A triangle A B C lying in the first quadrant has two vertices as A 1 , 2 and B 3 , 1 . If ∠ B A C = 90 o , and ar Δ ABC = 5 5 sq. units, then the abscissa of the vertex C is :

Step-by-step solution

m A C = β − 2 α − 1 m A B = 2 − 1 1 − 3 = − 1 2 A B ⊥ A C ∴ β − 2 α − 1 − 1 2 = − 1 β = 2 α − 2 + 2 β = 2 α Now area of Δ A B C = 5 5 = 1 2 A B . A C ⇒ 1 2 3 − 1 2 + 1 − 2 2 . α − 1 2 + β − 2 2 = 5 5 ⇒ α − 1 2 + 2 α − 2 2 = 10 ⇒ α − 1 2 5 = 10 ⇒ α − 1 = 2 5 ⇒ α = 1 ± 2 5 .