Straight Lines

🟢 Lite — Quick Review

Rapid summary for last-minute revision before your exam.

Straight Lines — Key Facts for JEE Main Equation forms: Slope-intercept y = mx + c; Point-slope y − y₁ = m(x − x₁); Two-point (y − y₁) = [(y₂ − y₁)/(x₂ − x₁)](x − x₁) General form: Ax + By + C = 0; slope = −A/B (when B ≠ 0) Distance from point (x₁, y₁) to line Ax + By + C = 0: |Ax₁ + By₁ + C|/√(A² + B²) Angle between two lines: tan θ = |m₁ − m₂|/(1 + m₁m₂)| ⚡ Exam tip: JEE Main tests distance of point from line, angle between lines, and family of lines — all are scoring topics!

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🟡 Standard — Core Study

Standard content for students with a few days to months.

Straight Lines — JEE Main Study Guide

Slope-intercept form: y = mx + c, where m = slope, c = y-intercept Point-slope form: y − y₁ = m(x − x₁) for line through (x₁, y₁) with slope m Two-point form: (y − y₁)/(y₂ − y₁) = (x − x₁)/(x₂ − x₁) for line through (x₁, y₁) and (x₂, y₂) Intercept form: x/a + y/b = 1 where a = x-intercept, b = y-intercept Normal form: x cos α + y sin α = p where p is perpendicular distance from origin, α is angle normal makes with x-axis

Slope of line through (x₁, y₁), (x₂, y₂): m = (y₂ − y₁)/(x₂ − x₁)

Angle between lines: For slopes m₁, m₂: tan θ = |(m₂ − m₁)/(1 + m₁m₂)|

• Parallel: m₁ = m₂; Perpendicular: m₁m₂ = −1
• If one line is vertical (x = constant): angle with horizontal = 90°

Distance formulas:

• Distance between points (x₁, y₁), (x₂, y₂) = √[(x₂−x₁)² + (y₂−y₁)²]
• Perpendicular distance from point (x₁, y₁) to line Ax + By + C = 0: |Ax₁ + By₁ + C|/√(A² + B²)
• Distance between two parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0: |C₁ − C₂|/√(A² + B²)

Section formula: Internal division of line joining A(x₁, y₁) and B(x₂, y₂) by point P in ratio m:n: P = [(mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n)] For midpoint: ((x₁+x₂)/2, (y₁+y₂)/2)

Area of triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃): Area = ½|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)| Area = 0 if points are collinear

Family of lines:

• Through intersection of L₁ = 0 and L₂ = 0: L₁ + λL₂ = 0
• Through point (x₁, y₁): y − y₁ = m(x − x₁) with any slope m
• Parallel to given line: Ax + By + k = 0 (same normal vector)
• Perpendicular to given line: Bx − Ay + k = 0

Concurrent lines: Three lines are concurrent (pass through common point) if determinant of coefficients = 0

• Key formula: Distance point to line: |Ax₁ + By₁ + C|/√(A² + B²); Section formula: P = [(mx₂+nx₁)/(m+n), …]
• Common trap: For perpendicular lines: m₁m₂ = −1 only when neither is vertical; if one is vertical, the other is horizontal
• Exam weight: 1 question per year (4 marks); frequently combined with circle or conic problems

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🔴 Extended — Deep Dive

Comprehensive coverage for students on a longer study timeline.

Straight Lines — Comprehensive JEE Main Notes

Shift of origin: When origin shifts to (h, k), a point P(x, y) in old coordinates has new coordinates (X, Y) where X = x − h, Y = y − k Useful for simplifying equations

Rotation of axes: For rotation by angle θ: x = X cos θ − Y sin θ; y = X sin θ + Y cos θ For removing xy term from second degree equations

Pair of straight lines through origin: ax² + 2hxy + by² = 0 represents two lines through origin if Δ = ab − h² > 0 Slopes given by m²a + 2hm + b = 0 Lines are at angle θ where tan θ = 2√(h²−ab)/(a+b)

Homogeneous pair of lines: ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents pair of lines if it factors Condition for pair of lines: abc + 2fgh − af² − bg² − ch² = 0

Bisectors of angles between lines: For lines L₁ = 0 and L₂ = 0, bisectors are given by (L₁)/√(A₁²+B₁²) = ± (L₂)/√(A₂²+B₂²)

• • gives bisector of angle containing origin (if constant terms positive) or acute angle bisector
• − gives the other bisector

Foot of perpendicular from point to line: From (x₁, y₁) to Ax + By + C = 0: x = (B²x₁ − AB y₁ − AC)/D; y = (A²y₁ − AB x₁ − BC)/D where D = A² + B²

Reflection of point across line: Find foot of perpendicular (F), then reflected point P’ is such that F is midpoint of PP’

Joint equation of pair of lines: For two lines through origin with slopes m₁, m₂: (y − m₁x)(y − m₂x) = 0 → y² − (m₁+m₂)xy + m₁m₂x² = 0

Angle bisector as locus: Locus of points with ratio of distances from two lines = constant k: |Ax + By + C|/√(A²+B²) = k·|A’x + B’y + C’|/√(A’²+B’²)

Properties of centroid, orthocenter: For triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃):

• Centroid: ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3)
• Area using determinant: ½|Σ(x_i(y_{i+1} − y_{i−1}))|

Incenter: Point equidistant from all sides; found using weighted average of vertices weighted by side lengths

Locus problems:

1. Write condition as equation in variables (x, y)
2. Eliminate parameter to get equation in x, y only

Line as linear combination: Line L: 2x + 3y = 5 can be written as L ≡ 0 Two lines intersect at point solving their equations simultaneously

Special points in triangle:

• Circumcenter: intersection of perpendicular bisectors

• Centroid: intersection of medians (divides each median in 2:1 ratio from vertex)

• Orthocenter: intersection of altitudes

• Remember: Distance point to line = |Ax₁ + By₁ + C|/√(A²+B²); parallel: m₁ = m₂; perpendicular: m₁m₂ = −1; section formula for ratio m:n internal division

• Previous years: “Find distance of point (3,4) from line 3x + 4y + 5 = 0” [2023]; “Find equation of line through (1,2) making 45° with x + y = 3” [2024]; “Show that lines 2x + 3y = 5 and 4x + 6y = 9 are parallel, find distance” [2024]

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📊 JEE Main Exam Essentials

Detail	Value
Questions	90 (30 per subject)
Time	3 hours
Marks	300 (90 per subject)
Section	Physics (30), Chemistry (30), Mathematics (30)
Negative	−1 for wrong answer
Mode	Computer-based

🎯 High-Yield Topics for JEE Main Mathematics

• Calculus (Differentiation + Integration) — ~35 marks combined
• Coordinate Geometry (straight lines, circles, conics) — ~20 marks
• Algebra (Complex Numbers, Quadratics, P&C, Probability) — ~25 marks
• Trigonometry + Inverse Trigonometry — ~15 marks
• Vector + 3D — ~15 marks

📝 Previous Year Question Patterns

• Straight Lines: 1 question per year, 4 marks
• Common patterns: distance point to line, angle between lines, locus, family of lines
• Weight: medium frequency, fundamental for coordinate geometry

💡 Pro Tips

• Always convert to standard form before finding distance or angle
• For family of lines problems, use L₁ + λL₂ = 0 for lines through intersection of L₁ and L₂
• Bisector formulas have two signs — pick the correct one based on which side of the line the origin lies
• For locus problems, eliminate the parameter to get equation purely in x and y
• Perpendicular distance formula works for any form — convert to Ax + By + C = 0 first
• When finding angle between two lines, check if lines are defined by their slopes or by their equations

🔗 Official Resources

• NTA Official JEE Main
• JEE Main Syllabus PDF

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