The equation of the pair of lines passing thr | vedclass

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(A) Let $m_1$ and $m_2$ be the slopes of the lines.Then,$m_1+m_2 = 	ext{arithmetic mean} = \frac{4+9}{2} = \frac{13}{2}$ and $m_1 m_2 = 	ext{geometr

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$12 x^2-13 x y+2 y^2=0$

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(A) Let $m_1$ and $m_2$ be the slopes of the lines.Then,$m_1+m_2 = 	ext{arithmetic mean} = \frac{4+9}{2} = \frac{13}{2}$ and $m_1 m_2 = 	ext{geometric mean} = \sqrt{4 	imes 9} = \sqrt{36} = 6$.Now,the equation of the pair of lines passing through the origin is given by $(y-m_1 x)(y-m_2 x) = 0$.This simplifies to $y^2 - (m_1+m_2)xy + m_1 m_2 x^2 = 0$.Substituting the values of $m_1+m_2$ and $m_1 m_2$,we get $y^2 - \frac{13}{2}xy + 6x^2 = 0$.Multiplying by $2$,we get $2y^2 - 13xy + 12x^2 = 0$,which is $12x^2 - 13xy + 2y^2 = 0$.

The equation of the pair of lines passing through the origin whose sum and product of slopes are respectively the arithmetic mean and geometric mean of $4$ and $9$ is

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