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The base AB of two equilateral triangles ABC and ABC- with side 2a lies along the x axis such that the mid point of AB is at the origin .find the

coordinates of the vertices C and C- of the triangles



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So the vertices of C will be (0.a√3)
and another C will be (0, - a√3)
0 0 0
Using Pythagoras theorem we get length of c on y axis = √(4a² - a²)
is '+' tive '-' tive direction of x axis
  so the vertices of C will be (0.a√3)
and another C will be (0, - a√3)
0 0 0
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