Q.Show that the relation R in the set A of points in a plane given by $R = \{(P, Q) :$ distance of the point P from the origin is same as the distance of the point Q from the origin$\}$, is an equivalence relation. Further, show that the set of all points related to a point $P \ne (0, 0)$ is the circle passing through P with origin as centre.

Concept: Relation Properties — checking reflexivity, symmetry, and transitivity.

Step 1 — Reflexive: For any point $P$, distance of $P$ from origin equals itself. So $(P,P) \in R$.

Step 2 — Symmetric: If $(P,Q) \in R$, then distance of $P$ from origin = distance of $Q$ from origin. Hence distance of $Q$ from origin = distance of $P$ from origin, so $(Q,P) \in R$.

Step 3 — Transitive: If $(P,Q) \in R$ and $(Q,R) \in R$, then distance of $P$ = distance of $Q$ and distance of $Q$ = distance of $R$. So distance of $P$ = distance of $R$, giving $(P,R) \in R$.

Thus $R$ is an equivalence relation. …

The key idea is simple: the relation $R$ is defined purely by equality of a number — the distance from the origin. Whenever a relation is defined by "two objects have the same value of some function", it is always an equivalence relation. Here the function is $f(P) = \text{distance of } P \text{ from } (0,0)$.

Let’s verify each property step by step.

1. Reflexivity

For any point $P$ in the plane, its distance from the origin is obviously equal to itself. So $(P,P) \in R$ for every $P$. Hence $R$ is reflexive.

2. Symmetry

If $(P,Q) \in R$, then the distance of $P$ from the origin equals the distance of $Q$ from the origin. That statement is symmetric: swapping $P$ and $Q$ gives the same equality. So $(Q,P) \in R$ whenever $(P,Q) \in R$. Hence $R$ is symmetric.

3. Transitivity

Suppose $(P,Q) \in R$ and $(Q,S) \in R$. Then

• distance of $P$ from origin = distance of $Q$ from origin
• distance of $Q$ from origin = distance of $S$ from origin

By transitivity of equality of numbers, the first and third distances are equal. So $(P,S) \in R$. Hence $R$ is transitive.

Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation.

Now the second part: the equivalence class of a point $P \neq (0,0)$.

The equivalence class of $P$ under $R$ is the set of all points $Q$ such that $(P,Q) \in R$. By definition, this means: …

Method: Equivalence Relation from Equal Distances, and Its Geometric Classes

Use this when a relation says two points share the same value of a geometric quantity (distance from the origin), and you must prove it is an equivalence relation and describe the classes.

Steps

Step 1: Identify the underlying "same value" attribute

Here $P\,R\,Q$ means $OP=OQ$ (equal distance from the origin), a "same value" relation, which is automatically an equivalence relation.

Step 2: Verify the three properties

Reflexive: $OP=OP$. Symmetric: $OP=OQ \Rightarrow OQ=OP$. Transitive: $OP=OQ$ and $OQ=OS \Rightarrow OP=OS$.

Step 3: Describe the equivalence class geometrically …

Common Mistakes

Mistake 1: Thinking equal distance means the points coincide

Why it's wrong: infinitely many points share a distance from the origin, forming a whole circle rather than one point. Correct approach: the equivalence class is the full circle of that radius.

Mistake 2: Describing the class as a disc or an arc

Why it's wrong: only points at exactly distance $r$ qualify, not those inside (a disc) or only part of the circle (an arc). Correct approach: the class is the complete circle of radius $OP$. …