Maths Q13 – JEE-MAIN 2025

A line passes through the origin and makes equal angles with the positive coordinate axes. It intersects the lines

L_{1} : 2 x + y + 6 = 0

and

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

Clue 1 of 3: Geometric Setup

First, determine the equation of the line passing through the origin and making equal angles with the positive coordinate axes.

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Ninja Strategy•Geometric Interpretation of Ratio

Recognize that $A M / B M$ in a right-angled triangle $A M B$ (where $M$ is the foot of the perpendicular from $A$ to $L_{2}$ ) is equal to $\tan θ$ , where $θ$ is the angle between the line $A B$ and $L_{2}$ . Calculate $\tan θ$ using the slopes of the two lines.

🧠 Full Solution Path

Step 1: Determine the line L and identify the geometric relationship.✦ Active

A line passing through the origin and making equal angles with the positive coordinate axes is $L : y = x$ . This line intersects $L_{1}$ at A and $L_{2}$ at B. M is the foot of the perpendicular from A to $L_{2}$ . This implies that $△ A M B$ is a right-angled triangle with the right angle at M. The line segment $A B$ is the hypotenuse of this triangle.

In $△ A M B$ , the angle $∠ A B M$ is the angle between the line $L$ (containing $A B$ ) and the line $L_{2}$ . Let this angle be $θ$ . Then, $A M = A B \sin θ$ and $B M = A B \cos θ$ . Therefore, the ratio $\frac{A M}{B M} = \frac{A B \sin θ}{A B \cos θ} = \tan θ$ .

Step 2: Calculate the slopes of the relevant lines.○ Expand

The line $L$ is $y = x$ , so its slope is $m_{L} = 1$ .

The line $L_{2}$ is $4 x + 2 y - p = 0$ . Rewriting it in slope-intercept form: $2 y = - 4 x + p \Rightarrow y = - 2 x + \frac{p}{2}$ . So, the slope of $L_{2}$ is $m_{L_{2}} = - 2$ .

Step 3: Calculate the tangent of the angle and the required ratio.○ Expand

The angle $θ$ between two lines with slopes $m_{1}$ and $m_{2}$ is given by $\tan θ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} |$ . Using $m_{L} = 1$ and $m_{L_{2}} = - 2$ :

\tan θ = | \frac{1 - (- 2)}{1 + (1) (- 2)} | = | \frac{3}{1 - 2} | = | \frac{3}{- 1} | = | - 3 | = 3

Therefore, $\frac{A M}{B M} = \tan θ = 3$ .

💡 Teacher's Secret Hint

Note that the value of $p$ and the length $A B$ are not required to find the ratio $\frac{A M}{B M}$ using this geometric approach.

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Maths Question 13 – JEE-MAIN 2025

Clue 1 of 3: Geometric Setup

🧠 Full Solution Path