Maths-
General
Easy
Question
The x-coordinates of the vertices of a square of unit area are the roots of the equation and the y-coordinates of the vertices are the roots of the equation then the possible vertices of the square is/are :
- (1, 1), (2, 1), (2, 2), (1, 2)
- (-1, 1), (-2, 1), (-2, 2), (-1, 2)
- (2, 1), (1, -1), (1, 2), (2, 2)
- (-2, 1), (-1, -1), (-1, 2), (-2, 2)
The correct answer is: (-1, 1), (-2, 1), (-2, 2), (-1, 2)
we are given that
The x-coordinates of the vertices of a square of unit area are the roots of the a equation and the y-coordinates of the vertices are the roots of another equation we have to find the possible vertices of the square are
x2−3∣x∣+2=0
⟹(∣x∣−1)(∣x∣−2)=0 ∴x=−1,1,2,−2
y2−3y+2=0
⟹(y−1)(y−2)=0⟹y=1,2.
The square is of unit radius. Hence, the coordinates of the square can be :
(1,1),(1,2),(2,1),(2,2)
(−1,1),(−1,2),(−2,2),(−2,1)
therefore the possible vertices of the square are (1,1),(1,2),(2,1),(2,2) and (−1,1),(−1,2),(−2,2),(−2,1)
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