The value of lambda for which the equation x^ | vedclass

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(B) The general equation of the second degree $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if $abc + 2fgh - af^2 - bg^

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$3$

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(B) The general equation of the second degree $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.Comparing the given equation $x^2 - \lambda xy + 2y^2 + 3x - 5y + 2 = 0$ with the general form:$a = 1, b = 2, c = 2, h = -\frac{\lambda}{2}, g = \frac{3}{2}, f = -\frac{5}{2}$.Substituting these values into the condition $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$:$(1)(2)(2) + 2(-\frac{5}{2})(\frac{3}{2})(-\frac{\lambda}{2}) - (1)(-\frac{5}{2})^2 - (2)(\frac{3}{2})^2 - (2)(-\frac{\lambda}{2})^2 = 0$.$4 + \frac{15\lambda}{4} - \frac{25}{4} - \frac{18}{4} - \frac{\lambda^2}{2} = 0$.Multiplying by $4$ to clear denominators:$16 + 15\lambda - 25 - 18 - 2\lambda^2 = 0$.$-2\lambda^2 + 15\lambda - 27 = 0$.$2\lambda^2 - 15\lambda + 27 = 0$.$(2\lambda - 9)(\lambda - 3) = 0$.Thus,$\lambda = 3$ or $\lambda = \frac{9}{2}$.

The value of $\lambda$ for which the equation $x^2 - \lambda xy + 2y^2 + 3x - 5y + 2 = 0$ represents a pair of straight lines is

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