7.1 Analytic Geometry: Straight Lines
Key Takeaways
- Slope of a non-vertical line through (x₁, y₁) and (x₂, y₂) is m = (y₂ − y₁)/(x₂ − x₁); a vertical line has undefined slope.
- Standard line forms: slope-intercept y = mx + c, point-slope y − y₁ = m(x − x₁), two-point, intercept x/a + y/b = 1, and general ax + by + c = 0.
- Parallel lines share equal slopes; perpendicular lines satisfy m₁m₂ = −1 (neither vertical).
- Distance from point (x₀, y₀) to line ax + by + c = 0 is |ax₀ + by₀ + c|/√(a² + b²).
- Efficient line practice relies on fluent slope, intercept, and distance formulas.
Why Straight-Line Geometry Matters
Analytic geometry of the straight line is central school-level mathematics and useful preparation for later engineering study. Build fluent recall of slope, intercept form, parallel/perpendicular tests, and the distance formula. Current public PAF sources do not establish this as an initial-test domain or publish a mathematics count or timing.
In an aeronautical-engineering pathway, lines model constant-rate change: climb gradients on charts, linearised force–displacement approximations, and simple kinematic plots. The examples stay at textbook recognition level and use the formulas below in short calculations.
Slope of a Straight Line
The slope (gradient) of a non-vertical line through points (x₁, y₁) and (x₂, y₂) is
m = (y₂ − y₁) / (x₂ − x₁).
| Situation | Slope m | Geometric meaning |
|---|---|---|
| Rising left → right | m > 0 | Acute angle with +x-axis |
| Falling left → right | m < 0 | Obtuse angle with +x-axis |
| Horizontal | m = 0 | Parallel to x-axis |
| Vertical | undefined | Parallel to y-axis; x = constant |
If the line makes an angle θ with the positive x-axis, then m = tan θ (where defined).
Worked Example 1 — Slope from two points
Find the slope of the line through A(2, −1) and B(6, 5).
Solution: m = (5 − (−1))/(6 − 2) = 6/4 = 3/2.
Trap: order of points does not matter—(y₁ − y₂)/(x₁ − x₂) gives the same ratio. Mixing numerator and denominator (Δx/Δy) is a common timed-test error.
Forms of the Equation of a Straight Line
Memorise these five forms, then practise choosing the equation that matches two given pieces of data.
| Form | Equation | When to use |
|---|---|---|
| Slope-intercept | y = mx + c | Slope m and y-intercept c known |
| Point-slope | y − y₁ = m(x − x₁) | Slope and one point known |
| Two-point | (y − y₁)/(x − x₁) = (y₂ − y₁)/(x₂ − x₁) | Two distinct points |
| Intercept | x/a + y/b = 1 | x-intercept a and y-intercept b (neither zero) |
| General | ax + by + c = 0 | Any line; a, b not both zero |
Vertical line: x = k. Horizontal line: y = k.
From the general form ax + by + c = 0 (with b ≠ 0), the slope is m = −a/b and the y-intercept is c_y = −c/b.
Worked Example 2 — Point-slope to general form
A line has slope −2 and passes through (3, 1). Write its equation in general form with integer coefficients.
Solution: y − 1 = −2(x − 3) → y − 1 = −2x + 6 → 2x + y − 7 = 0.
Worked Example 3 — Intercept form
A line cuts the axes at (4, 0) and (0, −3). Find its equation.
Solution: x/4 + y/(−3) = 1 → (−3)x + 4y = −12 → 3x − 4y − 12 = 0 (multiply by −1 for a positive leading coefficient if preferred: −3x + 4y + 12 = 0 is equally correct).
Check: x = 4, y = 0 → 12 − 0 − 12 = 0 ✓; x = 0, y = −3 → 0 + 12 − 12 = 0 ✓.
Parallel and Perpendicular Lines
Two non-vertical lines with slopes m₁ and m₂ satisfy:
- Parallel: m₁ = m₂ (and distinct intercepts, otherwise coincident).
- Perpendicular: m₁m₂ = −1.
Special cases: any vertical line is parallel to any other vertical line and perpendicular to every horizontal line.
Worked Example 4 — Perpendicular through a point
Find the equation of the line through (1, −2) perpendicular to 2x − 3y + 5 = 0.
Solution: Slope of given line: m = −(2)/(−3) = 2/3. Perpendicular slope: m⊥ = −3/2.
Point-slope: y + 2 = (−3/2)(x − 1) → 2(y + 2) = −3(x − 1) → 2y + 4 = −3x + 3 → 3x + 2y + 1 = 0.
Distance from a Point to a Line
Distance from point (x₀, y₀) to the line ax + by + c = 0 is
d = |ax₀ + by₀ + c| / √(a² + b²).
Distance between two parallel lines ax + by + c₁ = 0 and ax + by + c₂ = 0 is
d = |c₁ − c₂| / √(a² + b²) (same a, b after scaling).
Worked Example 5 — Point-to-line distance
Find the distance from P(−1, 3) to the line 3x − 4y + 5 = 0.
Solution: d = |3(−1) − 4(3) + 5| / √(9 + 16) = |−3 − 12 + 5|/5 = |−10|/5 = 2.
Worked Example 6 — Parallel-line distance
Find the distance between 2x + y − 4 = 0 and 4x + 2y + 6 = 0.
Solution: Divide the second equation by 2: 2x + y + 3 = 0. Now a = 2, b = 1, c₁ = −4, c₂ = 3.
d = |−4 − 3| / √(4 + 1) = 7/√5 = (7√5)/5 after rationalising.
Angle Between Two Lines
If slopes are m₁ and m₂, the acute/obtuse angles θ between the lines satisfy
tan θ = |(m₁ − m₂)/(1 + m₁m₂)|, provided 1 + m₁m₂ ≠ 0 (that case is perpendicular, θ = 90°).
On timed MCQs you rarely need both angles—usually you only check whether tan θ matches a given option or whether the lines are parallel/perpendicular.
Common Traps and Efficiency Tips
- Sign of c in ax + by + c = 0: write the constant on the same side before plugging into the distance formula.
- Parallel vs identical: equal slopes alone do not prove distinct parallel lines—compare intercepts or check whether one equation is a scalar multiple of the other.
- Undefined slope: never write m = ∞ in an equation; use x = constant.
- Unit consistency: distance answers are lengths; if options show rationals and surds, keep √(a² + b²) exact unless the stem asks for a decimal.
Quick Formula Card
| Goal | Formula |
|---|---|
| Slope | m = (y₂ − y₁)/(x₂ − x₁) |
| Point-slope | y − y₁ = m(x − x₁) |
| Parallel | m₁ = m₂ |
| Perpendicular | m₁m₂ = −1 |
| Point–line distance | |ax₀ + by₀ + c|/√(a² + b²) |
Mastering these five rows provides a compact foundation for straight-line recognition and calculation exercises.
What is the slope of the line through the points (1, 4) and (5, −2)?
A line has slope 1/2 and passes through (−2, 3). Which equation represents the line in general form?
The line 3x + 4y − 7 = 0 is perpendicular to which of the following lines?
What is the distance from the point (2, −1) to the line x − y + 4 = 0?