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The area (in square units) of the triangle formed by $x + y + 1 = 0$ and the pair of straight lines $x^{2} - 3 xy + 2 y^{2} = 0$ is

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$\frac{7}{12}$

$\frac{5}{12}$

$\frac{1}{12}$

$\frac{1}{6}$

answer is C.

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Detailed Solution

$u s e Δ = | \frac{n^{2} \sqrt{h^{2} - a b}}{a m^{2} - 2 h l m + b l^{2}} |$

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