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 | 2 years ago |  104

1 Answer

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 | 2 years ago

Related Questions

Find the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.

Class 11 Maths Straight Lines View Answer

Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.

Class 11 Maths Straight Lines View Answer

Find the acute angle between the lines 2x – y + 3 = 0 and x + y + 2 = 0.

Class 11 Maths Straight Lines View Answer

Find the angles between pairs of straight lines 3x – y + 5 = 0 and x – 3y + 1 = 0

Class 11 Maths Straight Lines View Answer

Find the angles between pairs of straight lines 3x + y + 12 = 0 and x + 2y – 1 = 0

Class 11 Maths Straight Lines View Answer