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 - 2 x = 2 such that ∆ A B C is an equilateral triangle. Then, the area of the ∆ A B C is

Step-by-step solution

On plotting the diagram of the given data we get, Now given, x = y and y - 2 x = 2 On solving we get, A ( - 2 , - 2 ) Now height h of ∆ A B C = Distance of point A to origin, h = - 2 2 + ( - 2 ) 2 ⇒ h = 2 2 Given ∆ A B C is an equilateral triangle, So, sin B = h A B ⇒ sin 60 ° = h A B       ( ∵ ∠ A = ∠ B = ∠ C = 60 ° ) Area of Δ = 3 4 A B 2 ⇒ Area of Δ = 3 4 h 2 sin 2 60 ∘ = 3 4 × 32 3 So, Area of ∆ A B C = 8 3