Suppose the slopes m_1 and m_2 of the lines r | vedclass

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(A) Given the equation $ax^2+2hxy+by^2=0$,the sum of slopes is $m_1+m_2 = -\frac{2h}{b}$ and the product of slopes is $m_1m_2 = \frac{a}{b}$.From $m_1

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$\frac{a}{12}=\frac{b}{6}=\frac{h}{\pm 11}$

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(A) Given the equation $ax^2+2hxy+by^2=0$,the sum of slopes is $m_1+m_2 = -\frac{2h}{b}$ and the product of slopes is $m_1m_2 = \frac{a}{b}$.From $m_1m_2-2=0$,we have $\frac{a}{b}=2$,so $a=2b$.From $3(m_1-m_2)-7=0$,we have $m_1-m_2 = \frac{7}{3}$.Using $(m_1-m_2)^2 = (m_1+m_2)^2 - 4m_1m_2$,we get $(\frac{7}{3})^2 = (-\frac{2h}{b})^2 - 4(\frac{a}{b})$.Substituting $a=2b$,we get $\frac{49}{9} = \frac{4h^2}{b^2} - 4(2) = \frac{4h^2}{b^2} - 8$.$\frac{4h^2}{b^2} = \frac{49}{9} + 8 = \frac{49+72}{9} = \frac{121}{9}$.Taking the square root,$\frac{2h}{b} = \pm \frac{11}{3}$,which implies $\frac{h}{b} = \pm \frac{11}{6}$.Thus,$\frac{h}{\pm 11} = \frac{b}{6}$.Since $a=2b$,we have $\frac{a}{2} = b$,so $\frac{a}{12} = \frac{b}{6}$.Therefore,$\frac{a}{12} = \frac{b}{6} = \frac{h}{\pm 11}$.

Suppose the slopes $m_1$ and $m_2$ of the lines represented by $ax^2+2hxy+by^2=0$ satisfy $3(m_1-m_2)-7=0$ and $m_1m_2-2=0$. Then,which of the following is true?

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