🎯 Key Points
- Parallel lines: m₁=m₂ (same slope); Perpendicular lines: m₁×m₂=-1 (negative reciprocal) — for general form, parallel needs a₁/a₂=b₁/b₂, perpendicular needs a₁a₂+b₁b₂=0
- Section formula: INTERNAL division uses + signs in both numerator and denominator; EXTERNAL division uses − signs in both — a small sign swap with a big effect on the answer
- Distance from point to line: |ax₀+by₀+c|/√(a²+b²) — note the absolute value, since distance is always non-negative regardless of which side of the line the point is on
- Area of triangle via coordinates: ½|x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)| — if this expression equals 0, the three points are COLLINEAR (no triangle forms)
The slope of a line is the ratio of vertical change (rise) to horizontal change (run) between any two points on it — equivalently, the tangent of the angle θ the line makes with the positive x-axis.
Coordinate Geometry
Also called analytical geometry: uses coordinates to represent and analyze geometric figures algebraically.
Distance, Midpoint, and Section Formula
- Distance: d = √[(x₂-x₁)² + (y₂-y₁)²]
- Midpoint: M = ((x₁+x₂)/2, (y₁+y₂)/2)
- Section formula (internal): P = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n))
- Centroid of triangle: G = ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3)
Straight Lines
- Slope: m = (y₂-y₁)/(x₂-x₁) = tan θ (θ = angle with x-axis)
- Point-slope form: y - y₁ = m(x - x₁)
- Slope-intercept: y = mx + c
- Two-intercept: x/a + y/b = 1
- General: ax + by + c = 0
- Parallel lines: same slope (m₁ = m₂)
- Perpendicular lines: m₁ × m₂ = -1
- Distance from point (x₀,y₀) to line ax+by+c=0: |ax₀+by₀+c|/√(a²+b²)
Circles
- Centre (h,k) radius r: (x-h)² + (y-k)² = r²
- Standard: x² + y² = r² (centre at origin)
- General: x² + y² + 2gx + 2fy + c = 0; centre (-g,-f), radius = √(g²+f²-c)
Conic Sections
- Parabola: y² = 4ax (opens right); x² = 4ay (opens up)
- Ellipse: x²/a² + y²/b² = 1 (a > b > 0)
- Hyperbola: x²/a² - y²/b² = 1
Area of Triangle
With vertices (x₁,y₁), (x₂,y₂), (x₃,y₃):
Area = ½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|
Worked Example: Distance and Section Formula
Let A(2, 3) and B(7, 15). The distance AB = sqrt[(7-2)^2 + (15-3)^2] = sqrt(25 + 144) = sqrt(169) = 13.
If point P divides AB internally in the ratio 2:1, then P = ((2x7 + 1x2)/(2+1), (2x15 + 1x3)/(2+1)) = (16/3, 11). Using A(1,2) and B(7,8) instead with the same ratio 2:1 internally gives P = ((2x7+1x1)/3, (2x8+1x2)/3) = (5, 6).
For external division in the same ratio 2:1 with A(1,2), B(7,8): P = ((2x7 - 1x1)/(2-1), (2x8 - 1x2)/(2-1)) = (13, 14). External division uses a minus sign in both numerator and denominator instead of a plus sign.
Worked Example: Equation of a Line
Find the line through A(2, 3) and B(6, 11).
Slope m = (11 - 3)/(6 - 2) = 8/4 = 2. Using point-slope form: y - 3 = 2(x - 2), which simplifies to y = 2x - 1.
The same line in intercept form: a line through (4, 0) and (0, 3) is x/4 + y/3 = 1 (check: substituting (4,0) gives 1, substituting (0,3) gives 1).
Parallel and Perpendicular Lines
- Two lines with slopes m1 and m2 are parallel if m1 = m2.
- Two lines are perpendicular if m1 x m2 = -1 (one slope is the negative reciprocal of the other).
- Worked example: The line y = 2x - 1 has slope 2, so any line perpendicular to it has slope -1/2. A perpendicular line through (2, 3) is y - 3 = -1/2(x - 2), i.e. x + 2y = 8.
- For lines given in general form a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0: parallel if a1/a2 = b1/b2; perpendicular if a1a2 + b1b2 = 0.
Angle Between Two Lines
If two lines have slopes m1 and m2, the angle θ between them satisfies tan θ = |(m1 - m2)/(1 + m1m2)|. This formula directly reduces to the parallel condition (θ = 0 when m1 = m2) and the perpendicular condition (θ = 90° when 1 + m1m2 = 0, making tan θ undefined).
More Forms of a Line: Two-Point and Normal Form
- Two-point form: the line through (x₁,y₁) and (x₂,y₂) is y - y₁ = [(y₂-y₁)/(x₂-x₁)](x - x₁). This is just the point-slope form with the slope written out as (y₂-y₁)/(x₂-x₁).
- Normal (perpendicular) form: x cos ω + y sin ω = p, where p (> 0) is the length of the perpendicular from the origin to the line and ω is the angle that this perpendicular makes with the positive x-axis. Here the coefficients of x and y satisfy cos²ω + sin²ω = 1.
- Converting general form: to write ax + by + c = 0 in normal form, divide throughout by ±√(a²+b²), choosing the sign that makes the constant term positive.
- Worked example: the line x + y = 4 divided by √2 gives x/√2 + y/√2 = 4/√2, i.e. x cos45° + y sin45° = 2√2, so its perpendicular distance from the origin is p = 2√2 and ω = 45°.
Distance Between Two Parallel Lines
Two parallel lines can be written with identical x and y coefficients as ax + by + c₁ = 0 and ax + by + c₂ = 0. The perpendicular distance between them is:
d = |c₁ - c₂| / √(a²+b²)
Worked example: for 3x + 4y + 7 = 0 and 3x + 4y - 3 = 0, d = |7 - (-3)|/√(9+16) = 10/5 = 2. (If the coefficients of the two lines are not identical, first scale one equation so both have the same a and b before applying the formula.)
Condition for Concurrency of Three Lines
Three lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0, and a₃x + b₃y + c₃ = 0 are concurrent (all pass through one common point) if and only if the determinant of their coefficients vanishes:
|a₁ b₁ c₁ ; a₂ b₂ c₂ ; a₃ b₃ c₃| = 0
Practical method: find the intersection point of any two of the lines, then check whether it satisfies the third line. If it does, the three lines are concurrent.
Foot of Perpendicular and Image of a Point in a Line
For the line ax + by + c = 0 and a point P(x₁,y₁), let k = (ax₁ + by₁ + c)/(a²+b²).
- Foot of perpendicular (the point M on the line nearest to P): M = (x₁ - a·k, y₁ - b·k).
- Image (reflection) of P in the line: P' = (x₁ - 2a·k, y₁ - 2b·k). The line is the perpendicular bisector of segment PP'.
- Worked example: reflect P(1,2) in the line x - y = 0. Here a=1, b=-1, c=0, so k = (1-2)/(1+1) = -1/2. Image = (1 - 2(1)(-1/2), 2 - 2(-1)(-1/2)) = (1+1, 2-1) = (2,1), matching the fact that reflection in y = x simply swaps coordinates.
Angle Bisectors of Two Lines
The two angle bisectors of the lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are given by equating the perpendicular distances of a point on a bisector from both lines:
(a₁x + b₁y + c₁)/√(a₁²+b₁²) = ± (a₂x + b₂y + c₂)/√(a₂²+b₂²)
The two bisectors are always perpendicular to each other. Taking the + sign gives one bisector and the - sign the other; after making both constant terms positive, the + sign gives the bisector of the angle containing the origin.
🚀 JEE Advanced Edge
Using the collinearity-via-area trick to verify or construct points: Since three points are collinear exactly when the triangle-area expression equals zero, this same formula doubles as a fast collinearity TEST — substitute the three points and check if the result is 0, without needing to compute slopes between pairs of points separately (which requires checking pairs match, a slower two-step process).
Family of lines through the intersection of two given lines: Any line passing through the intersection point of L1: a₁x+b₁y+c₁=0 and L2: a₂x+b₂y+c₂=0 can be written as L1 + λL2 = 0 for some parameter λ, WITHOUT actually solving for the intersection point's coordinates first — this technique is the standard JEE shortcut for "find the line through the intersection of these two lines satisfying [extra condition]" problems.
Worked problem: Find the equation of the line through the intersection of x+y-2=0 and 2x-y+1=0, which also passes through the point (1,2). Approach: Family of lines: (x+y-2)+λ(2x-y+1)=0. Substitute (1,2): (1+2-2)+λ(2-2+1)=0 → 1+λ=0 → λ=-1. Substituting back: (x+y-2)-(2x-y+1)=0 → -x+2y-3=0 → x-2y+3=0.