Let $\mathrm{A}=\{(x, y) \in \mathbf{R} \times \mathbf{R}:|x+y| \geqslant 3\}$ and $\mathrm{B}=\{(x, y) \in \mathbf{R} \times \mathbf{R}:|x|+|y| \leq 3\}$.
If $\mathrm{C}=\{(x, y) \in \mathrm{A} \cap \mathrm{B}: x=0$ or $y=0\}$, then $\sum_{(x, y) \in \mathrm{C}}|x+y|$ is :

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Aicharya · Accepted Answer

$\begin{aligned} & A=\{(x, y) \in \mathbf{R} \times \mathbf{R}:|x+y| \geq 3\} \\ & \text { and } B=\{(x, y) \in \mathbf{R} \times \mathbf{R}:|x|+|y| \leq 3\} \\ & C=\{(x, y) \in A \cap B: x=0 \text { or } y=0\}\end{aligned}$

$A \cap B$ will have only common points lying on the line $P Q$ and $R S$
Now, $C=\{(-3,0),(3,0),(0,3),(0,-3)\}$
$\sum_{(x, y) \in c}|x+y|=3+3+3+3=12$ *

Let $\mathrm{A}=\{(x, y) \in \mathbf{R} \times \mathbf{R}:|x+y| \geqslant 3\}$ and $\mathrm{B}=\{(x, y) \in…

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