The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find the vertices of the triangle.

Key points to solve the problem:


The idea of distance formula- Distance between two points P(x1,y1) and Q(x2,y2) is given by- PQ =


Equilateral triangle- triangle with all 3 sides equal.


Coordinates of the midpoint of a line segment – Let P(x1,y1 �) and Q(x2,y2) be the end points of line segment PQ. Then coordinated of the midpoint of PQ is given by –


Given, an equilateral triangle with base along y axis and midpoint at (0,0)


coordinates of triangle will be A(0,y1) B(0,y2) and C(x,0)


As midpoint is at origin y1+y2 = 0 y1 = -y2 …..eqn 1


Also length of each side = 2a (given)


AB = ….eqn 2


from eqn 1 and 2:


y1 = a and y2 = -a


2 coordinates are – A(0,a) and B(0,-a)


See the figure:



Clearly from figure:


DC = x


Also in ΔADC: cos 30° =



Squaring both sides:




Coordinates of C are (√3a,0) or (-√3a,0) ….ans


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