Given coordinates of triangle ABC, A(a,0),B(b,2b+1),C(0,b)
Area=21∣ab002b+1b111∣
21[a(b+1)+b2]=±1
ab+a+b2=±2
If 2 is positive
a=b+12−b2
If 2is negative
a=b+1−2−b2
Sum of possible values of a=b+12−b2+b+1−2−b2
=b+1−2b2
Let A(a,0),B(b,2b+1) and C(0,b),b=0,∣b∣=1, be points such that the area of triangle ABC is 1 sq. unit, then the sum of all possible values of a is:
Held on 27 Aug 2021 · Verified 6 Jul 2026.
b+1−2b
b+12b2
b+1−2b2
b+12b
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