JEE Main 2025 (April)

LEVELJEE Advanced

Animated Solution for Mathematics - Straight Lines: 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 $A B = \frac{9}{2}$ 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

Q: Why is the ratio $\frac{AM}{BM}$ equal to $\tan(\theta)$? I'm a bit confused about the triangle setup.

Imagine the triangle $\triangle ABM$ where $\angle M = 90^\circ$. The side $AM$ is opposite to the angle $\theta$ (the angle between the transversal and $L_2$), and $BM$ is the adjacent side. By the definition of tangent in a right-angled triangle, $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AM}{BM}$. It's that simple!

Select Answer:

Visualized Solution

Introduction to the Problem

Given lines:
$L_{1} : 2 x + y + 6 = 0$
$L_{2} : 4 x + 2 y - p = 0, p > 0$
Line $L$ passes through $(0, 0)$ and makes equal angles with positive axes.
Goal: Find the ratio $\frac{A M}{B M}$ .

Equation of the Transversal Line

Line makes equal angles with positive axes $\Rightarrow θ = 4 5^{\circ}$ .
Slope $m = tan (4 5^{\circ}) = 1$ .
Equation of line: $y - 0 = 1 (x - 0) \Rightarrow y = x$ .

Finding Intersection Point $A$

Intersection of $y = x$ and $2 x + y + 6 = 0$ :
$2 x + x + 6 = 0 \Rightarrow 3 x = - 6 \Rightarrow x = - 2$ .
Since $y = x$ , point $A$ is $(- 2, - 2)$ .

Finding Intersection Point $B$

Intersection of $y = x$ and $4 x + 2 y - p = 0$ :
$4 x + 2 x = p \Rightarrow 6 x = p \Rightarrow x = \frac{p}{6}$ .
Point $B$ is $(\frac{p}{6}, \frac{p}{6})$ .

Using the Distance $A B$

Distance $A B = (\frac{p}{6} - (- 2))^{2} + (\frac{p}{6} - (- 2))^{2}$
$A B = 2 (\frac{p}{6} + 2)^{2} = 2 (\frac{p}{6} + 2)$
Given $A B = \frac{9}{2}$ .

Solving for the Parameter $p$

$2 (\frac{p}{6} + 2) = \frac{9}{2}$
$\frac{p}{6} + 2 = \frac{9}{2} = 4.5$
$\frac{p}{6} = 2.5 \Rightarrow B (2.5, 2.5)$

Visualizing the Geometry

Slope of $L_{1}$ and $L_{2}$ is $- 2 \Rightarrow L_{1} ∥ L_{2}$ .
$A \in L_{1}$ and $B, M \in L_{2}$ .
$△ A B M$ is a right-angled triangle at $M$ ( $A M ⊥ L_{2}$ ).

The Trigonometric Relation

In $△ A B M$ , $tan (∠ A B M) = \frac{A M}{B M}$ .
Let $θ = ∠ A B M$ .
$θ$ is the angle between line $y = x$ and $L_{2}$ .

Slopes of the Lines

Slope of line $y = x$ is $m_{1} = 1$ .
Slope of line $L_{2} : 4 x + 2 y - p = 0$ is $m_{2} = - 2$ .
Formula: $tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}$

Calculating Tan Theta

$tan θ = \frac{1 - ( - 2 )}{1 + ( 1 ) ( - 2 )}$
$tan θ = \frac{3}{1 - 2} = \frac{3}{- 1} = 3$ .
Therefore, $\frac{A M}{B M} = 3$ .

Final Result and Summary

Key Takeaway:
The ratio $\frac{A M}{B M}$ in a right triangle is the tangent of the angle between the hypotenuse and the base.
Final Answer: $\frac{A M}{B M} = 3$ .

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The Sigma Insight: Angle Between Two Lines

Solution Diagram

Analyzing the Setup

The problem begins with a line passing through the origin making equal angles with the positive axes. This line is defined by the equation

y = x

, which has a slope of

m = tan (4 5^{\circ}) = 1

Next, we examine the lines

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

and

L_{2} : 4 x + 2 y - p = 0

. By rewriting

L_{1}

y = - 2 x - 6

and

L_{2}

y = - 2 x + \frac{p}{2}

, we observe that both lines share a slope of

- 2

This reveals that

L_{1}

and

L_{2}

are parallel. The entire geometry of the problem simplifies to two parallel lines being intersected by a transversal line

y = x

Visualizing the Triangle

The transversal intersects

L_{1}

at point

A

and

L_{2}

at point

B

. We are given that

M

is the foot of the perpendicular from

A

onto

L_{2}

This construction forms a right-angled triangle,

△ A B M

, where the angle at

M

9 0^{\circ}

. We are tasked with finding the ratio

\frac{A M}{B M}

In this right-angled triangle, the ratio of the side opposite to an angle to the side adjacent to that angle is the tangent of that angle. Specifically,

\frac{A M}{B M} = tan (θ)

, where

θ

is the angle between the transversal

y = x

and the line

L_{2}

The Master Equation

We do not need the specific coordinates of

M

or the value of

p

. We only require the angle between the transversal and

L_{2}

Given the slope of the transversal

m_{1} = 1

and the slope of

L_{2}

m_{2} = - 2

, the tangent of the angle

θ

between these two lines is given by:

tan (θ) = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}

Substituting our known slopes into the equation:

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

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

Final Result

By applying the geometric relationship between the slopes, we find that the ratio

\frac{A M}{B M}

is exactly 3.

The provided distance

A B = \frac{9}{2}

was a distractor, demonstrating that the most elegant path in JEE Advanced often involves identifying geometric relationships that render tedious coordinate calculations unnecessary.

Common Traps & FAQs

Wait, why are

L_{1}

and

L_{2}

parallel? I thought they were just random lines.

Great observation! Always check the slopes first. The slope of

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

- 2

. The slope of

L_{2} : 4 x + 2 y - p = 0

is also

- 2

. Since their slopes are identical, they are indeed parallel. Recognizing this early is the key to unlocking the geometric shortcut.

Do I really need to find the coordinates of point

M

? That looks like it will take forever.

Absolutely not! That is the trap the examiner wants you to fall into. By finding the coordinates of

M

, you would be doing unnecessary algebra. In JEE Advanced, we look for the 'geometric soul' of the problem. Since we have a right-angled triangle

△ A B M

, we can use trigonometry to find the ratio directly.

Why is the ratio

\frac{A M}{B M}

equal to

tan (θ)

? I'm a bit confused about the triangle setup.

Imagine the triangle

△ A B M

where

∠ M = 9 0^{\circ}

. The side

A M

is opposite to the angle

θ

(the angle between the transversal and

L_{2}

), and

B M

is the adjacent side. By the definition of tangent in a right-angled triangle,

tan (θ) = \frac{Opposite}{Adjacent} = \frac{A M}{B M}

. It's that simple!

Select Answer:

Visualized Solution

Introduction to the Problem

Equation of the Transversal Line

Finding Intersection Point $A$

Finding Intersection Point $B$

Using the Distance $A B$

Solving for the Parameter $p$

Visualizing the Geometry

The Trigonometric Relation

Slopes of the Lines

Calculating Tan Theta

Final Result and Summary

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

Visualizing the Triangle

The Master Equation

Final Result

Similar Questions

If the line $\frac{2 - x}{3} = \frac{3 y - 2}{4 λ + 1} = 4 - z$ makes a right angle with the line $\frac{x + 3}{3 μ} = \frac{1 - 2 y}{6} = \frac{5 - z}{7}$ , then $4 λ + 9 μ$ is equal to :

If the two lines $x + (a - 1) y = 1$ and $2 x + a^{2} y = 1 (a \in R - {0, 1})$ are perpendicular, then the distance of their point of intersection from the origin is :-

If one of the lines of $m y^{2} + (1 - m^{2}) x y - m x^{2} = 0$ is a bisector of the angle between the lines $x y = 0$ , then $m$ is

The distance between the two points $A$ and $A^{'}$ which lie on $y = 2$ such that both the line segments $A B$ and $A^{'} B$ (where $B$ is the point $(2, 3)$ ) subtend angle $\frac{π}{4}$ at the origin, is equal to :

If the two lines $l_{1} : \frac{x - 2}{3} = \frac{y + 1}{- 2}, z = 2$ and $l_{2} : \frac{x - 1}{1} = \frac{2 y + 3}{α} = \frac{z + 5}{2}$ are perpendicular, then an angle between the lines $l_{2}$ and $l_{3} : \frac{1 - x}{3} = \frac{2 y - 1}{- 4} = \frac{z}{4}$ is:

The combined equation of the two lines $a x + b y + c = 0$ and $a^{'} x + b^{'} y + c^{'} = 0$ can be written as $(a x + b y + c) (a^{'} x + b^{'} y + c^{'}) = 0$ . The equation of the angle bisectors of the lines represented by the equation $2 x^{2} + x y - 3 y^{2} = 0$ is

Select Answer:

Visualized Solution

Introduction to the Problem

Equation of the Transversal Line

Finding Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Finding Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Using the Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Solving for the Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Visualizing the Geometry

The Trigonometric Relation

Slopes of the Lines

Calculating Tan Theta

Final Result and Summary

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

Visualizing the Triangle

The Master Equation

Final Result

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the line 32−x​=4λ+13y−2​=4−z makes a right angle with the line 3μx+3​=61−2y​=75−z​, then 4λ+9μ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the two lines x+(a−1)y=1 and 2x+a2y=1(a∈R−{0,1}) are perpendicular, then the distance of their point of intersection from the origin is :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If one of the lines of my2+(1−m2)xy−mx2=0 is a bisector of the angle between the lines xy=0, then m is

Finding Intersection Point $A$

Finding Intersection Point $B$

Using the Distance $A B$

Solving for the Parameter $p$

If the line $\frac{2 - x}{3} = \frac{3 y - 2}{4 λ + 1} = 4 - z$ makes a right angle with the line $\frac{x + 3}{3 μ} = \frac{1 - 2 y}{6} = \frac{5 - z}{7}$ , then $4 λ + 9 μ$ is equal to :

If the two lines $x + (a - 1) y = 1$ and $2 x + a^{2} y = 1 (a \in R - {0, 1})$ are perpendicular, then the distance of their point of intersection from the origin is :-

If one of the lines of $m y^{2} + (1 - m^{2}) x y - m x^{2} = 0$ is a bisector of the angle between the lines $x y = 0$ , then $m$ is