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…

A triangle ABC lying in the first quadrant has two vertices as A1,2 and B3,1. If∠BAC=90o,and arΔABC=55 sq. units, then the abscissa of the vertex C is :
  1. 1+5
  2. 1+25 
  3. 2+5
  4. 25-1

Solution

mAC=β−2α−1

mAB=2−11−3=−12

AB⊥AC

∴ β−2α−1−12=−1

β=2α−2+2

β=2α

Now area of ΔABC=55=12AB.AC

⇒ 123−12+1−22.α−12+β−22=55

⇒ α−12+2α−22=10

⇒ α−125=10

⇒ α−1=25

⇒ α=1±25.

Asked in: JEE Main 2020 (04 Sep Shift 1)

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