Relation \( OP = OQ \) on Points in a Plane
📺 Video Explanation
📝 Question
Let \( O \) be the origin. Define a relation \( R \) between two points \( P \) and \( Q \) in a plane as:
\[ (P, Q) \in R \iff OP = OQ \]
Show that \( R \) is an equivalence relation.
✅ Solution
🔹 Step 1: Reflexive
For every point \( P \), \[ OP = OP \]
So, \[ (P, P) \in R \]
✔ Therefore, the relation is Reflexive.
🔹 Step 2: Symmetric
Assume: \[ (P, Q) \in R \Rightarrow OP = OQ \]
Then: \[ OQ = OP \]
So, \[ (Q, P) \in R \]
✔ Therefore, the relation is Symmetric.
🔹 Step 3: Transitive
Assume: \[ (P, Q) \in R,\ (Q, R) \in R \]
\[ OP = OQ,\quad OQ = OR \]
Thus, \[ OP = OR \]
So, \[ (P, R) \in R \]
✔ Therefore, the relation is Transitive.
🎯 Final Conclusion
✔ Reflexive: Yes
✔ Symmetric: Yes
✔ Transitive: Yes
\[ \therefore R \text{ is an equivalence relation} \]
🔹 Geometrical Meaning (Very Important)
All points having the same distance from the origin lie on a circle.
If \( OP = r \), then all related points satisfy: \[ x^2 + y^2 = r^2 \]
Thus, the equivalence class of a point \( P \neq (0,0) \) is:
\[ [P] = \{\text{all points on a circle with center } O \text{ and radius } OP\} \]
🚀 Exam Insight
- Same distance from origin ⇒ circle
- Each equivalence class = one circle
- Center always at origin