JEE Mathematics 2025 Question with Solution

Given: A line passes through the origin and makes equal angles with the positive coordinate axes. It meets $L_1 : 2x + y + 6 = 0$ at A and $L_2 : 4x + 2y - p = 0$ at B. Also, $AB = \frac{9}{\sqrt{2}}$ and M is the foot of the perpendicular from A to $L_2$.

Find: $\frac{AM}{BM}$.

The line through the origin making equal angles with the positive coordinate axes is

Intersecting it with $L_1$:

So, point A is $(-2,-2)$.

Now intersect $y=x$ with $L_2$:

Hence, point B is $\left(\frac{p}{6}, \frac{p}{6}\right)$.

Using the distance formula and the given value of $AB$,

Since $p>0$,

Therefore,

The slope of $L_2$ is $-2$ and the slope of $y=x$ is $1$. Let $\theta$ be the angle between these two lines. Then

From the geometry, $\tan\theta = \frac{AM}{BM}$.

Therefore, $\frac{AM}{BM} = 3$. The correct option is D.

The second approach shown in the source gives an inconsistent intermediate value $p=18$, but the solution itself declares the correct option as D and the first approach consistently gives $p=15$ and $\frac{AM}{BM}=3$.

Given: A and B lie on the line $y=x$, and M is the foot of the perpendicular from A to $L_2$.

Find: Why $\frac{AM}{BM}$ equals the tangent of the angle between $y=x$ and $L_2$.

Since M lies on $L_2$ and $AM \perp L_2$, triangle $AMB$ is right-angled at M. Also, points A and B lie on the line $y=x$, so segment $AB$ lies along $y=x$.

Thus, the angle at B in triangle $AMB$ is exactly the angle between the lines $y=x$ and $L_2$. If that angle is $\theta$, then in right triangle $AMB$,

Using slopes $1$ and $-2$,

Hence,

So the correct option is D.