left(z_1-z_3\right)\left(z_3-z_2\right)$
  3. $\left(z_1-z_2\right)^2=2\left(z_1-z_3\right)\left(z_3-z_2\right)$
  4. $z_1^2+z_2^2+z_3^2=z_1 z_2 z_3+2$

Solution

$\triangle A B C$ is an isosceles right angled triangle with $A B$ as hypotenuse.
$ \frac{z_1-z_2}{z_3-z_2}=\sqrt{2} e^{i \pi / 4} $ [Apply rotation about a point $B$ ] $ \begin{aligned} & \frac{z_2-z_1}{z_3-z_1}=\sqrt{2} e^{-\frac{i \pi}{4}} \\ & \text { [Apply rotation about a point } A \text { ] } \\ \Rightarrow & \left(\frac{z_1-z_2}{z_3-z_2}\right)\left(\frac{z_2-z_1}{z_3-z_1}\right)=2 \\ \Rightarrow & \frac{\left(z_1-z_2\right)\left(z_1-z_2\right)}{\left(z_3-z_2\right)\left(z_1-z_3\right)}=2 \\ \Rightarrow & \left(z_1-z_2\right)^2=2\left(z_1-z_3\right)\left(z_3-z_2\right) \end{aligned} $

Asked in: AP EAMCET 2022 (06 Jul Shift 2)

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