In a triangle A B C , coordinates of A are 1 , 2 and the equations of the medians through B and C are respectively, x + y = 5 and x = 4 . Then area of Δ A B C (in sq. units) is :

Step-by-step solution

Let C 4 , α be on x = 4 . Now, midpoint of AC is 5 2 , α + 2 2 which lies on x + y = 5 . ∴ α + 2 2 = 5 2 ⇒ α = 3 ∴ C 4 , 3 Similarly, let B β , 5 - β be on x + y = 5 . The midpoint of AB = β + 1 2 , 7 - β 2 which lies on x = 4 ⇒ β = 7 ⇒ B 7 , - 2 . ∴ Area of Δ = 1 2 1 2 1 7 - 2 1 4 3 1 = 1 2 1 - 2 - 3 - 2 7 - 4 + 1 21 + 8 = 1 2 18 = 9 sq. units