JEE Main 2021 · 26 February · morning · Single correct · +4 / −1
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 = { (x,y)|x² + y² = √2 }
- (b) S = { (x,y)|x² + y² = 2}
- (c) S = { (x,y)|x² + y² = 1}
- (d) S = { (x,y)|x² + y² = 4}
Answer and solution
Answer (b) S = { (x,y)|x² + y² = 2}
Solution
Maths sir · 3 steps
Same distance from the origin means same circle: find the radius of the point (1, −1) and you are done.
- What the class means. The equivalence class of (1, -1) is every point (x, y) related to it, that is, every point at the same distance from the origin as (1, -1).
- Distance of (1, −1). Its distance from the origin is √(1² + (-1)²) = √2.
√(x² + y²) = √2 ⇒ x² + y² = 2
- So. The equivalence class is the circle centred at the origin with radius √2, which is S = {(x, y) | x² + y² = 2}, option (b).
The idea to keep. For a relation like 'same distance from the origin', each equivalence class is a circle centred at the origin. Find the radius r of the given point and write x² + y² = r².
Answer from the NTA final answer key. Solution by PrepWiser.