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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.
Given:
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.
To do:
We have to find the coordinates of the vertices R and R’ of the triangles.
Solution:
\( \Delta \mathrm{PQR} \) and \( \Delta \mathrm{PQR}^{\prime} \) are equilateral triangles with each side being $2a$ units and base $PQ$ and mid-point of $PQ$ is $O (0, 0)$. $PQ$ lies along y-axis.
In the figure,
\( \mathrm{PR}=\mathrm{QR}=\mathrm{PR}^{\prime}=\mathrm{QR}^{\prime}=2 a \) \( \mathrm{PO}=\mathrm{OQ}=a \)
Therefore,
In right \( \Delta \mathrm{PRO} \)
\( \mathrm{PR}^{2}=\mathrm{PO}^{2}+\mathrm{OR}^{2} \quad \) (Pythagoras theorem)
\( \Rightarrow(2 a)^{2}=(a)^{2}+\mathrm{OR}^{2} \)
\( \Rightarrow 4 a^{2}=a^{2}+\mathrm{OR}^{2} \)
\( \mathrm{OR}^{2}=4 a^{2}-a^{2}=3 a^{2} \)
\( \therefore \mathrm{OR}=\sqrt{3} a \)
Similarly,
\( \mathrm{OR}^{\prime}=-\sqrt{3} a \)
Therefore, the co-ordinates of \( \mathrm{R} \) is \( (\sqrt{3} a, 0) \) the and co-ordinates of \( \mathrm{R}^{\prime} \) is \( (-\sqrt{3} a, 0) \).
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