If set A and B are defined as
`A = {(x,y)|y = 1/x, 0 ne x in R}, B = {(x,y)|y = -x , x in R,}`. Then (a) A ∩ B = A (b)A ∩ B = B (c)A ∩ B = ϕ (d)A ∪ B = A

Options

A. `A nn B = A`
B. `A nn B = B`
C. `A nn B = phi`
D. `A uu B = A`

Explanation

To solve the problem, we need to analyze the two sets A and B defined as follows:

- Set A: \( A = \{(x,y) | y = \frac{1}{x}, x \neq 0, x \in \mathbb{R}\} \)
- Set B: \( B = \{(x,y) | y = -x, x \in \mathbb{R}\} \)

### Step 1: Understand Set A
Set A consists of all points \((x,y)\) where \(y\) is defined as \(\frac{1}{x}\) for all real numbers \(x\) except for \(0\). This means:
- For \(x = 1\), \(y = 1\) → Point (1, 1)
- For \(x = 2\), \(y = \frac{1}{2}\) → Point (2, 0.5)
- For \(x = -1\), \(y = -1\) → Point (-1, -1)
- For \(x = -2\), \(y = -\frac{1}{2}\) → Point (-2, -0.5)
- The set A will have points that approach infinity as \(x\) approaches \(0\) from either side.

### Step 2: Understand Set B
Set B consists of all points \((x,y)\) where \(y\) is defined as \(-x\). This means:
- For \(x = 1\), \(y = -1\) → Point (1, -1)
- For \(x = 2\), \(y = -2\) → Point (2, -2)
- For \(x = -1\), \(y = 1\) → Point (-1, 1)
- For \(x = -2\), \(y = 2\) → Point (-2, 2)
- This set represents a straight line with a negative slope.

### Step 3: Find the Intersection \(A \cap B\)
To find the intersection \(A \cap B\), we need to find points that satisfy both conditions:
1. From Set A: \(y = \frac{1}{x}\)
2. From Set B: \(y = -x\)

Setting these equal to each other:
\[ \frac{1}{x} = -x \]
Multiplying both sides by \(x\) (noting \(x \neq 0\)):
\[ 1 = -x^2 \]
This leads to:
\[ x^2 = -1 \]
Since there are no real solutions to this equation (as \(x^2\) cannot be negative), we conclude that there are no common points between sets A and B.

Thus, the intersection \(A \cap B = \emptyset\) (the empty set).

### Step 4: Conclusion
Given the options:
(a) \(A \cap B = A\)
(b) \(A \cap B = B\)
(c) \(A \cap B = \emptyset\)
(d) \(A \cup B = A\)

The correct answer is (c) \(A \cap B = \emptyset\).