Answers

2016-03-17T17:17:38+05:30
Let an equilateral triangle Δ ABC : AB = BC = AC.
 vertices are A(3,2) and B(-3,2) and C(x,y).
The mid point of the side AB is M (0,2).
 (AB)² =  (3+3)² + (2-2)² = 6² + 0² = 36 
AB = 6 cm
AB = BC = AC = 6 unit.
AM = 3 unit.
As two vertices are A(-3,2) and B(3,0) so that Third vertex will be at
 Y-axis(x=0).
so that
Third vertex C(0,y) and it is located below the origin.
Hence it contains the origin.
Now,

    y² = (AC)² - (AM)² 
⇒ y² = 6² - 3² = 36 - 9 =27
    y = √27 = 3√3 unit
Hence Third vertices of the triangle are A(-3,2) , B(3,2) and C(0,3√3).
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2016-03-17T20:12:26+05:30
I think ths is the answer
c(0,3root3
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