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16/01/2022 5:22 pm
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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
(1) \(\frac{-2b}{b+1}\)
(2) \(\frac{2b}{b+1}\)
(3) \(\frac{2b^2}{b+1}\)
(4) \(\frac{-2b^2}{b+1}\)
1 Answer
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16/01/2022 5:32 pm
Correct answer: (4) \(\frac{-2b^2}{b+1}\)
Explanation:
\(\frac{1}{2}\) \(\begin{vmatrix} a & 0 & 1 \\[0.3em] b & 2b+1 & 1 \\[0.3em] 0 & b & 1 \end{vmatrix}\) = 1
⇒ \(\begin{vmatrix} a & 0 & 1 \\[0.3em] b & 2b+1 & 1 \\[0.3em] 0 & b & 1 \end{vmatrix}\) = ±2
⇒ a (2b + 1 – b) – 0 + 1 (b2 – 0) = ± 2
⇒ a = \(\frac{\pm 2 - b^2}{b+1}\)
∴ a = \(\frac{2 - b^2}{b+1}\) and a = \(\frac{-2 - b^2}{b+1}\)
sum of possible values of 'a' is
= \(\frac{-2b^2}{b+1}\)