If 3x^{2} + xy - y^{2} - 3x + 6y + k = 0 repr | vedclass

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(D) The general equation of a second-degree curve is $ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0$. Comparing this with the given equation $3x^{2} + xy

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$-9$

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(D) The general equation of a second-degree curve is $ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0$. Comparing this with the given equation $3x^{2} + xy - y^{2} - 3x + 6y + k = 0$,we get: $a = 3, b = -1, h = \frac{1}{2}, g = -\frac{3}{2}, f = 3, c = k$. For the equation to represent a pair of lines,the condition is $abc + 2fgh - af^{2} - bg^{2} - ch^{2} = 0$. Substituting the values: $3(-1)(k) + 2(3)(-\frac{3}{2})(\frac{1}{2}) - 3(3)^{2} - (-1)(-\frac{3}{2})^{2} - k(\frac{1}{2})^{2} = 0$. $-3k - \frac{9}{2} - 27 + \frac{9}{4} - \frac{k}{4} = 0$. Multiplying by $4$ to clear the denominators: $-12k - 18 - 108 + 9 - k = 0$. $-13k - 117 = 0$. $-13k = 117$. $k = -9$.

If $3x^{2} + xy - y^{2} - 3x + 6y + k = 0$ represents a pair of lines,then $k$ is equal to

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