If L{x^2} - 10xy + 12{y^2} + 5x - 16y - 3 = 0 | vedclass

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(B) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.Comparing the given equation $L{x^2} - 10xy + 12{y^2} +

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$2$

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(B) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.Comparing the given equation $L{x^2} - 10xy + 12{y^2} + 5x - 16y - 3 = 0$ with the general equation,we get:$a = L, h = -5, b = 12, g = \frac{5}{2}, f = -8, c = -3$.For the equation to represent a pair of straight lines,the condition is $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.Substituting the values:$L(12)(-3) + 2(-8)(\frac{5}{2})(-5) - L(-8)^2 - 12(\frac{5}{2})^2 - (-3)(-5)^2 = 0$$-36L + 200 - 64L - 12(\frac{25}{4}) + 3(25) = 0$$-36L + 200 - 64L - 75 + 75 = 0$$-100L + 200 = 0$$100L = 200$$L = 2$.

If $L{x^2} - 10xy + 12{y^2} + 5x - 16y - 3 = 0$ represents a pair of straight lines,then the value of $L$ is:

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