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 $3m^2 - 8m + 5 = 0$:$3m^2 - 3m - 5m + 5 = 0$$3m(m - 1) - 5(m - 1) = 0$$(3m - 5)(m - 1) = 0$Thus,the slopes are $m_1 = \frac{5}{3}$ and $m_2 = 1$ (since $m_1 > m_2$).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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