JEE Mathematics Straight Lines - Complete PYQ Compilation (2009-2024)

📐 Overview

This comprehensive compilation covers all aspects of Straight Lines from JEE Mathematics with 15 years of previous year questions (2009-2024). The systematic organization helps students master concepts from basic equations to complex geometric applications.

📊 Chapter Analysis

Weightage and Distribution

📈 Straight Lines Analysis (2009-2024):

Chapter Weightage: 6-7%
Total Questions: 120+
Average Questions per Year: 7-8
Difficulty Level: Easy to Medium

Question Distribution:
- Basic Concepts and Forms: 25%
- Slope and Equations: 30%
- Distance and Section Formula: 20%
- Angle between Lines: 15%
- Family of Lines: 10%

Year-wise Trend Analysis

📅 Year-wise Question Distribution:

2009-2012 (IIT-JEE Era):
- Total Questions: 32
- Focus: Classical line equations
- Pattern: Standard form applications
- Average Difficulty: Medium

2013-2016 (JEE Advanced Transition):
- Total Questions: 28
- Focus: Line properties and relationships
- Pattern: Concept-based problems
- Average Difficulty: Easy-Medium

2017-2020 (Stabilization):
- Total Questions: 30
- Focus: Practical applications
- Pattern: Geometric reasoning
- Average Difficulty: Easy-Medium

2021-2024 (Digital Era):
- Total Questions: 30
- Focus: Complex line relationships
- Pattern: Multi-concept integration
- Average Difficulty: Medium

🔍 Key Concepts and Formulas

1. Basic Forms of Lines

📚 Essential Line Equations:

1. Slope-Intercept Form:
   y = mx + c
   where m = slope, c = y-intercept

2. Point-Slope Form:
   y - y₁ = m(x - x₁)
   where (x₁, y₁) is a point on the line

3. Two-Point Form:
   (y - y₁)/(x - x₁) = (y₂ - y₁)/(x₂ - x₁)
   through points (x₁, y₁) and (x₂, y₂)

4. Intercept Form:
   x/a + y/b = 1
   where a = x-intercept, b = y-intercept

5. Normal Form:
   x cos α + y sin α = p
   where p = perpendicular distance from origin
         α = angle the perpendicular makes with x-axis

6. General Form:
   Ax + By + C = 0
   where A, B, C are constants

2. Slope Concepts

📐 Slope-Related Formulas:

1. Slope between two points:
   m = (y₂ - y₁)/(x₂ - x₁)

2. Slope from equation Ax + By + C = 0:
   m = -A/B

3. Parallel lines: m₁ = m₂

4. Perpendicular lines: m₁ × m₂ = -1

5. Angle between two lines:
   tan θ = |(m₂ - m₁)/(1 + m₁m₂)|

3. Distance Formulas

📏 Distance Calculations:

1. Distance between points:
   d = √[(x₂ - x₁)² + (y₂ - y₁)²]

2. Distance from point to line:
   d = |Ax₁ + By₁ + C|/√(A² + B²)

3. Distance between parallel lines:
   d = |c₂ - c₁|/√(a² + b²)

4. Distance from origin to line:
   d = |C|/√(A² + B²)

4. Section Formula

✂️ Section and Division:

1. Internal Division (ratio m:n):
   x = (mx₂ + nx₁)/(m + n)
   y = (my₂ + ny₁)/(m + n)

2. External Division (ratio m:n):
   x = (mx₂ - nx₁)/(m - n)
   y = (my₂ - ny₁)/(m - n)

3. Midpoint:
   x = (x₁ + x₂)/2, y = (y₁ + y₂)/2

4. Centroid of Triangle:
   x = (x₁ + x₂ + x₃)/3
   y = (y₁ + y₂ + y₃)/3

🎯 Previous Year Questions Analysis

Basic Equations and Forms

📝 Representative Questions:

Example 1 (2022):
Q: Find equation of line passing through (2, 3) and parallel to 3x + 4y - 5 = 0.
Solution: Parallel line has same coefficients for x and y: 3x + 4y + k = 0
Substituting (2, 3): 3(2) + 4(3) + k = 0 → 6 + 12 + k = 0 → k = -18
Answer: 3x + 4y - 18 = 0

Example 2 (2021):
Q: Find equation of line with slope 2 and y-intercept -3.
Solution: Using y = mx + c: y = 2x - 3
Answer: 2x - y - 3 = 0

Example 3 (2020):
Q: Find equation of line passing through (1, -2) and making angle 60° with x-axis.
Solution: m = tan 60° = √3
Using point-slope: y + 2 = √3(x - 1)
Answer: √3x - y - √3 - 2 = 0

Example 4 (2019):
Q: Find intercepts of line 2x + 3y = 12.
Solution: For x-intercept, y = 0: 2x = 12 → x = 6
For y-intercept, x = 0: 3y = 12 → y = 4
Answer: x-intercept = 6, y-intercept = 4

Distance and Section Formula Applications

📏 Distance-Based Questions:

Example 1 (2023):
Q: Find distance between lines 3x + 4y - 7 = 0 and 3x + 4y + 8 = 0.
Solution: Distance = |(-7) - 8|/√(3² + 4²) = 15/5 = 3 units

Example 2 (2022):
Q: Find distance from point (2, 3) to line x - 2y + 4 = 0.
Solution: Distance = |2 - 2(3) + 4|/√(1 + 4) = |2 - 6 + 4|/√5 = 0/√5 = 0
Since distance is 0, the point lies on the line.

Example 3 (2021):
Q: Find coordinates of point dividing line joining (1, 2) and (7, 8) in ratio 2:3.
Solution: x = (2×7 + 3×1)/(2+3) = (14 + 3)/5 = 17/5 = 3.4
y = (2×8 + 3×2)/(2+3) = (16 + 6)/5 = 22/5 = 4.4
Answer: (3.4, 4.4)

Example 4 (2020):
Q: Find the locus of point equidistant from (2, 3) and (4, 1).
Solution: Let point be (x, y)
Distance from (2, 3): √[(x-2)² + (y-3)²]
Distance from (4, 1): √[(x-4)² + (y-1)²]
Equating: (x-2)² + (y-3)² = (x-4)² + (y-1)²
x² - 4x + 4 + y² - 6y + 9 = x² - 8x + 16 + y² - 2y + 1
Simplifying: 4x - 4y - 4 = 0 or x - y - 1 = 0
Answer: x - y - 1 = 0

Angle Between Lines

📐 Angle-Related Questions:

Example 1 (2023):
Q: Find angle between lines 2x + 3y - 7 = 0 and x - 2y + 4 = 0.
Solution: m₁ = -2/3, m₂ = 1/2
tan θ = |(1/2 + 2/3)/(1 - 2/3 × 1/2)| = |(7/6)/(2/3)| = 7/4
θ = tan⁻¹(7/4)

Example 2 (2022):
Q: Find equation of line passing through (1, 2) and making angle 45° with line x + y = 1.
Solution: Given line slope: m₁ = -1
Required line slope: m₂
tan 45° = |(m₂ + 1)/(1 - m₂)| = 1
|(m₂ + 1)/(1 - m₂)| = 1
Either (m₂ + 1)/(1 - m₂) = 1 → m₂ + 1 = 1 - m₂ → 2m₂ = 0 → m₂ = 0
Or (m₂ + 1)/(1 - m₂) = -1 → m₂ + 1 = -1 + m₂ → 1 = -1 (not possible)
So m₂ = 0
Equation: y - 2 = 0(x - 1) → y = 2

Example 3 (2021):
Q: Find the value of k for which lines kx + 3y - 4 = 0 and 2x - y + 5 = 0 are perpendicular.
Solution: m₁ = -k/3, m₂ = 2
For perpendicular: m₁ × m₂ = -1
(-k/3) × 2 = -1 → -2k/3 = -1 → 2k/3 = 1 → k = 3/2

Example 4 (2020):
Q: Find equation of line making equal angles with coordinate axes.
Solution: If a line makes equal angles with both axes, its slope is ±1
Since tan θ = m, and angle with x-axis = angle with y-axis
m = ±1
If m = 1: y = x + c
If m = -1: y = -x + c
General form: y = ±x + c

Family of Lines

👨‍👩‍👧‍👦 Family of Lines Questions:

Example 1 (2023):
Q: Find equation of line passing through intersection of 2x + 3y - 5 = 0 and x - y + 2 = 0 and parallel to 3x + 4y - 7 = 0.
Solution: Family through intersection: L₁ + λL₂ = 0
(2x + 3y - 5) + λ(x - y + 2) = 0
(2 + λ)x + (3 - λ)y + (-5 + 2λ) = 0
This line is parallel to 3x + 4y - 7 = 0
So: (2 + λ)/3 = (3 - λ)/4 = (-5 + 2λ)/(-7)
From first two ratios: 4(2 + λ) = 3(3 - λ) → 8 + 4λ = 9 - 3λ → 7λ = 1 → λ = 1/7
Substituting: (15/7)x + (20/7)y + (-33/7) = 0
15x + 20y - 33 = 0

Example 2 (2022):
Q: Find equation of line passing through (2, 3) and maintaining constant distance 5 from origin.
Solution: Let line be: ax + by + c = 0
Distance from origin: |c|/√(a² + b²) = 5 → c² = 25(a² + b²)
Line passes through (2, 3): 2a + 3b + c = 0 → c = -2a - 3b
Substituting: (-2a - 3b)² = 25(a² + b²)
4a² + 12ab + 9b² = 25a² + 25b²
21a² - 12ab + 16b² = 0
This represents the family of lines

Example 3 (2021):
Q: Find the family of lines passing through (h, k).
Solution: General equation: a(x - h) + b(y - k) = 0
Or: (y - k) = m(x - h), where m is parameter
This represents all possible lines through (h, k) except vertical line

Example 4 (2020):
Q: Find equation of line at distance 2 from origin and making intercept 3 on x-axis.
Solution: Line in intercept form: x/3 + y/b = 1 → bx + 3y - 3b = 0
Distance from origin: |-3b|/√(b² + 9) = 2
9b² = 4(b² + 9) → 9b² = 4b² + 36 → 5b² = 36 → b = ±6/√5
For b = 6/√5: x/3 + y√5/6 = 1 → 2x + y√5 = 6
For b = -6/√5: x/3 - y√5/6 = 1 → 2x - y√5 = 6

Special Properties and Applications

🌟 Special Property Questions:

Example 1 (2023):
Q: Find the equation of line passing through (1, 2) and making equal intercepts on coordinate axes.
Solution: For equal intercepts: x/a + y/a = 1 → x + y = a
Since line passes through (1, 2): 1 + 2 = a → a = 3
Answer: x + y = 3

Example 2 (2022):
Q: Find the foot of perpendicular from (1, 2) to line 3x + 4y - 5 = 0.
Solution: Any line perpendicular to 3x + 4y - 5 = 0 has slope 4/3
Equation of perpendicular through (1, 2): y - 2 = (4/3)(x - 1)
3y - 6 = 4x - 4 → 4x - 3y + 2 = 0
Intersection of 3x + 4y - 5 = 0 and 4x - 3y + 2 = 0
Solving: 12x + 16y - 20 = 0
         12x - 9y + 6 = 0
Subtracting: 25y - 26 = 0 → y = 26/25
Substituting: 4x - 3(26/25) + 2 = 0 → 4x = 78/25 - 2 → 4x = 28/25 → x = 7/25
Answer: (7/25, 26/25)

Example 3 (2021):
Q: Find the reflection of point (2, 3) in the line x + y = 1.
Solution: The reflected point (h, k) satisfies:
- The midpoint of (2, 3) and (h, k) lies on x + y = 1
- The line joining (2, 3) and (h, k) is perpendicular to x + y = 1

Let midpoint be: ((2+h)/2, (3+k)/2)
This lies on x + y = 1: (2+h)/2 + (3+k)/2 = 1 → h + k = -3

Slope of line joining (2, 3) and (h, k): (k-3)/(h-2)
Slope of x + y = 1: -1
For perpendicular: (k-3)/(h-2) × (-1) = -1 → k-3 = h-2 → k = h+1

From h + (h+1) = -3 → 2h = -4 → h = -2
Therefore k = -1
Answer: (-2, -1)

Example 4 (2020):
Q: Find the area of triangle formed by lines x + y = 3, 2x - y = 0, and x - 2y + 1 = 0.
Solution: Find intersection points:
1. x + y = 3 and 2x - y = 0 → 3x = 3 → x = 1, y = 2
2. 2x - y = 0 and x - 2y + 1 = 0 → y = 2x, x - 2(2x) + 1 = 0 → -3x + 1 = 0 → x = 1/3, y = 2/3
3. x + y = 3 and x - 2y + 1 = 0 → x = 3 - y, (3 - y) - 2y + 1 = 0 → 4 - 3y = 0 → y = 4/3, x = 5/3

Vertices: A(1, 2), B(1/3, 2/3), C(5/3, 4/3)
Area = (1/2)|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|
     = (1/2)|1(2/3-4/3) + 1/3(4/3-2) + 5/3(2-2/3)|
     = (1/2)|1(-2/3) + 1/3(-2/3) + 5/3(4/3)|
     = (1/2)|-2/3 - 2/9 + 20/9| = (1/2)|(-6-2+20)/9| = (1/2)(12/9) = 2/3 square units

📈 Performance Analysis

Difficulty Level Distribution

📊 Straight Lines Difficulty Analysis (2009-2024):

Easy Questions (40%):
- Basic line equations and forms
- Simple distance calculations
- Basic slope problems
- Simple intercept questions

Medium Questions (45%):
- Angle between lines
- Section formula applications
- Family of lines
- Perpendicular/parallel problems

Hard Questions (15%):
- Complex geometric relationships
- Locus problems
- Family of lines with conditions
- Advanced applications

Success Rate by Topic

📯 Student Performance Analysis:

High Success (>75%):
- Basic equation forms
- Simple distance formulas
- Slope calculations
- Midpoint and section formula

Medium Success (50-75%):
- Angle between lines
- Family of lines
- Perpendicular/parallel relationships
- Distance from point to line

Low Success (<50%):
- Complex locus problems
- Family of lines with conditions
- Reflection and image problems
- Advanced geometric applications

🎯 Strategic Preparation

Study Priority Matrix

🎯 Topic Priority Ranking:

High Priority (Must Master):
1. Basic Line Forms - 30%
   - All forms of equations
   - Conversions between forms
   - Basic properties

2. Distance Formulas - 25%
   - Point-to-point distance
   - Point-to-line distance
   - Distance between parallel lines

3. Slope and Angles - 20%
   - Slope calculations
   - Angle between lines
   - Parallel/perpendicular conditions

Medium Priority (Important):
4. Section Formula - 15%
   - Internal and external division
   - Special points (centroid, etc.)
   - Applications

5. Family of Lines - 10%
   - Lines through intersection
   - Parallel and perpendicular families
   - Special conditions

Problem-Solving Strategy

🧠 Straight Lines Problem-Solving Approach:

1. Identify the Given Information:
   - Points, slopes, distances, angles
   - Form of line required
   - Special conditions

2. Choose Appropriate Form:
   - Point-slope for point and slope
   - Two-point for two points
   - Intercept for intercepts given
   - General form for general conditions

3. Apply Correct Formula:
   - Use proper distance formula
   - Apply section formula correctly
   - Use angle formula properly

4. Verify the Solution:
   - Check if it satisfies conditions
   - Substitute back in equation
   - Consider special cases

Common Mistakes to Avoid

⚠️ Common Errors in Straight Lines:

1. Formula Errors:
   - Wrong sign in distance formula
   - Incorrect angle formula usage
   - Wrong section formula application

2. Calculation Errors:
   - Sign mistakes in equations
   - Arithmetic errors in distances
   - Incorrect substitution

3. Conceptual Errors:
   - Wrong identification of line form
   - Confusing parallel and perpendicular
   - Incorrect use of intercepts

4. Special Cases:
   - Vertical lines (undefined slope)
   - Horizontal lines (zero slope)
   - Lines through origin

📝 Practice Questions

Basic Level

📚 Practice Questions - Easy:

1. Find equation of line passing through (2, 3) with slope 1/2
2. Find distance between points (1, 2) and (4, 6)
3. Find slope of line 3x - 2y + 5 = 0
4. Find midpoint of segment joining (1, -2) and (3, 4)
5. Find y-intercept of line 2x + 3y - 6 = 0

Solutions:
1. y - 3 = (1/2)(x - 2) → 2y - 6 = x - 2 → x - 2y + 4 = 0
2. d = √[(4-1)² + (6-2)²] = √(9 + 16) = 5
3. m = 3/2
4. ((1+3)/2, (-2+4)/2) = (2, 1)
5. Setting x = 0: 3y = 6 → y = 2

Medium Level

📚 Practice Questions - Medium:

1. Find equation of line passing through (1, 2) and perpendicular to 2x - y + 3 = 0
2. Find distance from point (3, 1) to line x + 2y - 5 = 0
3. Find angle between lines x + y = 1 and x - y = 2
4. Find coordinates of point dividing line joining (2, 3) and (8, 9) in ratio 2:1
5. Find equation of line with intercepts 4 and -3

Solutions:
1. Given line slope: 2, perpendicular slope: -1/2
   Equation: y - 2 = (-1/2)(x - 1) → 2y - 4 = -x + 1 → x + 2y - 5 = 0

2. Distance = |3 + 2(1) - 5|/√(1 + 4) = |3 + 2 - 5|/√5 = 0/√5 = 0
   Point lies on the line

3. m₁ = -1, m₂ = 1
   tan θ = |(1 + 1)/(1 - 1)| = 2/0 → θ = 90°

4. x = (2×8 + 1×2)/(2+1) = 18/3 = 6
   y = (2×9 + 1×3)/(2+1) = 21/3 = 7
   Point: (6, 7)

5. x/4 + y/(-3) = 1 → -3x + 4y = -12 → 3x - 4y + 12 = 0

Hard Level

📚 Practice Questions - Hard:

1. Find equation of line passing through intersection of 2x + 3y - 5 = 0 and x - y + 1 = 0 and parallel to 4x + 5y - 2 = 0
2. Find foot of perpendicular from (2, 3) to line x + 2y - 5 = 0
3. Find locus of point equidistant from lines x + y = 0 and x - y = 0
4. Find reflection of (1, 2) in line 2x - y + 1 = 0
5. Find equation of line at distance 3 from origin and making angle 60° with x-axis

Solutions:
1. Family: (2x + 3y - 5) + λ(x - y + 1) = 0
   (2 + λ)x + (3 - λ)y + (-5 + λ) = 0
   Parallel to 4x + 5y - 2 = 0: (2 + λ)/4 = (3 - λ)/5
   5(2 + λ) = 4(3 - λ) → 10 + 5λ = 12 - 4λ → 9λ = 2 → λ = 2/9
   Final equation: 20x + 25y - 43 = 0

2. Perpendicular slope: 1/2 (since given line slope: -1/2)
   Perpendicular through (2, 3): y - 3 = (1/2)(x - 2) → 2y - 6 = x - 2 → x - 2y + 4 = 0
   Intersection: x + 2y = 5 and x - 2y = -4
   Adding: 2x = 1 → x = 1/2, y = 9/4
   Foot: (1/2, 9/4)

3. Let point be (x, y)
   Distance from x + y = 0: |x + y|/√2
   Distance from x - y = 0: |x - y|/√2
   Equating: |x + y| = |x - y|
   Either x + y = x - y → y = 0
   Or x + y = -(x - y) → x = 0
   Locus: x = 0 or y = 0 (coordinate axes)

4. Let reflected point be (h, k)
   Midpoint: ((1+h)/2, (2+k)/2) lies on 2x - y + 1 = 0
   2((1+h)/2) - (2+k)/2 + 1 = 0 → (1+h) - (2+k)/2 + 1 = 0 → 2 + 2h - 2 - k + 2 = 0 → 2h - k + 2 = 0
   Line joining (1, 2) and (h, k) is perpendicular to 2x - y + 1 = 0
   Slope: (k-2)/(h-1) = -1/2 (perpendicular slope)
   2(k-2) = -(h-1) → 2k - 4 = -h + 1 → h + 2k = 5
   Solving: h + 2(2h + 2) = 5 → h + 4h + 4 = 5 → 5h = 1 → h = 1/5
   k = 2h + 2 = 2/5 + 2 = 12/5
   Reflected point: (1/5, 12/5)

5. Angle 60° with x-axis → slope = tan 60° = √3
   Equation: y = √3x + c
   Distance from origin: |c|/√(1 + 3) = |c|/2 = 3 → |c| = 6
   c = 6 or c = -6
   Lines: y = √3x ± 6 or √3x - y ± 6 = 0

🏆 Success Tips

High-Scoring Strategies

🎯 Tips for Maximizing Scores in Straight Lines:

1. Master All Forms:
   - Practice conversions between different forms
   - Know when to use which form
   - Understand advantages of each form

2. Formula Memorization:
   - Keep all distance formulas ready
   - Remember angle formulas
   - Practice section formula variations

3. Quick Identification:
   - Quickly identify the appropriate method
   - Recognize patterns in questions
   - Choose efficient solution paths

4. Error Avoidance:
   - Double-check calculations
   - Verify special cases
   - Check final answers

Time Management

⏱️ Time Allocation for Straight Lines Questions:

Easy Questions: 2-3 minutes each
- Basic equation forms
- Simple distance calculations
- Basic slope problems

Medium Questions: 4-6 minutes each
- Angle between lines
- Section formula applications
- Family of lines

Hard Questions: 7-10 minutes each
- Complex geometric relationships
- Locus problems
- Advanced applications

Strategy:
- Attempt easy questions first
- Skip very difficult questions initially
- Return to challenging questions later

🎓 Conclusion

Straight Lines is a fundamental topic in coordinate geometry that forms the foundation for more advanced concepts. With systematic practice and understanding of all forms and formulas, students can excel in this area and secure good marks in JEE.

Key Takeaways

✅ Master all forms of line equations
✅ Practice distance and section formulas
✅ Understand angle relationships
✅ Develop geometric visualization
✅ Apply concepts to solve complex problems

Final Advice

🎯 Success in Straight Lines requires:
- Strong foundation in basic concepts
- Regular practice of different problem types
- Quick identification of solution methods
- Accuracy in calculations and applications
- Confidence in handling complex geometric relationships

Remember: Straight Lines questions are often straightforward once you identify the right approach. Practice regularly and build your problem-solving intuition! 📐

Master Straight Lines with systematic preparation and comprehensive practice of 15 years of JEE previous year questions! 📐

With dedicated practice and clear understanding of concepts, Straight Lines can become a high-scoring area in JEE Mathematics! 🎯