A triangle ABC lying in the first quadrant has two vertices at A(1,2) and B(3,1). If ∠BAC=90∘and area Δ(ABC)=55 sq. units then the abscissa of the vertex C is

Let the coordinates of vertex C be (α,β). It is given that ∠BAC=90∘⇒ Slope of AB x Slope of AC = -1⇒ 1−23−1×β−2α−1=−1⇒ 2α=β              …(i)Area of Δ(ABC)=55⇒ 12AB×AC=55⇒ (3−1)2+(1−2)2×(α−1)2+(β−2)2=105⇒ 5(α−1)2+(2α−2)2=105⇒ 5|α−1|=10⇒ |α−1|=25⇒α−1=±25⇒α=1±25