Straight Lines

🎯 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)

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.

Slope of a Line

• The slope (gradient) m of a line measures its steepness and is defined as the tangent of the angle theta the line makes with the positive x-axis, m = tan(theta).
• For two points (x₁, y₁) and (x₂, y₂) on the line, m = (y₂ - y₁) / (x₂ - x₁), provided x₁ is not equal to x₂.
• A line parallel to the x-axis has slope 0; a line parallel to the y-axis has an undefined slope (vertical line).
• A positive slope rises left to right; a negative slope falls left to right.
• Two non-vertical lines are parallel when their slopes are equal, and perpendicular when the product of their slopes is -1.

Slope-Intercept and Point-Slope Forms

• The slope-intercept form of a line is y = mx + c, where m is the slope and c is the y-intercept (the value of y where the line meets the y-axis).
• The point-slope form, for a line of slope m passing through a known point (x₁, y₁), is y - y₁ = m(x - x₁).
• Slope-intercept form is ideal when the slope and where it cuts the y-axis are known; point-slope form is ideal when the slope and one point are known.
• Setting x = 0 in y = mx + c immediately gives the y-intercept c.
• Example: a line of slope 2 through (1, 3) is y - 3 = 2(x - 1), i.e. y = 2x + 1.

Intercept Form of a Line

• The intercept form of a line is x/a + y/b = 1, where a is the x-intercept and b is the y-intercept.
• Here the line crosses the x-axis at (a, 0) and the y-axis at (0, b), so both intercepts are read off directly.
• This form requires the line to pass through neither axis at the origin, since a and b must be non-zero.
• It is convenient for problems that specify where a line cuts the two axes, or for finding the area of the triangle a line forms with the axes, which is (1/2)|a b|.
• Example: the line x/3 + y/4 = 1 meets the axes at (3, 0) and (0, 4).

Distance of a Point from a Line

• The perpendicular distance from a point (x₁, y₁) to the line Ax + By + C = 0 is d = |A x₁ + B y₁ + C| / sqrt(A² + B²).
• The absolute value in the numerator ensures the distance is non-negative regardless of which side of the line the point lies on.
• If the numerator is zero, the point lies on the line, giving distance 0.
• The sign of A x₁ + B y₁ + C (before taking modulus) tells which side of the line the point is on, useful for region problems.
• Example: distance from (0, 0) to 3x + 4y - 10 = 0 is |-10| / sqrt(9 + 16) = 10/5 = 2.

General Equation of a Line and Its Reductions

• Every straight line can be written in the general form Ax + By + C = 0, where A and B are not both zero.
• Its slope is m = -A/B (when B is not 0), its x-intercept is -C/A, and its y-intercept is -C/B.
• It reduces to slope-intercept form y = (-A/B)x + (-C/B) by solving for y.
• It reduces to intercept form x/(-C/A) + y/(-C/B) = 1 when C is not zero.
• It reduces to normal form x cos(omega) + y sin(omega) = p by dividing through by sqrt(A² + B²), giving the perpendicular distance p from the origin.