Key Concepts and Formulas

• Equation of a line passing through the origin: $y = mx$
• Intersection of two lines: Solve the system of equations formed by the lines.
• Distance between parallel lines: $\frac{|c_2 - c_1|}{\sqrt{a^2 + b^2}}$ for lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$.
• Angle between two lines: $\tan \theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|$, where $m_1$ and $m_2$ are the slopes of the lines.
• Area of a triangle: $\frac{1}{2} \cdot base \cdot height$

Step-by-Step Solution

Step 1: Determine the Equation of Line L and the Coordinates of Point A

The line L passes through the origin $(0,0)$ and makes equal angles with the positive coordinate axes. This implies that the slope of the line is 1. Therefore, the equation of line L is $y=x$.

Point A is the intersection of line L ($y=x$) and line $\mathrm{L}_1: 2x+y+6=0$. Substituting $y=x$ into the equation of $\mathrm{L}_1$: $2x + x + 6 = 0$ $3x = -6$ $x = -2$ Since $y=x$, we have $y=-2$. So, the coordinates of point A are $(-2, -2)$.

Step 2: Determine the Coordinates of Point B

Point B is the intersection of line L ($y=x$) and line $\mathrm{L}_2: 4x+2y-p=0$. Substituting $y=x$ into the equation of $\mathrm{L}_2$: $4x + 2x - p = 0$ $6x = p$ $x = \frac{p}{6}$ Since $y=x$, we have $y=\frac{p}{6}$. So, the coordinates of point B are $\left(\frac{p}{6}, \frac{p}{6}\right)$.

Step 3: Use the distance AB to find p

We are given that $AB = \frac{9}{\sqrt{2}}$. Using the distance formula: $AB = \sqrt{\left(\frac{p}{6} - (-2)\right)^2 + \left(\frac{p}{6} - (-2)\right)^2} = \frac{9}{\sqrt{2}}$ $\sqrt{2\left(\frac{p}{6} + 2\right)^2} = \frac{9}{\sqrt{2}}$ $\sqrt{2}\left|\frac{p}{6} + 2\right| = \frac{9}{\sqrt{2}}$ $\left|\frac{p}{6} + 2\right| = \frac{9}{2}$

Since $p>0$, $\frac{p}{6} + 2$ must be positive. Therefore, $\frac{p}{6} + 2 = \frac{9}{2}$ $\frac{p}{6} = \frac{9}{2} - 2 = \frac{5}{2}$ $p = \frac{5}{2} \times 6 = 15$ So, the coordinates of point B are $\left(\frac{15}{6}, \frac{15}{6}\right) = \left(\frac{5}{2}, \frac{5}{2}\right)$.

Step 4: Find AM, the distance from A to L2

The equation of line $\mathrm{L}_2$ is $4x + 2y - 15 = 0$, or $2x + y - \frac{15}{2} = 0$. The distance AM from point A $(-2, -2)$ to line $\mathrm{L}_2$ is: $AM = \frac{|2(-2) + (-2) - \frac{15}{2}|}{\sqrt{2^2 + 1^2}} = \frac{|-4 - 2 - \frac{15}{2}|}{\sqrt{5}} = \frac{|-6 - \frac{15}{2}|}{\sqrt{5}} = \frac{|-\frac{12}{2} - \frac{15}{2}|}{\sqrt{5}} = \frac{|-\frac{27}{2}|}{\sqrt{5}} = \frac{27}{2\sqrt{5}}$

Step 5: Find BM, the distance from B to L1

Since L1 and L2 are parallel, and AM is perpendicular to L2, AM is also perpendicular to L1. Let N be the foot of the perpendicular from B onto L1. Then BN is the distance from B to L1. Since A, M, and N are collinear, $AM + MN = AN$. Also, since L1 and L2 are parallel, $AM = BN$.

The equation of line $\mathrm{L}_1$ is $2x + y + 6 = 0$. The distance from B $\left(\frac{5}{2}, \frac{5}{2}\right)$ to line $\mathrm{L}_1$ is: $BN = \frac{|2(\frac{5}{2}) + \frac{5}{2} + 6|}{\sqrt{2^2 + 1^2}} = \frac{|5 + \frac{5}{2} + 6|}{\sqrt{5}} = \frac{|11 + \frac{5}{2}|}{\sqrt{5}} = \frac{|\frac{22}{2} + \frac{5}{2}|}{\sqrt{5}} = \frac{\frac{27}{2}}{\sqrt{5}} = \frac{27}{2\sqrt{5}}$ Since L1 and L2 are parallel lines, AM is perpendicular to both L1 and L2. Also BN is perpendicular to L1. Therefore, A, M, and B are collinear. $AM + MB = AB$. Then $BM = AB \cos \theta$.

However, since L1 and L2 are parallel, $\frac{AM}{BN}=1$. The question asks for $\frac{AM}{BM}$. Since A, M, and B are collinear and M is the foot of the perpendicular from A to L2, $\triangle AMB$ is a right triangle. We know $AB = \frac{9}{\sqrt{2}}$ and $AM = \frac{27}{2\sqrt{5}}$.

We can use similar triangles: $\frac{AM}{AB} = \sin \theta = \frac{\frac{27}{2\sqrt{5}}}{\frac{9}{\sqrt{2}}} = \frac{27\sqrt{2}}{18\sqrt{5}} = \frac{3\sqrt{2}}{2\sqrt{5}} = \frac{3\sqrt{10}}{10}$. Since $\sin^2 \theta + \cos^2 \theta = 1$, $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{90}{100}} = \sqrt{\frac{10}{100}} = \frac{\sqrt{10}}{10}$. $BM = AB \cos \theta$, so $BM = \frac{9}{\sqrt{2}} \cdot \frac{\sqrt{10}}{10} = \frac{9\sqrt{5}}{10}$ Thus, $\frac{AM}{BM} = \frac{\frac{27}{2\sqrt{5}}}{\frac{9\sqrt{5}}{10}} = \frac{27}{2\sqrt{5}} \cdot \frac{10}{9\sqrt{5}} = \frac{3}{1} \cdot \frac{5}{5} = 3$.

Step 6: Re-evaluate and correct the solution

Since A lies on L1 and M lies on L2, AM is perpendicular to L2, we use the distance between parallel lines formula. Rewrite L1: $2x + y + 6 = 0$ and L2: $2x + y - \frac{15}{2} = 0$. Since AM is the perpendicular distance between L1 and L2: $AM = \frac{|6 - (-\frac{15}{2})|}{\sqrt{2^2 + 1^2}} = \frac{|6 + \frac{15}{2}|}{\sqrt{5}} = \frac{|\frac{12}{2} + \frac{15}{2}|}{\sqrt{5}} = \frac{27}{2\sqrt{5}}$

We know A, M, and B are collinear. $AB = AM + MB$, so $MB = AB - AM = \frac{9}{\sqrt{2}} - \frac{27}{2\sqrt{5}} = \frac{9\sqrt{2}}{2}-\frac{27\sqrt{5}}{10}$. $\frac{AM}{MB} = \frac{\frac{27}{2\sqrt{5}}}{\frac{9}{\sqrt{2}} - \frac{27}{2\sqrt{5}} }$. This is not the correct answer.

Since L and L2 intersect at B, the distance from B to L1 is not AM.

Let L1 be $2x+y+6=0$ and L2 be $4x+2y-15=0$, or $2x+y-\frac{15}{2}=0$. AM is the perpendicular distance from A to L2: $AM = \frac{|2(-2)+(-2)-\frac{15}{2}|}{\sqrt{5}} = \frac{|-4-2-\frac{15}{2}|}{\sqrt{5}} = \frac{|-6-\frac{15}{2}|}{\sqrt{5}} = \frac{27}{2\sqrt{5}}$ Let AN be the perpendicular distance from B to L1: $AN = \frac{|2(\frac{5}{2})+\frac{5}{2}+6|}{\sqrt{5}} = \frac{|5+\frac{5}{2}+6|}{\sqrt{5}} = \frac{27}{2\sqrt{5}}$ Let the distance between the parallel lines be $d = \frac{|6-(-\frac{15}{2})|}{\sqrt{5}} = \frac{27}{2\sqrt{5}}$. Thus, $AM = AN = d$.

Let $\theta$ be the angle between L and the parallel lines. Then $\tan \theta = |\frac{1-(-2)}{1+1(-2)}| = |\frac{3}{-1}| = 3$. Therefore $\sin \theta = \frac{3}{\sqrt{10}}$ and $\cos \theta = \frac{1}{\sqrt{10}}$. $AB = \frac{9}{\sqrt{2}}$. Also, $AM = AB \sin \theta$, so $\frac{27}{2\sqrt{5}} = \frac{9}{\sqrt{2}}\sin \theta$, so $\sin \theta = \frac{27\sqrt{2}}{18\sqrt{5}} = \frac{3\sqrt{2}}{2\sqrt{5}} = \frac{3\sqrt{10}}{10}$. $BM = AB \cos \theta = \frac{9}{\sqrt{2}} \cdot \frac{1}{\sqrt{10}} = \frac{9}{2\sqrt{5}}$. Then $\frac{AM}{BM} = \frac{\frac{27}{2\sqrt{5}}}{\frac{9}{2\sqrt{5}}} = 3$.

We have $\frac{Area(A,B,M)}{AB} = \text{height}=AM$, so $Area(A,B,M) = \frac{9}{\sqrt{2}} \frac{27}{2\sqrt{5}}$.

Distance between L1 and L2: $\frac{|6+15/2|}{\sqrt{5}} = \frac{27}{2\sqrt{5}}$, so AM is $\frac{27}{2\sqrt{5}}$. $AB = \frac{9}{\sqrt{2}}$. $\frac{AM}{BM} = \frac{3}{\sqrt{2}}$ and since $\theta = 3$, we have $BM = \frac{AM}{3}$ and $\frac{\frac{27}{2\sqrt{5}}}{3} = \frac{27}{6\sqrt{5}} = \frac{9}{2\sqrt{5}}$. Then $AM + BM = \frac{27}{2\sqrt{5}} + \frac{9}{2\sqrt{5}} = \frac{36}{2\sqrt{5}} = \frac{18}{\sqrt{5}}$. Since A, M, and B are collinear, $AB = AM + BM$, so $AM = \frac{AB \cdot Area(AMB)}{AB}= 5$.

Corrected Step 5: Since A, M, and B are collinear, and AM is perpendicular to L2, triangle AMB is a right triangle. Let $\theta$ be the angle between L and L2. Then $AM = AB \sin{\theta}$. We want $\frac{AM}{BM} = \frac{AB \sin{\theta}}{AB \cos{\theta}} = \tan{\theta}$. $\tan{\theta} = |\frac{m_1 - m_2}{1+m_1m_2}| = |\frac{1-(-2)}{1+1(-2)}| = |\frac{3}{-1}| = 3$. Therefore, $\frac{AM}{BM} = 3$.

Common Mistakes & Tips

• Confusing the angle between the lines with the angle the lines make with the x-axis.
• Incorrectly applying the distance formula or the distance between parallel lines formula.
• Not recognizing that lines L1 and L2 are parallel, making calculations harder.

Summary

We first found the equation of line L and the coordinates of points A and B. Then, we found AM by realizing that it's the distance between two parallel lines. Finally, we recognized that $\frac{AM}{BM}$ is the tangent of the angle between lines L and L2. The final calculation gives us $\frac{AM}{BM} = 3$.

Final Answer The final answer is \boxed{3}, which corresponds to option (B).