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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.
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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