How to Solve: Straight Lines (IIT JEE)

How to Solve: Straight Lines (IIT JEE)

Introduction

"Mastering straight lines unlocks 8–12 marks in JEE Main and 15+ in JEE Advanced—enough to push you from a 90 to a 120+ percentile. Whether it’s finding the angle between two roads, shifting the origin to simplify a problem, or writing the equation of a family of lines, this topic is your gateway to coordinate geometry."

What You Need To Know First

1. Coordinate Geometry Basics – Plotting points, distance formula, section formula.
2. Slope of a Line – Definition, positive/negative slope, parallel/perpendicular slopes.
3. Basic Algebra – Solving linear equations, substitution, and elimination.

Key Vocabulary

Formulas To Know

1. Equation Forms of a Straight Line

2. Angle Between Two Lines

• Formula: tanθ = |(m₁ – m₂)/(1 + m₁m₂)|
• m₁, m₂ = slopes of the two lines
• θ = angle between them
• Special Cases:
• If m₁m₂ = -1, lines are perpendicular (θ = 90°).
• If m₁ = m₂, lines are parallel (θ = 0°).
• MEMORISE THIS

3. Distance from Point to Line

• Formula: d = |Ax₁ + By₁ + C| / √(A² + B²)
• (x₁, y₁) = point
• Ax + By + C = 0 = line equation
• MEMORISE THIS

4. Distance Between Parallel Lines

• Formula: d = |C₁ – C₂| / √(A² + B²)
• Lines: Ax + By + C₁ = 0 and Ax + By + C₂ = 0
• MEMORISE THIS

5. Family of Lines

• Through Intersection of Two Lines:
• L₁ + λL₂ = 0 (where L₁ = 0 and L₂ = 0 are two lines)
• λ = parameter
• Through a Fixed Point (x₁, y₁):
• y – y₁ = m(x – x₁) (m varies)
• MEMORISE THIS

6. Shifting of Origin

• New Origin at (h, k):
• Replace x → x + h, y → y + k in the equation.
• MEMORISE THIS

Step-by-Step Method

How to Solve Any Straight Line Problem (JEE-Style)

1. Identify the Given Information – Points, slopes, intercepts, or equations.
2. Choose the Right Formula – Match the problem to the correct equation form.
3. Substitute Values – Plug in known values into the formula.
4. Simplify – Solve for the unknown (slope, intercept, angle, distance, etc.).
5. Check for Special Cases – Parallel, perpendicular, or coinciding lines.
6. Verify Units & Signs – Ensure distance is positive, angle is acute, etc.

Worked Example (Using Steps Above)

Problem: Find the equation of the line passing through (2, -3) and parallel to 3x – 4y + 5 = 0.

Step 1: Identify given info. - Point: (2, -3) - Line: 3x – 4y + 5 = 0 (parallel to our line)

Step 2: Choose the right formula. - Since lines are parallel, they have the same slope. - Use slope-intercept form or point-slope form.

Step 3: Find slope of given line. - Rewrite 3x – 4y + 5 = 0 in slope-intercept form: - 4y = 3x + 5 → y = (3/4)x + 5/4 - Slope (m) = 3/4

Step 4: Use point-slope form. - y – y₁ = m(x – x₁) - y – (-3) = (3/4)(x – 2) - y + 3 = (3/4)x – 3/2

Step 5: Simplify to general form. - Multiply by 4 to eliminate fractions: - 4y + 12 = 3x – 6 - 3x – 4y – 18 = 0

Step 6: Verify. - Check if (2, -3) satisfies the equation: - 3(2) – 4(-3) – 18 = 6 + 12 – 18 = 0 ✔️ - Check slope: 3x – 4y – 18 = 0 → y = (3/4)x – 9/2 → slope = 3/4 ✔️

Final Answer: 3x – 4y – 18 = 0

Worked Examples

Example 1 – Basic (Equation of a Line)

Problem: Find the equation of the line with slope 2 and y-intercept -5.

Solution: 1. Use slope-intercept form: y = mx + c 2. Given: m = 2, c = -5 3. Substitute: y = 2x – 5

What we did and why: - Direct substitution into the simplest form of a line equation.

Example 2 – Medium (Angle Between Lines)

Problem: Find the angle between the lines 2x + y – 3 = 0 and x – 3y + 2 = 0.

Solution: 1. Find slopes:
- Line 1: 2x + y – 3 = 0 → y = -2x + 3 → m₁ = -2
- Line 2: x – 3y + 2 = 0 → 3y = x + 2 → y = (1/3)x + 2/3 → m₂ = 1/3 2. Use angle formula: tanθ = |(m₁ – m₂)/(1 + m₁m₂)|
- tanθ = |(-2 – 1/3)/(1 + (-2)(1/3))| = |(-7/3)/(1/3)| = 7 3. θ = tan⁻¹(7) ≈ 81.87°

What we did and why: - Converted lines to slope-intercept form to find slopes. - Applied the angle formula directly.

Example 3 – Exam-Style (Family of Lines + Distance)

Problem: Find the equation of the line belonging to the family L: (2x + 3y – 1) + λ(x – y + 2) = 0 that is at a distance of 1 unit from the point (1, -2).

Solution: 1. Rewrite family in standard form:
- (2 + λ)x + (3 – λ)y + (-1 + 2λ) = 0 2. Distance formula: d = |Ax₁ + By₁ + C| / √(A² + B²)
- A = 2 + λ, B = 3 – λ, C = -1 + 2λ
- Point: (1, -2)
- Distance = 1 → |(2+λ)(1) + (3-λ)(-2) + (-1+2λ)| / √[(2+λ)² + (3-λ)²] = 1 3. Simplify numerator:
- (2 + λ) – 2(3 – λ) + (-1 + 2λ) = 2 + λ – 6 + 2λ – 1 + 2λ = 5λ – 5 4. Denominator:
- √[(2+λ)² + (3-λ)²] = √[4 + 4λ + λ² + 9 – 6λ + λ²] = √[2λ² – 2λ + 13] 5. Equation becomes:
- |5λ – 5| / √(2λ² – 2λ + 13) = 1
- Square both sides: (5λ – 5)² = 2λ² – 2λ + 13
- 25λ² – 50λ + 25 = 2λ² – 2λ + 13
- 23λ² – 48λ + 12 = 0 6. Solve quadratic:
- λ = [48 ± √(2304 – 1104)] / 46 = [48 ± √1200]/46 = [48 ± 20√3]/46
- λ = (24 ± 10√3)/23 7. Substitute λ back into family equation:
- For λ = (24 + 10√3)/23:
- (2 + (24 + 10√3)/23)x + (3 – (24 + 10√3)/23)y + (-1 + 2(24 + 10√3)/23) = 0
- Simplify to get final equation.

What we did and why: - Used the family of lines formula and distance condition. - Solved a quadratic in λ to find the required line.

Common Mistakes

Exam Traps

1-Minute Recap (Night Before Exam)

"Listen up—straight lines are 10+ marks in JEE, and you can’t afford to lose them. Here’s the crash course:

1. Equation Forms: Memorise y = mx + c, y – y₁ = m(x – x₁), and Ax + By + C = 0. If you have two points, use the two-point form.
2. Angle Between Lines: Use tanθ = |(m₁ – m₂)/(1 + m₁m₂)|. If lines are perpendicular, m₁m₂ = -1.
3. Distance from Point to Line: |Ax₁ + By₁ + C| / √(A² + B²)—never forget the absolute value!
4. Family of Lines: If two lines intersect, the family is L₁ + λL₂ = 0. If a fixed point is given, use y – y₁ = m(x – x₁).
5. Shifting Origin: New origin at (h,k)? Replace x → x – h, y → y – k in the equation.

Last tip: Always check if lines are parallel (same slope) or perpendicular (m₁m₂ = -1) before solving. Now go crush that exam!