MathematicsIf ‘ a  ‘ is the length of one of the sides of an equilateral triangle ABC,   base BC   lies on x-axis and vertex B   is at the origin, then from the following choices, the coordinates of the vertices of the triangle A,B,C   is:

If ' a  ' is the length of one of the sides of an equilateral triangle ABC,   base BC   lies on x-axis and vertex B   is at the origin, then from the following choices, the coordinates of the vertices of the triangle A,B,C   is:


  1. A
    a 2 , a 2  
  2. B
    a 2 , a 2  
  3. C
    a 2 , a 3 2  
  4. D
    a 2 , a 3 2   

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    Solution:

    Given that, ' a  ' is the length of one of the sides of an equilateral triangle ABC.
    AB=BC=CA=a  .
    The distance formula between two points ( x 1 , y 1 ),( x 2 , y 2 )   is given by,
    x 2 x 1 2 + y 2 y 1 2  
    Since the length of an equilateral triangle is ‘a’ and BC lies on x-axis, so the y-coordinate will be zero.
    And, x = x.
    Since, B is at the origin, C is the point C(x, 0).
    We know that, BC = a.
    Here,
    ( x 1 , y 1 )=(0,0) ( x 2 , y 2 )=(x,y)  
    x0 2 + 00 2 =a x=a  
    So, the coordinate of C is (a, 0).
    Put AB=a  .
    AB=a Here, ( x 1 , y 1 )=( x 1 , y 1 ) ( x 2 , y 2 )=(0,0) x 1 0 2 + y 1 0 2 = a 2 x 1 2 + y 1 2 = a 2             ......(1)  
    Put AC=a  , so that the coordinate of A be ( x 1 , y 1 )=( x 1 , y 1 )  and C is ( x 2 , y 2 )=(a,0)  , we get,
    a x 1 2 + 0 y 1 2 = a 2 a 2 + x 1 2 2a x 1 + y 1 2 = a 2 2a x 1 = a 2 x 1 = a 2                   ......(2)   Substitute the value of (2) in (1), we get,
    a 4 + y 1 2 = a 2 y 1 2 = 3 a 2 4 y 1 = a 3 2   A= a 2 , a 3 2  
    The coordinate is A= a 2 , a 3 2   .
    Hence, option 3) is correct.
     
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