the base of an equilateral triangle with side 2a lies along y-axis such that the midpoint of the base is at the origin.find the vertices of the triangle.
Since it is an equilateral triangle, each of its sides will be 2a.
Midpoint is origin....therefore,taking a vertice on +ve side of x axis,
taking (0,0) as midpoint, length of +ve y=a [half of 2a, upwards from midpoint]
therefore, coordinates of A=(0,a)
length of -ve y=-a [half of 2a,taken downwards from midpoint]
Therefore, coordinates of B=(0,-a)
Now, we find C by applying the pythagoras theorem,
(2a)2 = (a)2 + (x)2
4a2 = a2 +x2
4a2-a2=x2
.:x2 = 3a2
.:x =( root 3) a
Similarily taking C on the -ve side of x axis, x = -(root 3)a
Therefore, the vertices are A(0,a),B(0,-a) and C( [root 3]a) or A(0,a), B(0,-a) and C(-[root 3]a)