Question

If a point in argand plane $A(3,2)$ rotated through $B(1,1)$ about $\frac{\pi }{4}$ to obtain $C$, then area of $△ABC$ is

• A. $\frac{5}{4}$ sq. unit
• B. $\frac{5}{2}$ sq. unit
• C. $\frac{\sqrt{50}}{2}$ sq. unit
• D. $\frac{\sqrt{50}}{4}$ sq. unit

Answer 1

The correct option is D $\frac{\sqrt{50}}{4}$ sq. unit

Let $z_A=3+2i,z_B=1+i$ and after rotating $z_A$ the complex number be $z_C$ then
$|z_A-z_B|=|z_C-z_B|$
Using rotation property we have
$\frac{z_C-z_B}{z_A-z_B}=\frac{|z_C-z_B|}{|z_A-z_B|}{e}^{i\pi /4}\Rightarrow \frac{z_C-z_B}{2+i}=\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\Rightarrow z_C=(1+i)+(\frac{1}{\sqrt{2}}+\frac{3i}{\sqrt{2}})\therefore z_C=\frac{1+\sqrt{2}}{\sqrt{2}}+\frac{(3+\sqrt{2})i}{\sqrt{2}}$
Now $AC=|z_A-z_C|=\frac{1}{\sqrt{2}}\sqrt{(2\sqrt{2}-1{)}^2+(\sqrt{2}-3{)}^2}$
$=\sqrt{10-5\sqrt{2}}$ unit

Now $BD=|z_B-\left(\frac{z_A+z_C}{2}\right)|[\because AB=CB]$
$BD=\frac{\sqrt{10+5\sqrt{2}}}{2}$ unit
Hence required area $=\frac{AC\times BD}{2}$
$=\frac{\sqrt{50}}{4}$ sq. unit
(Here we have considers anticlockwise direction however clock wise direction will also give same answer)

Alternte Solution:
Using $\text{sine}$ rule
required area will be $=\frac{1}{2}\left(AB\right)\left(BC\right)\sin \left(\pi /4\right)$
$=\frac{1}{2\sqrt{2}}(AB{)}^2=\frac{1}{2\sqrt{2}}[2^2+1^2]$
$=\frac{\sqrt{50}}{4}$ sq. unit