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JEE Math Practice Question

A triangle ABC lying in the first quadrant has two vertices as A(1, 2) and B(3, 1). If $$\angle BAC = {90^o}$$ and area$$\left( {\Delta ABC} \right) = 5\sqrt 5 $$ s units, then the abscissa of the vertex C is :

  1. A.$$1 + 2\sqrt 5 $$
  2. B.$$ 2\sqrt 5 - 1$$
  3. C.$$1 + \sqrt 5 $$
  4. D.$$2 + \sqrt 5 $$

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