JEE Main 2025 (Online) 3rd April Morning ShiftQuestion 1
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 $L_2:4x+2y-p=0$, $p>0$, 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
Points and Straight Lines
Solution
Step 1 - The line. Equal angles with the positive axes means the line $y=x$. Step 2 - Points $A$ and $B$. $$A:\ 2x+x+6=0\Rightarrow A=(-2,-2),\qquad B:\ 4x+2x=p\Rightarrow B=\left(\tfrac p6,\tfrac p6\right).$$Step 3 - Find $p$. $$AB=\sqrt2\left(\tfrac p6+2\right)=\frac{9}{\sqrt2}\ \Longrightarrow\ \tfrac p6+2=\tfrac92\ \Longrightarrow\ p=15 .$$ Step 4 - $AM$. $L_1:\ 2x+y+6=0$ and $L_2:\ 2x+y-\tfrac{15}{2}=0$ are parallel, so $AM$ (from $A\in L_1$ perpendicular to $L_2$) is the distance between them: $$AM=\frac{\left|6+\tfrac{15}{2}\right|}{\sqrt5}=\frac{27}{2\sqrt5}.$$ Step 5 - $BM$. $\triangle AMB$ is right-angled at $M$: $$BM=\sqrt{AB^2-AM^2}=\sqrt{\frac{81}{2}-\frac{729}{20}}=\sqrt{\frac{81}{20}}=\frac{9}{2\sqrt5}.$$ Step 6 - Ratio. $$\frac{AM}{BM}=\frac{27/(2\sqrt5)}{9/(2\sqrt5)}=3 .$$ Answer: (B)