Straight Lines
The exam tests slope, the several forms of a line's equation, angle and perpendicularity conditions, and perpendicular distance from a point to a line.
Slope and forms
m = (y₂−y₁)/(x₂−x₁). For ax + by + c = 0, m = −a/b (b ≠ 0). Standard forms:
- Slope-intercept: y = mx + c
- Point-slope: y − y₁ = m(x − x₁)
- Two-point: (y − y₁)/(x − x₁) = (y₂ − y₁)/(x₂ − x₁)
- Intercept: x/a + y/b = 1
Angles and distances
Angle between two lines: tan θ = |(m₁ − m₂)/(1 + m₁m₂)|. Parallel ⇔ m₁ = m₂; perpendicular ⇔ m₁m₂ = −1. Distance from (x₀, y₀) to ax + by + c = 0 is |ax₀ + by₀ + c|/√(a² + b²). Distance between parallel lines ax+by+c₁=0 and ax+by+c₂=0 is |c₁ − c₂|/√(a²+b²).
Worked example
Find the line through (1,2) perpendicular to 3x + 4y = 7. The given slope is −3/4, so the required slope is 4/3. Then y − 2 = (4/3)(x − 1), i.e. 4x − 3y + 2 = 0. Substituting (1,2) gives 4 − 6 + 2 = 0, confirming the point lies on it.
Traps that cost marks
- Writing the perpendicular slope as the negative reciprocal of the wrong quantity, or using m₁ + m₂ = −1.
- Expecting a slope for a vertical line - it is undefined, so use x = a.
- Applying the distance formula before clearing the line into general form, or dropping the modulus sign.
- Losing the constant term when rearranging between point-slope and general form.