JEE Advanced 2008

LEVELJEE Main

Animated Solution for Mathematics - Straight Lines: Consider the lines given by $L_{1} : x + 3 y - 5 = 0$ ; $L_{2} : 3 x - k y - 1 = 0$ ; $L_{3} : 5 x + 2 y - 12 = 0$ . Match the Statements / Expressions in Column I with the Statements / Expressions in Column II and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the ORS.

Q: Is there a trick to remember the slope condition for $L_2$?

Yes! For any line $ax + by + c = 0$, the slope is $-\frac{a}{b}$. For $L_2: 3x - ky - 1 = 0$, the slope is $-\frac{3}{-k} = \frac{3}{k}$. Just be careful with the signs—that is where most students lose marks!

List-I

(P)

L_{1}, L_{2}, L_{3}

are concurrent, if

(Q)

One of

L_{1}, L_{2}, L_{3}

is parallel to at least one of the other two, if

(R)

L_{1}, L_{2}, L_{3}

form a triangle, if

(S)

L_{1}, L_{2}, L_{3}

do not form a triangle, if

List-II

(1)

k = - 9

(2)

k = - \frac{6}{5}

(3)

k = \frac{5}{6}

(4)

k = 5

Select Matching Pairs:

* Multiple Allowed

PMatches

QMatches

RMatches

SMatches

Visualized Solution

Introduction to the Three Lines $L_{1}$ , $L_{2}$ , and $L_{3}$

We are given three straight lines in a two-dimensional plane:
Line $L_{1} : x + 3 y - 5 = 0$
Line $L_{2} : 3 x - k y - 1 = 0$ (where $k$ is a variable parameter)
Line $L_{3} : 5 x + 2 y - 12 = 0$

Finding the Intersection of $L_{1}$ and $L_{3}$

To find the intersection point, we solve $L_{1}$ and $L_{3}$ simultaneously.
From $L_{1}$ , we get: $x = 5 - 3 y$
Substitute $x$ into $L_{3}$ : $5 (5 - 3 y) + 2 y = 12$
Simplifying: $25 - 15 y + 2 y = 12 ⟹ - 13 y = - 13 ⟹ y = 1$
Substituting $y = 1$ back: $x = 5 - 3 (1) = 2$
The unique intersection point is $(2, 1)$ .

Understanding Concurrency

Three lines are concurrent if they all pass through a single common point.
Since $L_{1}$ and $L_{3}$ intersect at $(2, 1)$ , line $L_{2}$ must also pass through $(2, 1)$ for concurrency.
This means the coordinates $(2, 1)$ must satisfy the equation of $L_{2}$ .

Solving for Concurrency ( $k = 5$ )

Substitute $(2, 1)$ into $L_{2} : 3 x - k y - 1 = 0$ :
$3 (2) - k (1) - 1 = 0$
$6 - k - 1 = 0 ⟹ 5 - k = 0 ⟹ k = 5$
Therefore, the lines are concurrent when $k = 5$ .
This matches Statement (A) in Column I with option (s) in Column II.

Checking for Parallelism ( $L_{1} ∥ L_{2}$ )

Two lines are parallel if their slopes are equal.
Slope of $L_{1}$ : $m_{1} = - \frac{1}{3}$
Slope of $L_{2}$ : $m_{2} = \frac{3}{k}$
Equating slopes: $- \frac{1}{3} = \frac{3}{k} ⟹ k = - 9$
Thus, $L_{1}$ is parallel to $L_{2}$ when $k = - 9$ .

Checking for Parallelism ( $L_{2} ∥ L_{3}$ )

Now check if $L_{2}$ is parallel to $L_{3}$ .
Slope of $L_{3}$ : $m_{3} = - \frac{5}{2}$
Slope of $L_{2}$ : $m_{2} = \frac{3}{k}$
Equating slopes: $\frac{3}{k} = - \frac{5}{2} ⟹ k = - \frac{6}{5}$
Thus, $L_{2}$ is parallel to $L_{3}$ when $k = - \frac{6}{5}$ .
Therefore, Statement (B) matches with both (p) and (q).

Condition to Form a Triangle

Three lines form a triangle if and only if:
1. No two lines are parallel to each other.
2. The three lines are not concurrent.
Mathematically, this means $k \neq = - 9$ , $k \neq = - \frac{6}{5}$ , and $k \neq = 5$ .

Matching Triangle Formation ( $k = \frac{5}{6}$ )

Looking at Column II, the only value of $k$ that is not $- 9$ , $- \frac{6}{5}$ , or $5$ is $k = \frac{5}{6}$ .
For $k = \frac{5}{6}$ , the lines are neither parallel nor concurrent, forming a valid triangle.
Thus, Statement (C) matches with option (r).

Condition for Not Forming a Triangle

Three lines do not form a triangle if:
1. At least two lines are parallel ( $k = - 9$ or $k = - \frac{6}{5}$ )
2. The lines are concurrent ( $k = 5$ )
Thus, the values of $k$ for which no triangle is formed are $k = - 9, - \frac{6}{5}, 5$ .
This matches Statement (D) with options (p), (q), and (s).

Final Matrix Match Summary

Let's summarize the final matching matrix:
$(A) \to (s)$ (Concurrent at $k = 5$ )
$(B) \to (p), (q)$ (Parallel at $k = - 9, - \frac{6}{5}$ )
$(C) \to (r)$ (Forms a triangle at $k = \frac{5}{6}$ )
$(D) \to (p), (q), (s)$ (Does not form a triangle at $k = - 9, - \frac{6}{5}, 5$ )

00:00 / 00:00

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

Imagine you are standing in a coordinate plane, watching three lines perform a dance. Two of these lines,

L_{1} : x + 3 y - 5 = 0

and

L_{3} : 5 x + 2 y - 12 = 0

, are the stage managers.

They are fixed, unmoving, and reliable. The third line,

L_{2} : 3 x - k y - 1 = 0

, is the dancer, changing its posture based on the value of the parameter

k

Finding the Meeting Point

Before we can understand how

L_{2}

interacts with the others, we must know where the stage managers meet. We solve the system of equations for

L_{1}

and

L_{3}

By expressing

x

x = 5 - 3 y

from

L_{1}

and substituting it into

L_{3}

, we obtain:

5 (5 - 3 y) + 2 y = 12

This simplifies beautifully to

- 13 y = - 13

, giving us

y = 1

. Substituting back into the expression for

x

, we find

x = 2

The intersection point is

(2, 1)

. This point is the anchor of our entire problem.

The Three Scenarios

Now, let's look at the conditions for the lines. Concurrency is the moment when all three lines meet at the same point.

For this to happen, our dancer

L_{2}

must pass through

(2, 1)

. Substituting these coordinates into

L_{2}

, we get:

3 (2) - k (1) - 1 = 0

This yields $k = 5$ . When

k = 5

, the lines are concurrent.

Next, we consider parallelism, which is the refusal of lines to meet.

L_{1}

is parallel to

L_{2}

when their slopes match:

\frac{- 1}{3} = \frac{3}{k} \Rightarrow k = - 9

Similarly,

L_{2}

is parallel to

L_{3}

when their slopes match:

\frac{- 5}{2} = \frac{3}{k} \Rightarrow k = - \frac{6}{5}

The Goldilocks Zone

Finally, we ask: when do these lines form a triangle? A triangle is a closed shape, a sanctuary of area.

It cannot exist if the lines are parallel (they never meet) or concurrent (they meet at a single point). Therefore, the triangle exists only in the 'Goldilocks zone'—where

k

is not

- 9

- \frac{6}{5}

, or

5

For any other value, the lines will intersect at three distinct points, creating a beautiful, enclosed triangle. You have just mastered the fundamental interplay of lines in coordinate geometry.

Common Traps & FAQs

Why do we solve

L_{1}

and

L_{3}

first? Can't we just use the determinant method for concurrency?

You absolutely can! The determinant method is a powerful, elegant tool. However, solving

L_{1}

and

L_{3}

first gives you the geometric intersection point, which builds intuition. In JEE, intuition often helps you catch silly calculation errors that a blind determinant calculation might hide.

Why do we say a triangle doesn't form when lines are parallel? Isn't a triangle just three lines?

A triangle is defined by three vertices. If two lines are parallel, they never intersect, meaning you lose a vertex. If all three are concurrent, the vertices collapse into one point. In both cases, the 'enclosed area' vanishes, so no triangle is formed.

Is there a trick to remember the slope condition for

L_{2}

Yes! For any line

a x + b y + c = 0

, the slope is

- \frac{a}{b}

. For

L_{2} : 3 x - k y - 1 = 0

, the slope is

- \frac{3}{- k} = \frac{3}{k}

. Just be careful with the signs—that is where most students lose marks!

List-I

List-II

Select Matching Pairs:

Visualized Solution

Introduction to the Three Lines $L_{1}$ , $L_{2}$ , and $L_{3}$

Finding the Intersection of $L_{1}$ and $L_{3}$

Understanding Concurrency

Solving for Concurrency ( $k = 5$ )

Checking for Parallelism ( $L_{1} ∥ L_{2}$ )

Checking for Parallelism ( $L_{2} ∥ L_{3}$ )

Condition to Form a Triangle

Matching Triangle Formation ( $k = \frac{5}{6}$ )

Condition for Not Forming a Triangle

Final Matrix Match Summary

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

Finding the Meeting Point

The Three Scenarios

The Goldilocks Zone

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List-I

List-II

Select Matching Pairs:

Visualized Solution

Finding the Intersection of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L3​

Understanding Concurrency

Solving for Concurrency ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=5)

Checking for Parallelism ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​∥L2​)

Checking for Parallelism ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​∥L3​)

Condition to Form a Triangle

Matching Triangle Formation ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=65​)

Condition for Not Forming a Triangle

Final Matrix Match Summary

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

Finding the Meeting Point

The Three Scenarios

The Goldilocks Zone

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