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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
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Verified Answer
The correct answer is:
$\left(z_1-z_2\right)^2=2\left(z_1-z_3\right)\left(z_3-z_2\right)$
$\triangle A B C$ 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 $B$ ]
$$
\begin{aligned}
& \frac{z_2-z_1}{z_3-z_1}=\sqrt{2} e^{-\frac{i \pi}{4}} \\
& \text { [Apply rotation about a point } A \text { ] } \\
\Rightarrow & \left(\frac{z_1-z_2}{z_3-z_2}\right)\left(\frac{z_2-z_1}{z_3-z_1}\right)=2 \\
\Rightarrow & \frac{\left(z_1-z_2\right)\left(z_1-z_2\right)}{\left(z_3-z_2\right)\left(z_1-z_3\right)}=2 \\
\Rightarrow & \left(z_1-z_2\right)^2=2\left(z_1-z_3\right)\left(z_3-z_2\right)
\end{aligned}
$$
$$
\frac{z_1-z_2}{z_3-z_2}=\sqrt{2} e^{i \pi / 4}
$$
[Apply rotation about a point $B$ ]
$$
\begin{aligned}
& \frac{z_2-z_1}{z_3-z_1}=\sqrt{2} e^{-\frac{i \pi}{4}} \\
& \text { [Apply rotation about a point } A \text { ] } \\
\Rightarrow & \left(\frac{z_1-z_2}{z_3-z_2}\right)\left(\frac{z_2-z_1}{z_3-z_1}\right)=2 \\
\Rightarrow & \frac{\left(z_1-z_2\right)\left(z_1-z_2\right)}{\left(z_3-z_2\right)\left(z_1-z_3\right)}=2 \\
\Rightarrow & \left(z_1-z_2\right)^2=2\left(z_1-z_3\right)\left(z_3-z_2\right)
\end{aligned}
$$
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