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 :

Solution

Maths sir · 3 steps

Area 1 means the determinant is +2 or -2, so two values of a; add them.

1. Area formula. For A(a, 0), B(b, 2b+1), C(0, b), use Area = 1/2|x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|.

   1/2|a(2b+1-b) + b(b-0) + 0| = 1/2|a(b+1) + b²| = 1

2. Remove the modulus. So a(b+1) + b² = 2 or a(b+1) + b² = -2. Since |b| ≠ 1, b + 1 ≠ 0, and we can divide. This gives a = (2 - b²)/(b+1) or a = (-2 - b²)/(b+1).
3. So. The sum is ((2 - b²) + (-2 - b²))/(b+1) = -2b²/(b+1), which is option (d).

The idea to keep. When an area is given, the modulus hides two cases, +2Δ and -2Δ. Keep both, because the question is asking for every possible value.

Answer from the NTA final answer key. Solution by PrepWiser.