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Question
The base of an equilateral triangle with side 2a lies along they y-axis such that the mid point of the base is at the origin. Find vertices of the triangle.
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Solution
Let ABC be the given equilateral triangle with side 2a.
Accordingly, AB = BC = CA = 2a
Assume that base BC lies along the y-axis such that the mid-point of BC is at the origin.
i.e., BO = OC = a, where O is the origin.
Now, it is clear that the coordinates of point C are (0, a), while the coordinates of point B are (0, –a).
It is known that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular.
Hence, vertex A lies on the y-axis.
On applying Pythagoras theorem to ΔAOC, we obtain
(AC)2 = (OA)2 + (OC)2
⇒ (2a)2 = (OA)2 + a2
⇒ 4a2 – a2 = (OA)2
⇒ (OA)2 = 3a2
⇒ OA = `sqrt3`
∴ Coordinates of point A = `(± sqrt(3a),0)`
Thus, the vertices of the given equilateral triangle are (0, -a) and `(sqrt(3a),0)` or (0, a), (0, -a) and `(-sqrt(3a),0)`.
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