Angles Between Lines and Concurrent Lines
Straight Lines: Angles Between Lines and Concurrent Lines
Angles Between Lines and Concurrent Lines
Angles Between Lines and Concurrent Lines
What you'll learn
- Angle between two lines: tan θ = |(m₁ − m₂)/(1 + m₁m₂)|.
- Parallel lines: m₁ = m₂ and perpendicular lines: m₁m₂ = −1.
- Angle bisectors of two intersecting lines.
- Family of lines through intersection: L₁ + λL₂ = 0.
- Concurrency condition using a determinant.
Key concepts
Level 1 — Angle between lines and parallel/perpendicular conditions
Angle between lines with slopes m₁ and m₂: tan θ = |(m₁ − m₂)/(1 + m₁m₂)| where θ is the acute angle. The two possible angles between lines are θ and π − θ (supplementary pair); tan gives the acute one via absolute value.
Parallel lines: Same slope m₁ = m₂ (and different y-intercepts; otherwise same line). Lines y = 2x + 1 and y = 2x − 3 are parallel.
Perpendicular lines: m₁ · m₂ = −1. If m₁ = 2, then perpendicular slope = −1/2. For horizontal line (m₁ = 0): perpendicular is vertical (undefined slope). Exceptions: vertical and horizontal lines.
Special angles:
- θ = 45°: tan θ = 1 → |m₁ − m₂| = |1 + m₁m₂|.
- θ = 60°: tan θ = √3.
- θ = 90°: tan θ → ∞ → 1 + m₁m₂ = 0.
Level 2 — Angle bisectors, family of lines, concurrency
Angle bisectors of L₁: A₁x + B₁y + C₁ = 0 and L₂: A₂x + B₂y + C₂ = 0: (A₁x + B₁y + C₁)/√(A₁² + B₁²) = ±(A₂x + B₂y + C₂)/√(A₂² + B₂²). Two bisectors — the '+' equation gives one pair, '−' gives the other (perpendicular bisectors).
Which bisector contains the acute angle? Check: if A₁A₂ + B₁B₂ < 0, the '+' equation is the acute bisector; if > 0, the '−' equation is the acute bisector.
Family of lines through intersection of L₁ = 0 and L₂ = 0: L₁ + λL₂ = 0 for parameter λ ∈ ℝ. Every value of λ gives a line through the intersection point of L₁ and L₂. To find a specific line in this family (e.g., passing through a third point P), substitute P to find λ.
Concurrency of three lines: L₁: a₁x + b₁y + c₁ = 0, L₂: a₂x + b₂y + c₂ = 0, L₃: a₃x + b₃y + c₃ = 0. Concurrent iff: |a₁ b₁ c₁; a₂ b₂ c₂; a₃ b₃ c₃| = 0 (determinant of coefficient matrix = 0).
Geometric meaning: Three lines concurrent = they all meet at a single point (triangle formed is degenerate).
JEE pattern — finding the line through intersection: To find the line through the intersection of L₁ = 0 and L₂ = 0 that also passes through P(h, k):
- Write L₁ + λL₂ = 0.
- Substitute (h, k) → solve for λ.
- Substitute λ back.
Angle bisector of x and y axes: The angle bisectors of x-axis (y = 0) and y-axis (x = 0) are y = x and y = −x (lines at 45° and 135°).
NCERT spotlight
The family of lines concept L₁ + λL₂ = 0 avoids solving simultaneously for the intersection point first. Concurrency determinant = 0 means the third line's equation is a linear combination of the first two. Perpendicularity condition m₁m₂ = −1 breaks down when either line is vertical — handle separately.
Worked example
Find the angle between lines 2x − 3y + 5 = 0 and x + 4y − 7 = 0.
Step 1 — Slopes: From 2x − 3y + 5 = 0: m₁ = 2/3.
From x + 4y − 7 = 0: m₂ = −1/4.
Step 2 — tan θ = |(m₁ − m₂)/(1 + m₁m₂)| = |(2/3 − (−1/4))/(1 + (2/3)(−1/4))|.
Step 3 — Numerator: 2/3 + 1/4 = 8/12 + 3/12 = 11/12.
Denominator: 1 − 2/12 = 1 − 1/6 = 5/6.
Step 4 — tan θ = |(11/12)/(5/6)| = |(11/12) × (6/5)| = |11/10| = 11/10.
Step 5 — θ = arctan(11/10) ≈ 47.7°. (Acute angle between the lines.)
Find the equation of the line through the intersection of x + y − 4 = 0 and 2x − y − 1 = 0 that passes through (2, −3).
Step 1 — Family of lines: L₁ + λL₂ = 0 → (x + y − 4) + λ(2x − y − 1) = 0.
Step 2 — Substitute (2, −3): (2 − 3 − 4) + λ(4 + 3 − 1) = 0.
−5 + 6λ = 0 → λ = 5/6.
Step 3 — Required line: (x + y − 4) + (5/6)(2x − y − 1) = 0.
Multiply by 6: 6(x + y − 4) + 5(2x − y − 1) = 0.
Step 4 — Expand: 6x + 6y − 24 + 10x − 5y − 5 = 0 → 16x + y − 29 = 0.
Step 5 — Verify (2, −3): 16(2) + (−3) − 29 = 32 − 3 − 29 = 0. ✓
Common mistakes
| Mistake | Why it happens | Fix |
|---|---|---|
| tan θ = (m₁−m₂)/(1+m₁m₂) without absolute value → negative angle | Formula gives signed value | Angle between lines is always acute (0 < θ ≤ 90°); use absolute value |
| Perpendicular condition: m₁ + m₂ = 0 (wrong) | Mixing up parallel (m₁ = m₂) and perpendicular | Perpendicular: m₁·m₂ = −1 (product, not sum) |
| Family of lines L₁ + λL₂ = 0 excludes L₂ itself | When λ → ∞, the line approaches L₂ but isn't covered | Write the family as μL₁ + λL₂ = 0 to include all lines through the intersection, or check L₂ separately |
| Concurrency determinant: using wrong column order | Mixing up which column is a, b, c | Standard: rows are lines, columns are [x-coefficient, y-coefficient, constant] |
Quick check
- Are the lines 3x − 4y + 5 = 0 and 4x + 3y − 7 = 0 perpendicular?
- Find the angle between y = √3 x + 1 and y = x − 2.
- Check if 3x − 2y + 5 = 0, x + y − 3 = 0, and 2x − y + 1 = 0 are concurrent.
- Find the angle bisectors of x + y = 0 and x − y = 0.
- Stretch: Find the equation of the line through the intersection of 3x + y − 1 = 0 and x − 2y + 3 = 0, and perpendicular to the line x − y + 2 = 0.
Open the Practice tab for graded questions on Straight Lines — Angles.
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Key Takeaways (TL;DR)
- What you'll learn
- Key concepts
- Worked example
- Common mistakes
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