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 problem, we need to determine the equivalence class of the point $(1, -1)$ under the given relation $R$ defined as: two points are related if they are at the same distance from the origin.

First, we calculate the distance of the point $(1, -1)$ from the origin, which can be found using the distance formula:

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

Substituting the coordinates of the point $(1, -1)$:

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

$d = \sqrt{1 + 1}$

$d = \sqrt{2}$

The distance of the point $(1, -1)$ from the origin is $\sqrt{2}$.

According to the relation $R$, the equivalence class contains all points $(x, y)$ that are at the same distance $\sqrt{2}$ from the origin. Thus, it is the set of all points that satisfy:

$x^2 + y^2 = 2$

Therefore, the equivalence class of $(1, -1)$ is the set:

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

This matches with the provided correct answer option:

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

Thus, the correct answer is the set where the distance from the origin is constant and equals to \sqrt{2}, forming a circle in the Cartesian plane.