Two points A and B in a plane are related if OA = OB, where O is a fixed point. This relation is

Options

A. partial order relation
B. equivalence relation
C. reflexive but not symmetric
D. reflexive but not transitive

Answer

To determine the nature of the relation defined by the condition \( OA = OB \) where \( O \) is a fixed point in the plane, we will analyze the relation based on the properties of reflexivity, symmetry, and transitivity.

### Step-by-Step Solution:

1. **Understanding the Relation**:
- We have two points \( A \) and \( B \) in a plane.
- The relation states that \( OA = OB \), meaning the distance from point \( O \) to point \( A \) is equal to the distance from point \( O \) to point \( B \).

2. **Checking Reflexivity**:
- A relation is reflexive if every element is related to itself.
- Here, we check if \( A \) is related to \( A \):
\[ OA = OA \]
- This is always true since the distance from \( O \) to \( A \) is equal to itself.
- **Conclusion**: The relation is reflexive.

3. **Checking Symmetry**:
- A relation is symmetric if whenever \( A \) is related to \( B \), then \( B \) is also related to \( A \).
- Given \( OA = OB \), we need to check if \( OB = OA \):
\[ OA = OB \implies OB = OA \]
- This is also true because equality is symmetric.
- **Conclusion**: The relation is symmetric.

4. **Checking Transitivity**:
- A relation is transitive if whenever \( A \) is related to \( B \) and \( B \) is related to \( C \), then \( A \) is related to \( C \).
- Assume \( OA = OB \) and \( OB = OC \). We need to show \( OA = OC \):
\[ OA = OB \quad \text{and} \quad OB = OC \implies OA = OC \]
- This follows from the transitive property of equality.
- **Conclusion**: The relation is transitive.

5. **Final Conclusion**:
- Since the relation satisfies reflexivity, symmetry, and transitivity, it is classified as an **equivalence relation**.

### Final Answer:
The relation defined by \( OA = OB \) is an **equivalence relation**.