Question -
Answer -
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, тАУa).
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 ╬ФAOC, we obtain
(AC)2┬а=(OA)2┬а+ (OC)2
тЗТ (2a)2┬а=(OA)2┬а+┬аa2
тЗТ 4a2┬атАУ┬аa2┬а=(OA)2
тЗТ (OA)2┬а=3a2
тЗТ OA =
тИ┤Coordinates of point A =
Thus, the vertices ofthe given equilateral triangle are (0,┬аa), (0, тАУa),and┬а
┬аor (0,┬аa), (0, тАУa),and┬а
.