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.

Asked by Abhisek | 1 year ago |  80

##### Solution :-

Let us consider ABC be the given equilateral triangle with side 2a.

Where, AB = BC = AC = 2a

In the above figure, by assuming that the base BC lies on the x axis such that the mid-point of BC is at the origin i.e. BO = OC = a, where O is the origin.

The co-ordinates of point C are (0, a) and that of B are (0,-a)

Since the line joining a vertex of an equilateral ∆ with the mid-point of its opposite side is perpendicular. So, vertex A lies on the y –axis

By applying Pythagoras theorem

(AC)2 = OA2 + OC2

(2a)2= a2 + OC2

4a2 – a2 = OC2

3a= OC2

OC =$$\sqrt{3a}$$

Co-ordinates of point C = ± $$\sqrt{3a}$$, 0

The vertices of the given equilateral triangle are (0, a), (0, -a), ($$\sqrt{3a}$$, 0)

Or (0, a), (0, -a) and ($$- \sqrt{3a}$$, 0)

Answered by Pragya Singh | 1 year ago

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