MathematicsMedium19×since 2002Q5027A triangle ABC lying in the first quadrant has two vertices as A(1, 2) and B(3, 1). If ∠BAC=90o\angle BAC = {90^o}∠BAC=90o and area(ΔABC)=55\left( {\Delta ABC} \right) = 5\sqrt 5(ΔABC)=55 s units, then the abscissa of the vertex C is :A1+251 + 2\sqrt 51+25B25−12\sqrt 5 - 125−1C1+51 + \sqrt 51+5D2+52 + \sqrt 52+5Check answerSkip