If m_1 and m_2 (m_1 m_2) are the slopes of | vedclass

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(C) The given equation is $5x^2 - 8xy + 3y^2 = 0$.Dividing by $x^2$,we get $3(\frac{y}{x})^2 - 8(\frac{y}{x}) + 5 = 0$.Let $m = \frac{y}{x}$ be the sl

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$5:3$

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(C) The given equation is $5x^2 - 8xy + 3y^2 = 0$.Dividing by $x^2$,we get $3(\frac{y}{x})^2 - 8(\frac{y}{x}) + 5 = 0$.Let $m = \frac{y}{x}$ be the slope of the lines. Then $3m^2 - 8m + 5 = 0$.Solving the quadratic equation $3m^2 - 8m + 5 = 0$ using the quadratic formula $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$m = \frac{8 \pm \sqrt{64 - 4(3)(5)}}{2(3)} = \frac{8 \pm \sqrt{64 - 60}}{6} = \frac{8 \pm \sqrt{4}}{6} = \frac{8 \pm 2}{6}$.The two slopes are $m_1 = \frac{8+2}{6} = \frac{10}{6} = \frac{5}{3}$ and $m_2 = \frac{8-2}{6} = \frac{6}{6} = 1$.Since $m_1 > m_2$,we have $m_1 = \frac{5}{3}$ and $m_2 = 1$.Therefore,the ratio $m_1 : m_2 = \frac{5}{3} : 1 = 5 : 3$.

If $m_1$ and $m_2$ $(m_1 > m_2)$ are the slopes of the lines represented by $5x^2 - 8xy + 3y^2 = 0$,then $m_1 : m_2$ equals

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