Question

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:

• A. $S=\left\{(x, y) | x^{2}+y^{2}=4\right\}$
• B. $S =\left\{( x , y )| x ^{2}+ y ^{2}=1\right\}$
• C. $S =\left\{( x , y ) | x ^{2}+ y ^{2}=\sqrt{2}\right\}$
• D. $S =\left\{( x , y ) \mid x ^{2}+ y ^{2}=2\right\}$

Answer 1

The Correct Option is D

To solve the given problem, we need to determine the equivalence class of the point \((1, -1)\) using the relation \(R =\{( P , Q ) | P\) and \(Q\) are at the same distance from the origin \(\}\).

1. First, we calculate the distance of the point \((1, -1)\) from the origin. The distance from the origin is given by the formula: \(d = \sqrt{x^2 + y^2}\). For the point \((1, -1)\):

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

1. The equivalence class of a point under this relation consists of all points that are equidistant from the origin as the given point. Therefore, we need all points \((x, y)\) such that the distance from the origin is \(\sqrt{2}\):

\(\sqrt{x^2 + y^2} = \sqrt{2}\)

1. Squaring both sides, we obtain:

\(x^2 + y^2 = 2\)

1. Thus, the set of all points \((x, y)\) satisfying this equation is:

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

Therefore, the equivalence class of the point \((1, -1)\) is the set of points \((x, y)\) such that \(x^2 + y^2 = 2\). This corresponds to the correct answer:

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