
mAC=α−1β−2
mAB=1−32−1=−21
AB⊥AC
∴ α−1β−2(−21)=−1
β=2α−2+2
β=2α
Now area of ΔABC=55=21AB.AC
⇒ 21(3−1)2+(1−2)2.(α−1)2+(β−2)2=55
⇒ (α−1)2+(2α−2)2=10
⇒ (α−1)25=10
⇒ ∣α−1∣=25
⇒ α=1±25.
JEE Main 2020 — Mathematics Coordinate Geometry
A triangle ABC lying in the first quadrant has two vertices as A(1,2) and B(3,1). If∠BAC=90o,and ar(ΔABC)=55 sq. units, then the abscissa of the vertex C is :
Held on 4 Sept 2020 · Verified 6 Jul 2026.
1+5
1+25
2+5
25−1
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