#### Question

The base PQ of two equilateral triangles PQR and PQR' with side 2a lies along y-axis such that the mid-point of PQ is at the origin. Find the coordinates of the vertices R and R' of the triangles.

#### Solution

In an equilateral triangle, the height ‘*h*’ is given by

`h =(sqrt3`("Side of the equilateral triangle"))/2`

Here it is given that '*PQ'* forms the base of two equilateral triangles whose side measures '*2a'* units.

The height of these two equilateral triangles has got to be

`h = (sqrt3("Side of the equilateral triangle"))/2`

`= (sqrt3(2a))/2`

`h = asqrt3`

In an equilateral triangle, the height drawn from one vertex meets the midpoint of the side opposite this vertex.

So here we have ‘*PQ*’ being the base lying along the *y*-axis with its midpoint at the origin, that is at *(0, 0)*

So the vertices ‘R’ and ‘R’’ will lie perpendicularly to the y-axis on either side of the origin at a distance of `asqrt3` units

Hence the co-ordinates of ‘*R*’ and ‘*R’*’ are

`R(asqrt3,0)`

`R'(-a sqrt3, 0)`