Let and be the two points on the line such that and are symmetric with respect to the origin. Suppose is a point on such that is an equilateral triangle. Then, the area of the is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: A
Solution
Since B and C are symmetric about the origin on y + x = 0, let B = (a, -a) and C = (-a, a) for some a > 0. The midpoint of BC is the origin O = (0,0), and BC has length 2a√2.
For an equilateral triangle, A lies on the perpendicular bisector of BC passing through the origin. The line BC has slope -1, so the perpendicular bisector has slope 1, i.e., y = x.
A is the intersection of y = x and y - 2x = 2: substituting y = x gives x - 2x = 2, so x = -2, y = -2. Thus A = (-2, -2).
The distance from A to the origin (midpoint of BC) is OA = √(4+4) = 2√2.
For an equilateral triangle with side s, the distance from the centroid to a vertex is s/√3, but here we use the median length. The median from A to midpoint of BC equals OA = 2√2, and for equilateral triangle the median = (√3/2)s, so s = 2·OA/√3 = 4√2/√3.
Area = (√3/4)s² = (√3/4)·(32/3) = 8√3/3 = 8/√3.
For an equilateral triangle, A lies on the perpendicular bisector of BC passing through the origin. The line BC has slope -1, so the perpendicular bisector has slope 1, i.e., y = x.
A is the intersection of y = x and y - 2x = 2: substituting y = x gives x - 2x = 2, so x = -2, y = -2. Thus A = (-2, -2).
The distance from A to the origin (midpoint of BC) is OA = √(4+4) = 2√2.
For an equilateral triangle with side s, the distance from the centroid to a vertex is s/√3, but here we use the median length. The median from A to midpoint of BC equals OA = 2√2, and for equilateral triangle the median = (√3/2)s, so s = 2·OA/√3 = 4√2/√3.
Area = (√3/4)s² = (√3/4)·(32/3) = 8√3/3 = 8/√3.
[JEE Main 2023 · 29 Jan · Shift 1]