On the set of all points in a plane, the relation defined by the phrase 'at the same distance from the origin' is an equivalence relation.

Answer

To determine whether the relation defined by "at the same distance from the origin" on the set of all points in a plane is an equivalence relation, we need to verify three properties: reflexivity, symmetry, and transitivity.

### Step 1: Define the Relation
Let \( P \) be the relation defined on the set of all points in a plane such that for any two points \( A \) and \( B \), \( A P B \) if and only if the distance from point \( A \) to the origin is equal to the distance from point \( B \) to the origin.

### Step 2: Check Reflexivity
A relation is reflexive if every element is related to itself.

- For any point \( A \), the distance from \( A \) to the origin is equal to itself:
\[ d(A, O) = d(A, O) \]
- Therefore, \( A P A \) holds true for all points \( A \).

**Conclusion**: The relation is reflexive.

### Step 3: Check Symmetry
A relation is symmetric if whenever \( A P B \), then \( B P A \).

- Assume \( A P B \). This means:
\[ d(A, O) = d(B, O) \]
- By the property of equality, we can reverse this:
\[ d(B, O) = d(A, O) \]
- Thus, \( B P A \) holds true.

**Conclusion**: The relation is symmetric.

### Step 4: Check Transitivity
A relation is transitive if whenever \( A P B \) and \( B P C \), then \( A P C \).

- Assume \( A P B \) and \( B P C \). This means:
\[ d(A, O) = d(B, O) \quad \text{and} \quad d(B, O) = d(C, O) \]
- From these two equalities, we can conclude:
\[ d(A, O) = d(C, O) \]
- Therefore, \( A P C \) holds true.

**Conclusion**: The relation is transitive.

### Final Conclusion
Since the relation \( P \) satisfies reflexivity, symmetry, and transitivity, we conclude that the relation defined by "at the same distance from the origin" is indeed an equivalence relation.

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