Question:
If sets $A$ and $B$ are defined as $A=\left\{(x, y): y=\frac{1}{x}, 0 \neq x \in R\right\}, B=\{(x, y): y=-x, x \in R\}$, then
(a) $A \cap B=A$
(b) $A \cap B=B$
(c) $A \cap B=\phi$
(d) $A \cup B=A$
Solution:
$A=\left\{(x, y): y=\frac{1}{x} ; 0 \neq x \in R\right\}$
$B=\{(x, y): y=-x ; x \in R\}$
then $(x, y) \in A \cap B$
i. e $(x, y) \in A$ and $(x, y) \in B$
i. e $y=\frac{1}{x}$ and $y=-x$
i. $\mathrm{e} \frac{1}{x}=-x ; \quad x \neq 0 \in R$
i. e $1=-x^{2} ; \quad x \neq 0 \in R$
i. e $x^{2}+1=0 ; x \neq 0 \in R$
No such real x exist such that x2 + 1 = 0
$\Rightarrow A \cap B=\phi$
Hence, the correct answer is option C.