Let `f(x, y)=|x+y| and g(x, y)=|x-y|`, then how many ordered pairs of the form (x, y) would satisfy `f(x, y)=g(x, y)`

Options

A. 1
B. 2
C. 4
D. infinitely many

Answer

To solve the problem, we need to find the ordered pairs \((x, y)\) that satisfy the equation \(f(x, y) = g(x, y)\), where \(f(x, y) = |x + y|\) and \(g(x, y) = |x - y|\).

### Step-by-step Solution:

1. **Set up the equation**:
We start with the equation:
\[ |x + y| = |x - y| \]

2. **Consider the cases for absolute values**:
The absolute value equation can be split into different cases based on the signs of the expressions inside the absolute values. We will consider the following cases:

- Case 1: \(x + y = x - y\)
- Case 2: \(x + y = -(x - y)\)
- Case 3: \(-(x + y) = x - y\)
- Case 4: \(-(x + y) = -(x - y)\)

3. **Solve Case 1**:
\[ x + y = x - y \]
Simplifying gives:
\[ 2y = 0 \implies y = 0 \]
This means for any \(x\), if \(y = 0\), we have the ordered pairs \((x, 0)\).

4. **Solve Case 2**:
\[ x + y = - (x - y) \]
Simplifying gives:
\[ x + y = -x + y \implies 2x = 0 \implies x = 0 \]
This means for any \(y\), if \(x = 0\), we have the ordered pairs \((0, y)\).

5. **Solve Case 3**:
\[ -(x + y) = x - y \]
Simplifying gives:
\[ -x - y = x - y \implies -2x = 0 \implies x = 0 \]
This case leads us back to the ordered pairs \((0, y)\).

6. **Solve Case 4**:
\[ -(x + y) = -(x - y) \]
Simplifying gives:
\[ -x - y = -x + y \implies -2y = 0 \implies y = 0 \]
This case leads us back to the ordered pairs \((x, 0)\).

7. **Conclusion**:
From the analysis of all four cases, we find that:
- For any \(x\), if \(y = 0\), we have the pairs \((x, 0)\).
- For any \(y\), if \(x = 0\), we have the pairs \((0, y)\).

Since both \(x\) and \(y\) can take any real number value, there are infinitely many ordered pairs of the form \((x, 0)\) and \((0, y)\).

### Final Answer:
Thus, the number of ordered pairs \((x, y)\) that satisfy \(f(x, y) = g(x, y)\) is **infinitely many**.

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