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

The slope of the line AB is 1-23-1=-12 Hence the slope of AC is 2It gives b−2a−1=2b−2=2a−2b=2a Suppose that C is (a,2a) The area of the triangle is 55 square units It implies 12ABAC=55125AC=55AC=10Hence, a−12+2a−22=25a−12+4a−12=105a−1=10a−1=25a=1+25 Therefore, abscissa of C is 1+25