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Question

The x-coordinates of the vertices of a square of unit area are the roots of the equation x to the power of 2 end exponent minus 3 vertical line x vertical line plus 2 equals 0 and the y-coordinates of the vertices are the roots of the equation y to the power of 2 end exponent minus 3 y plus 2 equals 0 then the possible vertices of the square is/are :

  1. (1, 1), (2, 1), (2, 2), (1, 2)    
  2. (-1, 1), (-2, 1), (-2, 2), (-1, 2)    
  3. (2, 1), (1, -1), (1, 2), (2, 2)    
  4. (-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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