The portion of the line 4x + 5y = 20 in the f | vedclass

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(D) The line $4x + 5y = 20$ intersects the axes at $P(5, 0)$ and $Q(0, 4)$.Let the points of trisection be $A$ and $B$ such that $PA = AB = BQ$.Using

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$\frac{30}{41}$

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(D) The line $4x + 5y = 20$ intersects the axes at $P(5, 0)$ and $Q(0, 4)$.Let the points of trisection be $A$ and $B$ such that $PA = AB = BQ$.Using the section formula,the coordinates of $A$ divide $PQ$ in the ratio $1:2$:$A = \left( \frac{1(0) + 2(5)}{1+2}, \frac{1(4) + 2(0)}{1+2} ight) = \left( \frac{10}{3}, \frac{4}{3} ight)$.The coordinates of $B$ divide $PQ$ in the ratio $2:1$:$B = \left( \frac{2(0) + 1(5)}{2+1}, \frac{2(4) + 1(0)}{2+1} ight) = \left( \frac{5}{3}, \frac{8}{3} ight)$.Slope of line $OA$ $(m_1)$ = $\frac{4/3}{10/3} = \frac{4}{10} = \frac{2}{5}$.Slope of line $OB$ $(m_2)$ = $\frac{8/3}{5/3} = \frac{8}{5}$.The tangent of the angle $	heta$ between $L_1$ and $L_2$ is given by:$	an 	heta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} ight| = \left| \frac{\frac{8}{5} - \frac{2}{5}}{1 + (\frac{8}{5})(\frac{2}{5})} ight| = \left| \frac{\frac{6}{5}}{1 + \frac{16}{25}} ight| = \left| \frac{\frac{6}{5}}{\frac{41}{25}} ight| = \frac{6}{5} 	imes \frac{25}{41} = \frac{30}{41}$.

The portion of the line $4x + 5y = 20$ in the first quadrant is trisected by the lines $L_1$ and $L_2$ passing through the origin. The tangent of an angle between the lines $L_1$ and $L_2$ is:

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