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(B) The given equation is $2x^2 - 3xy - 2y^2 + 10x + 5y = 0$. Factoring the homogeneous part $2x^2 - 3xy - 2y^2$,we get $(2x + y)(x - 2y) = 0$. Thus,t

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$\frac{-1}{2}$

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(B) The given equation is $2x^2 - 3xy - 2y^2 + 10x + 5y = 0$. Factoring the homogeneous part $2x^2 - 3xy - 2y^2$,we get $(2x + y)(x - 2y) = 0$. Thus,the lines are $2x + y + c_1 = 0$ and $x - 2y + c_2 = 0$. Comparing with the original equation,we find the lines are $(2x + y)(x - 2y + 5) = 0$,which gives $2x + y = 0$ and $x - 2y + 5 = 0$. The intersection point of these two lines is found by solving $2x + y = 0$ and $x - 2y + 5 = 0$. From $y = -2x$,substituting into the second equation: $x - 2(-2x) + 5 = 0$ $\Rightarrow 5x = -5$ $\Rightarrow x = -1$. Then $y = -2(-1) = 2$. The intersection point is $(-1, 2)$. The line $L$ joins the origin $(0, 0)$ to $(-1, 2)$. The slope of $L$ is $m_1 = \frac{2 - 0}{-1 - 0} = -2$. Since $L$ is perpendicular to $kx + y + 3 = 0$,the product of their slopes is $-1$. The slope of $kx + y + 3 = 0$ is $m_2 = -k$. Therefore,$m_1 	imes m_2 = -1$ $\Rightarrow (-2) 	imes (-k) = -1$ $\Rightarrow 2k = -1$ $\Rightarrow k = -\frac{1}{2}$.

Let $L$ be the line joining the origin to the point of intersection of the lines represented by $2x^2 - 3xy - 2y^2 + 10x + 5y = 0$. If $L$ is perpendicular to the line $kx + y + 3 = 0$,then $k$ is equal to

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