**Question:**

The base *PQ* of two equilateral triangles *PQR* and *PQR'* with side 2*a* 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

$h=\frac{\sqrt{3} \text { (Side of the equilateral triangle) }}{2}$

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

$h=\frac{\sqrt{3} \text { (Side of the equilateral triangle) }}{2}$

$=\frac{\sqrt{3}(2 a)}{2}$

$h=a \sqrt{3}$

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 sides of the origin at a distance of ' $a \sqrt{3}$ ' units.

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

$R(a \sqrt{3}, 0)$

$R^{\prime}(-a \sqrt{3}, 0)$