# Relation and Function

## NCERT Solution

### Exercise 1.1 Part 5

Question 10: Give an example of a relation, which is

- Symmetric but neither reflexive nor transitive
- Transitive but neither reflexive nor symmetric
- Reflexive and symmetric but not transitive
- Reflexive and transitive but not symmetric
- Symmetric and transitive but not reflexive.

**Solution:**

(i) Let R = {(1, 2), (2, 1)}

(ii) Let R = {(1, 2), (2, 3), (1, 3)}

(iii) R = {(1, 1) (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}

(iv) R ={(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (, 2), (2, 3), (3, 2)}

(v) R = {(2, 3), (3, 2), (1, 2), (1, 3), (3, 1)}

Thus, R is symmetric.

Thus, R is transitive.

Question 11: 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 ≠ (0, 0) is the circle passing through P with origin as centre.

**Solution:** Let O is the origin

Given, R = {(P, Q): distance of the point P from the origin is same as the distance of the point Q from the origin}

Therefore, R = {(P, Q): OP = PQ}

Now, let OP = y

Thus, R is reflexive relation.

Let OP = OQ = y

Thus, R is symmetric.

Again, Let OP = OQ = y and OQ = OR = y

Thus, R is transitive.

Thus, all distance related to P from the origin is same as OP. As a circle is the locus of all points having same distance from a point, in the given case from O, therefore, the set of the points related to P is a circle passing through P with O as the centre, a fixed point.