Q.If sets $A$ and $B$ are defined as $A = \left\{(x, y) \mid y = \dfrac{1}{x},\ 0 \ne x \in \mathbb{R}\right\}$, $B = \{(x, y) \mid y = -x,\ x \in \mathbb{R}\}$, then
(A) $A \cap B = A$
(B) $A \cap B = B$
(C) $A \cap B = \phi$
(D) $A \cup B = A$

Concept: Cartesian product interpretation — each set is a collection of ordered pairs $(x,y)$ satisfying a given equation. Their intersection is the set of points that satisfy both equations simultaneously.

Step 1: For a point to be in $A \cap B$, it must satisfy both $y = \frac{1}{x}$ and $y = -x$.

Step 2: Equate the two: $\frac{1}{x} = -x \implies 1 = -x^2 \implies x^2 = -1$. …

We start with the Cartesian product idea: each set is a collection of ordered pairs $(x, y)$ in the plane. $A$ is the set of all points on the hyperbola $y = 1/x$ (excluding $x=0$), and $B$ is the set of all points on the line $y = -x$. The intersection $A \cap B$ consists of points that lie on both curves simultaneously.

1. Set up the condition for intersection. A point $(x, y)$ belongs to $A \cap B$ if and only if it satisfies both equations:

Since $x \neq 0$ in $A$, we must also have $x \neq 0$ for any candidate point.

2. Equate the two expressions for $y$. From $y = 1/x$ and $y = -x$, we get:

Multiply both sides by $x$ (which is allowed because $x \neq 0$):

3. Check for real solutions.

   The equation $x^2 = -1$ has no real solution — the square of a real number is never negative. Therefore, there is no real $x$ that satisfies the condition.

4. Conclude about the intersection.

   Since no real $x$ exists, there are no points $(x, y)$ common to both $A$ and $B$. Hence: