JEE MAIN - Mathematics (2025 - 3rd April Morning Shift - No. 19)
A line passes through the origin and makes equal angles with the positive coordinate axes. It intersects the lines $\mathrm{L}_1: 2 x+y+6=0$ and $\mathrm{L}_2: 4 x+2 y-p=0, p>0$, at the points A and B , respectively. If $A B=\frac{9}{\sqrt{2}}$ and the foot of the perpendicular from the point $A$ on the line $L_2$ is $M$, then $\frac{A M}{B M}$ is equal to
5
3
2
4
Explanation
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$$\begin{aligned} & \tan \theta=\left|\frac{-2-1}{1+(-2)(1)}\right|=\frac{3}{1} \\ & \sin \theta=\frac{3}{\sqrt{10}} \\ & A M=\frac{3}{\sqrt{10}} \times \frac{9}{\sqrt{2}}=\frac{27}{2 \sqrt{5}} \end{aligned}$$
Dist. between $I_1 ~\& ~l_2, \quad A M=\left|\frac{\frac{P}{2}+6}{\sqrt{5}}\right|=\frac{27}{2 \sqrt{5}}$
$$\begin{aligned} & \frac{P}{2}+6= \pm \frac{27}{2} \\ & \frac{P}{2}=\frac{27}{2}-6 \Rightarrow P=15, \text { As } P>0 \\ & B M=\sqrt{\frac{81}{2}-\left(\frac{27}{2 \sqrt{5}}\right)^2} \\ & =\sqrt{\frac{810-729}{20}}=\sqrt{\frac{81}{20}}=\frac{9}{2 \sqrt{5}} \\ & \text { Now, } \frac{A M}{B M}=\frac{\frac{27}{\frac{2 \sqrt{5}}{9}}}{\frac{2 \sqrt{5}}{2 \sqrt{2}}}=3 \end{aligned}$$
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