Let R = {(P, Q) | P and Q are at the same distance from the origin } be a relation, then the equival

Q: Why is the equivalence class a circle and not just a single point?

An equivalence class includes *all* elements that satisfy the relation. Since infinitely many points are at the same distance $\sqrt{2}$ from the origin, they form a continuous path—a circle. It is the set of all points that are 'equivalent' to $(1, -1)$ under the distance relation.

Q: Do I need to prove the relation is an equivalence relation first?

While the problem asks for the equivalence class, implying it *is* an equivalence relation, it is good practice to verify reflexivity, symmetry, and transitivity. Here, the distance from the origin is an invariant, and equality of distance is reflexive, symmetric, and transitive, so it is indeed an equivalence relation.

Q: Why did we square the equation $x^2 + y^2 = 2$? Is it necessary?

It is not strictly necessary, but it is standard practice to express the locus in the form $x^2 + y^2 = r^2$, which is the canonical equation of a circle. It makes the geometric interpretation much clearer.

Question

Let $R = \{(P, Q) \mid P 	ext{ and } Q 	ext{ are at the same distance from the origin }\}$ be a relation, then the equivalence class of $(1,-1)$ is the set:

Options:

(A) $S = \{(x, y) \mid x^2 + y^2 = 1\}$

(B) $S = \{(x, y) \mid x^2 + y^2 = 4\}$

(C) $S = \{(x, y) \mid x^2 + y^2 = \sqrt{2}\}$

(D) $S = \{(x, y) \mid x^2 + y^2 = 2\}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical landscape. Today, we stand before a problem that might seem like a simple coordinate geometry question, but it holds the key to understanding the beautiful concept of equivalence classes.

Imagine you are standing at the origin of a vast, infinite plane. You have a friend, let's call them point $P$, located at $(1, -1)$.

The relation $R$ defines a 'club' of points. To be in this club, you must be at the same distance from the origin as $P$. This is not just about a single point; it is about a family of points that share a common bond.

### The Distance Invariant

To understand who belongs to this club, we must first understand the defining characteristic of $P$. We need to calculate the distance of $P$ from the origin $O(0, 0)$.

Using the distance formula, we find that the distance $d$ is given by $d = \sqrt{x^2 + y^2}$. For our point $P(1, -1)$, the calculation is:

$$d = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}$$

This value, $\sqrt{2}$, is the 'membership fee' for our club. Any point $Q(x, y)$ that wants to join must also be at a distance of $\sqrt{2}$ from the origin.

### The Master Equation

Now, let us consider a general point $Q(x, y)$. For $Q$ to be in the equivalence class of $P$, it must satisfy the condition that its distance from the origin is also $\sqrt{2}$.

Mathematically, this is expressed as:

$$\sqrt{x^2 + y^2} = \sqrt{2}$$

This equation is the heart of our problem. It tells us that for any point $(x, y)$ in the equivalence class, the sum of the squares of its coordinates must be constant.

To make this look more familiar, we square both sides of the equation, yielding:

$$x^2 + y^2 = 2$$

### The Geometric Revelation

Look closely at the equation $x^2 + y^2 = 2$. It is the standard equation of a circle centered at the origin with a radius of $\sqrt{2}$.

This is the beauty of mathematics! What started as an abstract relation has revealed itself to be a perfect, symmetric circle.

Every single point on the circumference of this circle is at a distance of $\sqrt{2}$ from the origin. Therefore, every point on this circle is a member of the equivalence class of $(1, -1)$.

### Final Conclusion

So, the equivalence class of $(1, -1)$ is not just a point; it is the entire set of points:

$$S = \{(x, y) \mid x^2 + y^2 = 2\}$$

You have successfully navigated the logic of relations and the geometry of circles. Remember, in JEE Advanced, the most complex-looking problems often boil down to simple, elegant geometric truths.

Keep visualizing, keep calculating, and most importantly, keep falling in love with the process.

Animated Solution for Mathematics - Sets and Relations: Let $R = {(P, Q) ∣ P and Q are at the same distance from the origin}$ be a relation, then the equivalence class of $(1, - 1)$ is the set:

Select Answer:

Visualized Solution

Coordinate System and Origin

Plotting the Given Point

Understanding the Relation $R$

Distance of $P$ from Origin

The Distance Formula

Substituting Coordinates of $P$

Calculating the Distance

Defining the Equivalence Class

Visualizing a General Point $Q$

Equation for the Equivalence Class

Simplifying the Equation

The Locus is a Circle

The Final Set $S$

The Sigma Insight: Equivalence Relations

Analyzing the Setup

The Distance Invariant

The Master Equation

The Geometric Revelation

Final Conclusion

Similar Questions

Let a set $A = A_{1} \cup A_{2} \cup \dots \cup A_{k}$ , where $A_{i} \cap A_{j} = ϕ$ for $i \neq = j, 1 \leq i, j \leq k$ . Define the relation $R$ from $A$ to $A$ by $R = {(x, y) : y \in A_{i} if and only if x \in A_{i}, 1 \leq i \leq k}$ . Then, $R$ is:

Let $R_{1}$ and $R_{2}$ be two relations defined on $R$ by $a R_{1} b ⟺ ab \geq 0$ and $a R_{2} b ⟺ a \geq b$ , then

Let $P (S)$ denote the power set of $S = {1, 2, 3, \dots, 10}$ . Define the relations $R_{1}$ and $R_{2}$ on $P (S)$ as $A R_{1} B$ if $(A \cap B^{c}) \cup (B \cap A^{c}) = \emptyset$ and $A R_{2} B$ if $A \cup B^{c} = B \cup A^{c}, \forall A, B \in P (S)$ . Then:

The relation $R = {(x, y) : x, y \in Z and x + y is even}$ is:

Let $R = {(1, 2), (2, 3), (3, 3)}$ be a relation defined on the set ${1, 2, 3, 4}$ . Then the minimum number of elements, needed to be added in R so that R becomes an equivalence relation, is:

Define a relation R over a class of $n \times n$ real matrices A and B as "ARB iff there exists a non-singular matrix P such that $P A P^{- 1} = B$ ". Then which of the following is true?

The minimum number of elements that must be added to the relation $R = {(a, b), (b, c), (b, d)}$ on the set ${a, b, c, d}$ so that it is an equivalence relation, is ______.

Let $R$ be a relation defined on $N$ as $a R b$ is $2 a + 3 b$ is a multiple of 5, $a, b \in N$ . Then $R$ is

Let $A = {2, 3, 4, 5, ..., 30}$ and ' $\sim$ ' be an equivalence relation on $A \times A$ , defined by $(a, b) \sim (c, d)$ , if and only if $a d = b c$ . Then the number of ordered pairs which satisfy this equivalence relation with ordered pair $(4, 3)$ is equal to