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

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(C) The given equation is $2x^2 + kxy - 6y^2 + 3x + y + 1 = 0$.For this to represent a pair of straight lines,the determinant of the matrix of coeffic

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

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(C) The given equation is $2x^2 + kxy - 6y^2 + 3x + y + 1 = 0$.For this to represent a pair of straight lines,the determinant of the matrix of coefficients must be zero:$\left|\begin{array}{ccc} 2 & k/2 & 3/2 \ k/2 & -6 & 1/2 \ 3/2 & 1/2 & 1 \end{array}ight| = 0$Multiplying rows to simplify: $\frac{1}{4} \left|\begin{array}{ccc} 4 & k & 3 \ k & -12 & 1 \ 3 & 1 & 2 \end{array}ight| = 0$$4(-24 - 1) - k(2k - 3) + 3(k + 36) = 0$$-100 - 2k^2 + 3k + 3k + 108 = 0$$-2k^2 + 6k + 8 = 0 \Rightarrow k^2 - 3k - 4 = 0$$(k - 4)(k + 1) = 0$. Since $k > 0$,we have $k = 4$.Substituting $k = 4$,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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