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(B) The given equation is $2x^2 - 3xy - 2y^2 + 10x + 5y = 0$.To find the point of intersection,we factor the quadratic expression. The homogeneous par

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$-\frac{1}{2}$

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(B) The given equation is $2x^2 - 3xy - 2y^2 + 10x + 5y = 0$.To find the point of intersection,we factor the quadratic expression. The homogeneous part is $2x^2 - 3xy - 2y^2 = (2x + y)(x - 2y)$.Let the lines be $(2x + y + c_1) = 0$ and $(x - 2y + c_2) = 0$.Expanding $(2x + y + c_1)(x - 2y + c_2) = 2x^2 - 3xy - 2y^2 + (2c_2 + c_1)x + (c_2 - 2c_1)y + c_1c_2 = 0$.Comparing coefficients with $2x^2 - 3xy - 2y^2 + 10x + 5y = 0$,we get $2c_2 + c_1 = 10$ and $c_2 - 2c_1 = 5$.Solving these equations: $c_1 = 0$ and $c_2 = 5$.So the lines are $2x + y = 0$ and $x - 2y + 5 = 0$.The intersection point is found by solving $2x + y = 0$ and $x - 2y + 5 = 0$. Substituting $y = -2x$ into the second equation gives $x - 2(-2x) + 5 = 0$,so $5x = -5$,which means $x = -1$ and $y = 2$.The line $L$ passes through $(0, 0)$ and $(-1, 2)$.The slope of $L$ is $m_1 = \frac{2 - 0}{-1 - 0} = -2$.Since $L$ is perpendicular to $kx + y + 3 = 0$,the slope of the second line is $m_2 = -k$.For perpendicular lines,$m_1 	imes m_2 = -1$.$-2 	imes (-k) = -1 \Rightarrow 2k = -1 \Rightarrow k = -\frac{1}{2}$.

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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