The base PQ of two equilateral triangles PQR and PQR’ with side 2a lies
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

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

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

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

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

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