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Q11:
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 $P \neq (0, 0)$ is the circle passing through $P$ with origin as centre.

Solution :

Step 1: Defining the Relation and Set

Let $A$ be the set of all points in a Cartesian plane. The relation $R$ is defined on $A$ as: $R = \{(P, Q) \in A \times A : dist(O, P) = dist(O, Q)\}$, where $O$ is the origin $(0, 0)$. Let $d(X)$ denote the Euclidean distance of a point $X$ from the origin, $d(X) = \sqrt{x^2 + y^2}$. Thus, $R = \{(P, Q) : d(P) = d(Q)\}$.

Step 2: Proving Reflexivity

For any point $P \in A$, the distance of $P$ from the origin is $d(P)$. Clearly, $d(P) = d(P)$. [By the Reflexive Property of Equality]. Therefore, $(P, P) \in R$ for all $P \in A$. Hence, $R$ is reflexive.

Step 3: Proving Symmetry

Suppose $(P, Q) \in R$. By definition, $d(P) = d(Q)$. By the Symmetric Property of Equality, $d(Q) = d(P)$. This implies that $(Q, P) \in R$. Since $(P, Q) \in R \implies (Q, P) \in R$, $R$ is symmetric.

Step 4: Proving Transitivity

Suppose $(P, Q) \in R$ and $(Q, S) \in R$. Then, $d(P) = d(Q)$ and $d(Q) = d(S)$. By the Transitive Property of Equality, $d(P) = d(S)$. Therefore, $(P, S) \in R$. Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation [Per the Definition of Equivalence Relation].

Step 5: Geometric Interpretation of the Equivalence Class

Let $P = (x_0, y_0)$ be a point such that $P \neq (0, 0)$. The set of all points $Q(x, y)$ related to $P$ is the set $\{Q : d(Q) = d(P)\}$. Let $k = d(P) = \sqrt{x_0^2 + y_0^2}$. The condition $d(Q) = k$ becomes $\sqrt{x^2 + y^2} = k$, which squares to $x^2 + y^2 = k^2$. [By the standard equation of a circle with center $(0,0)$ and radius $r = k$]. This is a circle centered at the origin passing through $P$.

P O Set of points related to P

Step 6: Conclusion

The set of all points $Q$ related to $P$ is the set of points equidistant from the origin. Geometrically, this locus describes a circle with radius $r = \sqrt{x_0^2 + y_0^2}$ centered at the origin, which necessarily passes through $P$.

Final Solution: The relation R is an equivalence relation because it satisfies reflexivity, symmetry, and transitivity. The set of all points related to $P \neq (0,0)$ is a circle centered at the origin with radius equal to the distance of $P$ from the origin.


More Questions from Class 12 Mathematics Relations and Functions EXERCISE 1.1


CBSE Solutions for Class 12 Mathematics Relations and Functions


Chapters in CBSE - Class 12 Mathematics


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