Question

$A\left(z_1\right),B\left(z_2\right),C\left(z_3\right)$ are the vertices of the triangle $ABC$ (in anticlockwise order). If $\angle ABC=\frac{\pi }{4}$ and $AB=\sqrt{2}\left(BC\right)$, then

• A. $z_2=(z_1-z_3)+i{z}_3$
• B. $z_2=z_3+i(z_1-z_3)$
• C. $z_2=z_1+z_3$
• D. None of these

Answer 1

The correct option is B $z_2=z_3+i(z_1-z_3)$


Rotating about the point $B$, we get
$\frac{z_1-z_2}{z_3-z_2}=\left(\frac{a\sqrt{2}}{a}\right)e^{i\pi /4}=\sqrt{2}(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})=(1+i)$
$\Rightarrow z_1-z_2=(z_3-z_2)(1+i)$
$\Rightarrow z_2=z_1-(z_3-z_2)(1+i)$
$\Rightarrow z_2(1-(1+i\left)\right)=z_1-z_3(1+i)$
$\Rightarrow z_2=\frac{z_1}{-i}-\frac{z_3}{-i}(1+i)=i{z}_1-i{z}_3(1+i)$
$=z_3+i(z_1-z_3)$

Alternate Solution:
Given $\angle ABC=\frac{\pi }{4}\Rightarrow \arg \left(\frac{z_1-z_2}{z_3-z_2}\right)=\frac{\pi }{4}\cdots \left(1\right)$
and $\frac{AB}{BC}=\sqrt{2}\Rightarrow \frac{|z_1-z_2|}{|z_3-z_2|}=\sqrt{2}\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$ we can say
$\frac{z_1-z_2}{z_3-z_2}=\sqrt{2}{e}^{i\pi /4}=1+i\Rightarrow z_2=z_1-(1+i)(z_3-z_2)\Rightarrow z_2=\frac{z_3(1+i)-z_1}{i}\Rightarrow z_2=z_3+i(z_1-z_3)$