Question

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

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

Answer 1

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

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, the area of triangle is given by
$△=\frac{1}{2}\left|\begin{array}{cc}{x}_1& y_1\\ x_2& y_2\\ x_3& y_3\\ x_1& y_1\end{array}\right|$

$\Rightarrow △=\frac{1}{2}\left|\begin{array}{cc}3& 2\\ 1& 1\\ \frac{1+\sqrt{2}}{\sqrt{2}}& \frac{(3+\sqrt{2})}{\sqrt{2}}\\ 3& 2\end{array}\right|$

$\Rightarrow △=\left|\frac{1}{2}\left[\right(3-2)+(\frac{(3+\sqrt{2})}{\sqrt{2}}-\frac{1+\sqrt{2}}{\sqrt{2}})+(\frac{2(1+\sqrt{2})}{\sqrt{2}}-\frac{3(3+\sqrt{2})}{\sqrt{2}})]\right|$
$\Rightarrow △=\frac{\sqrt{50}}{4}$ sq. units