Key Concepts and Formulas

• Distance from Origin: The distance of a point $(x, y)$ from the origin $(0,0)$ is given by the formula $d = \sqrt{x^2 + y^2}$.
• Equivalence Class: For an equivalence relation $R$ on a set $A$, the equivalence class of an element $a \in A$, denoted by $[a]$, is the set of all elements $b \in A$ such that $(a, b) \in R$. Mathematically, $[a] = \{b \in A \mid (a, b) \in R\}$.
• Equivalence Relation Properties: An equivalence relation must be reflexive ($aRa$), symmetric ($aRb \implies bRa$), and transitive ($aRb \text{ and } bRc \implies aRc$).

Step-by-Step Solution

Step 1: Understand the Given Relation R The relation $R$ is defined as $R = \{(P, Q) \mid P \text{ and } Q \text{ are at the same distance from the origin}\}$. Let $P = (x_P, y_P)$ and $Q = (x_Q, y_Q)$. The distance of a point $(x, y)$ from the origin $(0,0)$ is $\sqrt{x^2+y^2}$. Thus, the condition for $(P, Q) \in R$ is: $\sqrt{x_P^2 + y_P^2} = \sqrt{x_Q^2 + y_Q^2}$ Squaring both sides, we get: $x_P^2 + y_P^2 = x_Q^2 + y_Q^2$ This means two points are related if and only if the sum of the squares of their coordinates (which is the square of their distance from the origin) is the same.

Step 2: Identify the Element for which the Equivalence Class is to be Found We need to find the equivalence class of the point $(1, -1)$. Let this point be $P_0 = (1, -1)$.

Step 3: Calculate the Characteristic Property of the Given Element The equivalence class of $P_0 = (1, -1)$ will consist of all points $Q = (x, y)$ that are related to $P_0$. According to the definition of $R$, this means $P_0$ and $Q$ must be at the same distance from the origin. First, calculate the square of the distance of $P_0 = (1, -1)$ from the origin: $d(P_0, \text{origin})^2 = (1)^2 + (-1)^2 = 1 + 1 = 2$ So, any point $(x, y)$ in the equivalence class of $(1, -1)$ must have the same square of distance from the origin, which is 2.

Step 4: Formulate the Equivalence Class The equivalence class of $(1, -1)$, denoted as $[(1, -1)]$, is the set of all points $(x, y)$ such that their square of distance from the origin is equal to 2. $[(1, -1)] = \{ (x,y) \mid x^2 + y^2 = 2 \}$

Step 5: Match the Result with the Given Options We have found that the equivalence class is the set $S = \{ (x,y) \mid x^2 + y^2 = 2 \}$. Let's compare this with the given options: (A) $S = \{ (x,y)|{x^2} + {y^2} = \sqrt 2 \}$ (B) $S = \{ (x,y)|{x^2} + {y^2} = 2\}$ (C) $S = \{ (x,y)|{x^2} + {y^2} = 1\}$ (D) $S = \{ (x,y)|{x^2} + {y^2} = 4\}$

Our derived set matches option (B).

Common Mistakes & Tips

• Confusing Distance and Squared Distance: The relation equates distances, but it's often easier to work with the square of the distance ($x^2 + y^2$) as it avoids square roots and potential sign errors. Ensure you equate the correct quantity.
• Misinterpreting the Relation: The relation is about points being at the same distance. Do not confuse this with points being at a specific distance, unless that specific distance is derived from the element whose equivalence class is being sought.
• Geometric Interpretation: The equivalence class of a point $(a,b)$ under this relation is always a circle centered at the origin passing through $(a,b)$. The equation of this circle is $x^2 + y^2 = a^2 + b^2$.

Summary

The relation $R$ states that two points are related if they are at the same distance from the origin. To find the equivalence class of a point $(1, -1)$, we first determine its distance from the origin. The square of this distance is $(1)^2 + (-1)^2 = 2$. The equivalence class then consists of all points $(x, y)$ that have the same square of distance from the origin, which is $x^2 + y^2 = 2$. This corresponds to a circle centered at the origin with radius $\sqrt{2}$.

The final answer is \boxed{B}.