A line passes through the origin and makes equal angles with the positive coordinate axes. It intersects the lines $$L_1 : 2x + y + 6 = 0$$ and $$L_2 : 4x + 2y - p = 0$$, $$p \gt 0$$, at the points A and B, respectively. If $$AB = \frac{9}{\sqrt{2}}$$ and the foot of the perpendicular from the point A on the line $$L_2$$ is M, then $$\frac{AM}{BM}$$ is equal to

The line through the origin that makes equal angles with the positive x- and y-axes must have slope $$1$$, because the angle it makes with the x-axis (say $$\theta$$) equals the angle it makes with the y-axis, which is $$90^\circ-\theta$$. Hence $$\theta = 45^\circ$$ and the required line is

$$y = x \qquad -(1)$$

--------------------------------------------------------------------

Case 1: Intersection with $$L_1$$

$$L_1 : 2x + y + 6 = 0$$. Substituting $$y = x$$ from $$(1)$$ gives
$$2x + x + 6 = 0 \;\Longrightarrow\; 3x = -6 \;\Longrightarrow\; x = -2$$.
Thus $$A(-2,\,-2)$$.

--------------------------------------------------------------------

Case 2: Intersection with $$L_2$$

$$L_2 : 4x + 2y - p = 0,\; p \gt 0$$. Again substituting $$y = x$$ gives
$$4x + 2x - p = 0 \;\Longrightarrow\; 6x = p \;\Longrightarrow\; x = \dfrac{p}{6}$$.
Hence $$B\!\left(\dfrac{p}{6},\,\dfrac{p}{6}\right)$$.

--------------------------------------------------------------------

Case 3: Using the given length $$AB = \dfrac{9}{\sqrt{2}}$$ to find $$p$$

The distance formula gives
$$AB = \sqrt{\left(\dfrac{p}{6}+2\right)^2 + \left(\dfrac{p}{6}+2\right)^2} = \sqrt{2}\!\left(\dfrac{p}{6}+2\right).$$

Equating to the given value:
$$\sqrt{2}\!\left(\dfrac{p}{6}+2\right) = \dfrac{9}{\sqrt{2}} \;\Longrightarrow\; 2\!\left(\dfrac{p}{6}+2\right) = 9 \;\Longrightarrow\; \dfrac{p}{6} = \dfrac{5}{2} \;\Longrightarrow\; p = 15.$$

So $$B\!\left(\dfrac{15}{6},\,\dfrac{15}{6}\right)=\left(\dfrac{5}{2},\,\dfrac{5}{2}\right).$$

--------------------------------------------------------------------

Case 4: Foot M of the perpendicular from A to $$L_2$$

The normal form for $$L_2$$ is $$4x + 2y - 15 = 0,$$ so $$A = 4,\; B = 2,\; C = -15.$$ For a point $$(x_0,y_0)$$ the foot $$(x',y')$$ on $$Ax + By + C = 0$$ is
$$x' = x_0 - A\dfrac{Ax_0 + By_0 + C}{A^2 + B^2},\quad y' = y_0 - B\dfrac{Ax_0 + By_0 + C}{A^2 + B^2}.$$ Here $$(x_0,y_0) = (-2,-2).$$

$$Ax_0 + By_0 + C = 4(-2) + 2(-2) -15 = -27,\\[4pt] A^2 + B^2 = 4^2 + 2^2 = 16 + 4 = 20.$$ Hence
$$x_M = -2 - 4\!\left(\dfrac{-27}{20}\right) = -2 + \dfrac{108}{20} = -2 + \dfrac{27}{5} = \dfrac{17}{5},\\[6pt] y_M = -2 - 2\!\left(\dfrac{-27}{20}\right) = -2 + \dfrac{54}{20} = -2 + \dfrac{27}{10} = \dfrac{7}{10}.$$ Thus $$M\!\left(\dfrac{17}{5},\,\dfrac{7}{10}\right).$$

--------------------------------------------------------------------

Case 5: Lengths $$AM$$ and $$BM$$

1. Vector $$\overrightarrow{AM} = \left(\dfrac{17}{5}+2,\;\dfrac{7}{10}+2\right) = \left(\dfrac{27}{5},\;\dfrac{27}{10}\right).$$
$$AM^2 = \left(\dfrac{27}{5}\right)^2 + \left(\dfrac{27}{10}\right)^2 = 729\!\left(\dfrac{1}{25} + \dfrac{1}{100}\right) = 729\!\left(\dfrac{5}{100}\right) = \dfrac{729}{20},\\ AM = \dfrac{27}{\sqrt{20}} = \dfrac{27}{2\sqrt{5}}.$$

2. Vector $$\overrightarrow{BM} = \left(\dfrac{17}{5}-\dfrac{5}{2},\;\dfrac{7}{10}-\dfrac{5}{2}\right) = \left(\dfrac{9}{10},\;-\dfrac{18}{10}\right).$$
$$BM^2 = \left(\dfrac{9}{10}\right)^2 + \left(-\dfrac{18}{10}\right)^2 = \dfrac{81}{100} + \dfrac{324}{100} = \dfrac{405}{100} = \dfrac{81}{20},\\ BM = \dfrac{9}{\sqrt{20}} = \dfrac{9}{2\sqrt{5}}.$$

--------------------------------------------------------------------

Case 6: Required ratio

$$\dfrac{AM}{BM}= \dfrac{\dfrac{27}{2\sqrt{5}}}{\dfrac{9}{2\sqrt{5}}}=3.$$

--------------------------------------------------------------------

Therefore $$\dfrac{AM}{BM}=3$$, which corresponds to Option D (3).