Skip to main content

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.



Class 12, NCERT Chapter 1,  Exercise 1.1, Q11


It is given that
R = {(P, Q): distance of the point P from the origin is the same as the distance of the point Q from the origin},
(P, P) ϵ R since the distance of point P from the origin is always the same as the distance of the same point P from the origin.
Therefore, R is reflexive.
Now, Let us take (P, Q) ϵ R,
⇒ The distance of point P from the origin is always the same as the distance of the same point Q from the origin.
⇒ The distance of point Q from origin is always the same as the distance of the same point P from the origin.
⇒ (Q, P) ϵ R
Therefore, R is symmetric.
Now, Let (P, Q), (Q, S) ϵ R
⇒ The distance of point P and Q from origin is always the same as the distance of the same point Q and S from the origin.
⇒ The distance of points P and S from the origin is the same.
⇒ (P, S) ϵ R
Therefore, R is transitive.
Therefore, R is an equivalence relation.
The set of all points related to P ≠ (0,0) will be those points whose distance from the origin is the same as the distance of point P from the origin.
So, if O(0,0) is the origin and OP = k, then the set of all points related to P is at a distance of k from the origin.
Therefore, this set of points forms a circle with the center as the origin and this circle passes through point P.
Labels:

Comments

Post a Comment

Popular posts from this blog

In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer. (i) f : R → R defined by f(x) = 3 – 4x (ii) f : R → R defined by f(x) = 1 + x^2

Class 12, NCERT Chapter 1,  Exercise 1.2, Q7 (i)  Here f(x) = 3 – 4x Then f(x 1 ) = 3 – 4x 1 and f (x 2 ) = 3 – 4x 2 Now f(x 1 ) = f (x 2 ) ⇒ 3 – 4x 1  = 3 – 4x 2 ⇒ – 4x 1  = –4x 2 ⇒ x 1  = x 2 ∴ f(x 1 ) = f(x 2 ) & x 1  = x 2 So, f is one-one function. Let f(x) = y ∈ R Then y=3 - 4x ⇒ x= 3-y/4 (ii) Let x 1  = 2 and x 2  = –2 ∈ R Here f(x) = 1 + x 2 Then f(x 1 ) = f (2) = 1 + (2)  2  = 5 and f (x 2 ) = f(–2) = 1 + (–2)  2  = 5 ∴ f (x 1 ) = f(x 2 ) but x 1  ≠ x 2 So, f is not one-one function. Let f(x) = – 2 ⇒R Then 1 + x 2  = –2 ⇒ x 2  = –3 = ± √–3 ∈R So, f is not onto function.
Post a Comment
Read more

Let f : R → R be defined as f(x) = x 4 . Choose the correct answer. (A) f is one-one onto (B) f is many-one onto (C) f is one-one but not onto (D) f is neither one-one nor onto.

Class 12, NCERT Chapter 1,  Exercise 1.2, Q11 f : R → R is defined as  f (x) = x 4 Let  x ,  y  ∈  R  such that  f ( x ) =  f ( y ). ∴ does not imply that x 1 =x 2 . For instance, f (1) =  f (-1) = 1 ∴  f  is not one-one. Consider an element 2 in co-domain  R . It is clear that there does not exist any  x  in domain  R  such that  f ( x ) = 2. ∴  f  is not onto. Hence, function  f  is neither one-one nor onto. The correct answer is D
Post a Comment
Read more

Let L be the set of all lines in a plane and R be the relation in L defined as R = {(L1 , L2 ) : L1 is perpendicular to L2 }. Show that R is symmetric but neither reflexive nor transitive.

RELATIONS AND FUNCTIONS Class 12, NCERT Chapter 1,  Example 3 Solution Figure R is not reflexive, as a line L1 can not be perpendicular to itself, i.e., (L1 , L1 ) ∉ R. R is symmetric as (L1 , L2 ) ∈ R ⇒ L1 is perpendicular to L2 ⇒ L2 is perpendicular to L1 ⇒ (L2 , L1 ) ∈ R. R is not transitive. Indeed, if L1 is perpendicular to L2 and L2 is perpendicular to L3 , then L1 can never be perpendicular to L3 . In fact, L1 is parallel to L3 , i.e., (L1 , L2 ) ∈ R, (L2 , L3 ) ∈ R but (L1 , L3 ) ∉ R. Example1       Example2 Example3 ←you are here
Post a Comment
Read more