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).
- PAF CAE maths items on lines are typically 30-second recognition or one-step computation—memorize slope, intercept, and distance formulas cold.
Why This Matters for the PAF CAE Academic Paper
Analytic geometry of the straight line sits near the centre of FSc Pre-Engineering maths and appears regularly on Pakistan Air Force College of Aeronautical Engineering (CAE) and related engineering-branch initial academic papers. Coaching and candidate reports often describe a computer-based maths block of roughly 50 MCQs in about 25 minutes (confirm your cycle’s slip on joinpaf.gov.pk). That pace leaves little room for long algebraic rearrangements—you need instant recall of slope, intercept form, parallel/perpendicular tests, and the distance formula.
In an aeronautical-engineering pathway, lines model constant-rate change: climb gradients on charts, linearised force–displacement approximations, and simple kinematic plots. The exam itself stays at textbook recognition level, but every formula below is fair game as a one-step MCQ.
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; PAF items often give two pieces of data and ask which equation matches.
| 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.
Exam Traps and Speed 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) |
Master these five rows and you can clear almost every straight-line MCQ that appears at FSc recognition depth on the CAE academic paper.
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?