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

Solution:

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

Accordingly, AB = BC = CA = 2a

Assume that base BC (the base of the triangle) 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 (0, a), while the coordinates of point B (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 DAOC , we obtain

AC² = OA² + OC²

OA² = AC² - OC²

OA² = (2a)² + (a)²

OA² = 4a² - a²

OA² = 3a²

OA = ± √3 a

Therefore, coordinates of point A (± √3 a, 0)

Thus, the vertices of the given equilateral triangle are (0, a), (0, - a) and (√3 a, 0) or (0, a), (0, - a) and (- √3 a, 0)

NCERT Solutions Class 11 Maths Chapter 10 Exercise 10.1 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

Summary:

We found out that the vertices of the triangle whose base is 2a and lies along y-axis such that the midpoint of the base is at the origin are (0, a), (0, - a) and (√3 a, 0) or (0, a), (0, - a) and (- √3 a, 0)