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(C) The line $L$ is $9x + 5y = 45$. The intercepts are $(5, 0)$ and $(0, 9)$.The lines $l_1$ and $l_2$ pass through the origin $(0, 0)$ and trisect th

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$y - x = 5$

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(C) The line $L$ is $9x + 5y = 45$. The intercepts are $(5, 0)$ and $(0, 9)$.The lines $l_1$ and $l_2$ pass through the origin $(0, 0)$ and trisect the segment between $(5, 0)$ and $(0, 9)$.Using the section formula,the points of trisection are:$P_1 = \left( \frac{2(0) + 1(5)}{3}, \frac{2(9) + 1(0)}{3} ight) = \left( \frac{5}{3}, 6 ight)$$P_2 = \left( \frac{1(0) + 2(5)}{3}, \frac{1(9) + 2(0)}{3} ight) = \left( \frac{10}{3}, 3 ight)$The slopes are $m_1 = \frac{6}{5/3} = \frac{18}{5}$ and $m_2 = \frac{3}{10/3} = \frac{9}{10}$.The sum of slopes is $m_1 + m_2 = \frac{18}{5} + \frac{9}{10} = \frac{36+9}{10} = \frac{45}{10} = \frac{9}{2}$.The line is $y = \frac{9}{2}x$,or $9x - 2y = 0$.To find the intersection with $L: 9x + 5y = 45$,subtract the equations:$(9x + 5y) - (9x - 2y) = 45 - 0 \implies 7y = 45 \implies y = \frac{45}{7}$.Then $9x = 2y = \frac{90}{7} \implies x = \frac{10}{7}$.The point is $(\frac{10}{7}, \frac{45}{7})$.Checking the options:$A: 6(\frac{10}{7}) + \frac{45}{7} = \frac{60+45}{7} = 15 
eq 10$$B: 6(\frac{10}{7}) - \frac{45}{7} = \frac{60-45}{7} = \frac{15}{7} 
eq 15$$C: \frac{45}{7} - \frac{10}{7} = \frac{35}{7} = 5$. This matches.Thus,the point lies on $y - x = 5$.

The straight lines $l_1$ and $l_2$ pass through the origin and trisect the line segment of the line $L: 9x + 5y = 45$ between the axes. If $m_1$ and $m_2$ are the slopes of the lines $l_1$ and $l_2$,then the point of intersection of the line $y = (m_1 + m_2)x$ with $L$ lies on

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