Question

2. The base of an equilateral triangle with side 2a lies along the y-axis such that the
mid-point of the base is at the origin. Find vertices of the triangle.
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Answer 1

Step-by-step explanation:

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 + a 2 ⇒ 4a 2 – a 2 = (OA)2 ⇒ (OA)2 = 3a 2 ⇒ OA = ∴Coordinates of point A = Thus, the vertices of the given equilateral triangle are (0, a), (0, –a), and or (0, a), (0,- a)

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