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.

Answer 1

Since midpoint of PQ is the origin and PQ=2a
Therefore, OP=OQ=a

Here the coordinates of P and Q are (0,a) &(0,-a) respectively.
Since ∆PQR and ∆PQR' are equilateral.
Therefore, their third vertice R and R' lie on the perpendicular bisector of PQ.
X'OX is perpendicular bisector of base PQ.
Thus R and R' lie on X-axis
therefore their Y coordinate is zero.


In ∆ROP

$O{R}^2+O{P}^2=P{R}^2$

$O{R}^2-a^2=2a^2$

$O{R}^2=3a^2$

$OR=\sqrt{3}a$

Similarly ,$O{R}^{\prime }=\sqrt{3}a$
Thus coordinate of vertices R and R' are(√3a,0) and (-√3a,0) respectively