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(A) The given equation is $2x^2 + 3xy + Ky^2 = 0$. Dividing by $Ky^2$,we get the quadratic in terms of slope $m = \frac{y}{x}$: $K(\frac{y}{x})^2 + 3(

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$\frac{\pi}{2}$

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(A) The given equation is $2x^2 + 3xy + Ky^2 = 0$. Dividing by $Ky^2$,we get the quadratic in terms of slope $m = \frac{y}{x}$: $K(\frac{y}{x})^2 + 3(\frac{y}{x}) + 2 = 0$,which is $Km^2 + 3m + 2 = 0$. Since one slope is $m_1 = 2$,it must satisfy the equation: $K(2)^2 + 3(2) + 2 = 0$ $\Rightarrow 4K + 8 = 0$ $\Rightarrow K = -2$. Substituting $K = -2$ into the equation $Km^2 + 3m + 2 = 0$: $-2m^2 + 3m + 2 = 0 \Rightarrow 2m^2 - 3m - 2 = 0$. Factoring the quadratic: $(2m + 1)(m - 2) = 0$. Thus,the slopes are $m_1 = 2$ and $m_2 = -\frac{1}{2}$. Since $m_1 	imes m_2 = 2 	imes (-\frac{1}{2}) = -1$,the product of the slopes is $-1$. Therefore,the lines are perpendicular to each other,and the angle between them is $	heta = \frac{\pi}{2}$.

If the slope of one of the pair of lines represented by $2x^2 + 3xy + Ky^2 = 0$ is $2$,then the angle between the pair of lines is

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