Question

Area of the region bounded by curves
$\left(i\right)|z-1-i|=|z+3-3i|$
$\left(ii\right)|Re\left(Z\right)-1|=|Re\left(Z\right)+3|$
$\left(iii\right)|z-2-i|-|z-1-i|=1$ is $\Delta ,$ then $4\Delta$ is

Answer 1

Curve$:\left(i\right)$
$|z-1-i|=|z+3-3i|$
$\Rightarrow |(x-1)+i(y-1)|=|(x+3)+i(y-3)|$
$\Rightarrow (x-1{)}^2+(y-1{)}^2=(x+3{)}^2+(y-3{)}^2$
$\Rightarrow 8x-4y+16=0$
$\Rightarrow 2x-y+4=0\text{(represented as}AC\text{in the diagram)}$
Curve$:\left(ii\right)$
$|x-1|=|x+3|$
$\Rightarrow x=-1\text{(represented as}AB\text{in the diagram)}$
Curve$:\left(iii\right)$
$|(x-2)+i(y-1)|-|(x-1)+i(y-1)|=1$
$\Rightarrow \sqrt{(x-2{)}^2+(y-1{)}^2}-\sqrt{(x-1{)}^2+(y-1{)}^2}=1$
$\Rightarrow \sqrt{(x-2{)}^2+(y-1{)}^2}=1+\sqrt{(x-1{)}^2+(y-1{)}^2}$
$\Rightarrow (x-2{)}^2=1+(x-1{)}^2+2\sqrt{(x-1{)}^2+(y-1{)}^2}$
$\Rightarrow 1-x=\sqrt{(x-1{)}^2+(y-1{)}^2}$
$\Rightarrow y=1\text{(represented as}BC\text{in the diagram)}$

Bounded region is the $△ABC$
$\therefore \text{area}\left(\Delta \right)=\frac{1}{2}\times BC\times AB=\frac{1}{2}\times \frac{1}{2}\times 1=\frac{1}{4}$
So, $4\Delta =4\times \frac{1}{4}=1$