If the equation 2x^2 + kxy - 6y^2 + 3x + y + | vedclass

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(C) The general equation of a pair of straight lines $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of lines if the determinant $\Delta =

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$\left(-\frac{5}{8}, -\frac{1}{8}\right)$

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(C) The general equation of a pair of straight lines $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of lines if the determinant $\Delta = \begin{vmatrix} a & h & g \ h & b & f \ g & f & c \end{vmatrix} = 0$.Comparing with $2x^2 + kxy - 6y^2 + 3x + y + 1 = 0$,we have $a=2, h=k/2, b=-6, g=3/2, f=1/2, c=1$.The condition $\Delta = 0$ gives $\begin{vmatrix} 2 & k/2 & 3/2 \ k/2 & -6 & 1/2 \ 3/2 & 1/2 & 1 \end{vmatrix} = 0$.Multiplying rows by $2$,we get $\frac{1}{8} \begin{vmatrix} 4 & k & 3 \ k & -12 & 1 \ 3 & 1 & 2 \end{vmatrix} = 0$.Expanding the determinant: $4(-24 - 1) - k(2k - 3) + 3(k + 36) = 0$.$-100 - 2k^2 + 3k + 3k + 108 = 0$ $\Rightarrow -2k^2 + 6k + 8 = 0$ $\Rightarrow k^2 - 3k - 4 = 0$.$(k - 4)(k + 1) = 0$. Since $k > 0$,we have $k = 4$.The equation becomes $2x^2 + 4xy - 6y^2 + 3x + y + 1 = 0$.Factoring the expression: $(2x - 2y + 1)(x + 3y + 1) = 0$.Solving the system $2x - 2y + 1 = 0$ and $x + 3y + 1 = 0$:From the second equation,$x = -3y - 1$. Substituting into the first: $2(-3y - 1) - 2y + 1 = 0$ $\Rightarrow -6y - 2 - 2y + 1 = 0$ $\Rightarrow -8y = 1$ $\Rightarrow y = -1/8$.Then $x = -3(-1/8) - 1 = 3/8 - 8/8 = -5/8$.The point of intersection is $\left(-\frac{5}{8}, -\frac{1}{8}ight)$.

If the equation $2x^2 + kxy - 6y^2 + 3x + y + 1 = 0$ $(k > 0)$ represents a pair of straight lines,then their point of intersection is

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