The value of k for which the equation x^2-4xy | vedclass

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(C) The given equation is $x^2-4xy-y^2+6x+2y+k=0$. Comparing this with the general second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get: $a=1, h

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$\frac{-4}{5}$

Vedclass · Answer

(C) The given equation is $x^2-4xy-y^2+6x+2y+k=0$. Comparing this with the general second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get: $a=1, h=-2, b=-1, g=3, f=1, c=k$. The condition for the equation to represent a pair of straight lines is $\Delta = abc+2fgh-af^2-bg^2-ch^2=0$. Substituting the values: $(1)(-1)(k) + 2(1)(3)(-2) - (1)(1)^2 - (-1)(3)^2 - k(-2)^2 = 0$. $-k - 12 - 1 + 9 - 4k = 0$. $-5k - 4 = 0$. $5k = -4$. $k = -\frac{4}{5}$. Thus,the correct option is $C$.

The value of $k$ for which the equation $x^2-4xy-y^2+6x+2y+k=0$ represents a pair of straight lines is equal to ........

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