Let B and C be the two points on the line y+x=0 such that B and C are symmetric with respect to the origin. Suppose A is a point on y-2x=2 such that △ ABC is an equilateral triangle. Then, the area of the △ ABC is :

Solution

Maths sir · 4 steps

The origin is the midpoint of BC, so A sits on the perpendicular bisector y = x.

1. Midpoint of BC. B and C are symmetric about the origin, so the origin O is the midpoint of BC. In an equilateral triangle the median from A is also the altitude, so AO ⊥ BC.
2. Locate A. BC lies on y = -x, with slope -1. The perpendicular through O is y = x. Meeting it with y = 2x + 2 gives x = 2x + 2, so x = -2 and A = (-2, -2).
3. Altitude length. The altitude is h = AO = √(4 + 4) = 2√2, so h² = 8.

   Area = h²/√3 = 8/√3

4. So. With h = √3/2a, the area √3/4a² becomes h²/√3. The area is 8/√3, option (d).

The idea to keep. When a fixed point is the midpoint of the base of an equilateral triangle, the apex lies on the perpendicular through it. Find the altitude h and use area = h²/√3.

Answer from the NTA final answer key. Solution by PrepWiser.