If the sets A and B are defined as A = { (x, | vedclass

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(C) To find the intersection $A \cap B$,we need to find the points $(x, y)$ that satisfy both equations simultaneously.Given equations are $y = \frac{

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$A \cap B = \phi$

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(C) To find the intersection $A \cap B$,we need to find the points $(x, y)$ that satisfy both equations simultaneously.Given equations are $y = \frac{1}{x}$ and $y = -x$.Substituting the value of $y$ from the second equation into the first,we get $-x = \frac{1}{x}$.This simplifies to $-x^2 = 1$,or $x^2 = -1$.Since $x$ must be a real number $(x \in R)$,the equation $x^2 = -1$ has no real solutions because the square of any real number is non-negative.Therefore,there are no points $(x, y)$ that belong to both sets $A$ and $B$.Hence,$A \cap B = \phi$.

If the sets $A$ and $B$ are defined as $A = \{ (x, y) : y = \frac{1}{x}, x \in R, x \neq 0 \}$ and $B = \{ (x, y) : y = -x, x \in R \}$,then:

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