JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Straight Lines

The line through the origin making equal angles with the positive axes is $y=x$. Its meet with $L_1:2x+y+6=0$ is $A(-2,-2)$, and with $L_2:4x+2y-p=0$ is $B\left(\frac{p}{6},\frac{p}{6}\right)$. The distance $AB=\left|\frac{p}{6}+2\right|\sqrt{2}=\frac{9}{\sqrt{2}}$ gives $\frac{p}{6}+2=\frac{9}{2}$, so $p=15$ and $B\left(\frac{5}{2},\frac{5}{2}\right)$. As $M$ is the foot of the perpendicular from $A$ to $L_2$, triangle $ABM$ is right-angled at $M$, and the angle at $B$ is the angle between $y=x$ (slope $1$) and $L_2$ (slope $-2$): $\tan \theta =\left|\frac{1-(-2)}{1+(1)(-2)}\right|=3$. Therefore $\frac{AM}{BM}=\tan \theta =3$.