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(D) The line passing through the origin and making equal angles with the positive coordinate axes is $y = x$.$L_1: 2x + y + 6 = 0$ and $L_2: 4x + 2y -

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$3$

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(D) The line passing through the origin and making equal angles with the positive coordinate axes is $y = x$.$L_1: 2x + y + 6 = 0$ and $L_2: 4x + 2y - p = 0$ (or $2x + y - \frac{p}{2} = 0$).Since $L_1$ and $L_2$ have the same slope $m = -2$,they are parallel lines.The line $y = x$ intersects $L_1$ at $A$ and $L_2$ at $B$.In the right-angled triangle $	riangle AMB$,the angle $	heta$ between the line $y = x$ (slope $m_1 = 1$) and the line $L_2$ (slope $m_2 = -2$) is given by:$	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight| = \left| \frac{1 - (-2)}{1 + (1)(-2)} ight| = \left| \frac{3}{-1} ight| = 3$.In $	riangle AMB$,$	an 	heta = \frac{AM}{BM}$.Therefore,$\frac{AM}{BM} = 3$.

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