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(A) Let the equation be represented as $(a y^2+p x y+e x^2)(y^2+q x y+k x^2) = 0$. For two lines to be perpendicular,their slopes $m_1$ and $m_2$ must

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$(b+d)(a d+b e)+(e-a)^2(a+c+e)=0$

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(A) Let the equation be represented as $(a y^2+p x y+e x^2)(y^2+q x y+k x^2) = 0$. For two lines to be perpendicular,their slopes $m_1$ and $m_2$ must satisfy $m_1 m_2 = -1$. By comparing the coefficients of the given homogeneous equation $a y^4+b x y^3+c x^2 y^2+d x^3 y+e x^4=0$ with the product of two quadratic factors,we derive the condition. Specifically,if we assume the factors are $(y^2 - m_1 x y - k_1 x^2)(y^2 - m_2 x y - k_2 x^2) = 0$ where $m_1 m_2 = -1$,the expansion leads to the relation: $(b+d)(a d+b e)+(e-a)^2(a+c+e)=0$.

Two of the lines represented by the equation $a y^4+b x y^3+c x^2 y^2+d x^3 y+e x^4=0$ will be perpendicular,then

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