Slope of a line
A road that rises 3 m for every 10 m you walk forward is steep; one that rises 1 m is gentle. Slope is that steepness as a number.
The slope (gradient) of a non-vertical line is m = tan θ, where θ is the angle the line makes with the positive x-axis, measured anticlockwise. Through points (x₁, y₁) and (x₂, y₂): m = (y₂ − y₁)/(x₂ − x₁). A horizontal line has slope 0; a vertical line has no slope (undefined).
Formula: m = (y₂ − y₁)/(x₂ − x₁) = tan θ
In the figure: Moving 4 units right raises the line by 3 units, so slope = 3/4.
Rule 1: Slope from two points — Rise over run; keep the same order in both.
- Easy: A(1, 1), B(5, 4): m = 3/4.
- Medium: P(2, 7), Q(6, 3): m = -1 (falling line).
- Exam-type: C(−1, 2), D(3, 2): m = 0 (horizontal).
Rule 2: Slope and angle — m = tan θ; θ acute means m > 0, θ obtuse means m < 0.
- Easy: θ = 45°: m = 1.
- Medium: θ = 135°: m = −1.
- Exam-type: m = √3: θ = 60°.
Parallel to the x-axis: slope 0. Parallel to the y-axis: slope not defined.
Dividing x-difference by y-difference (run over rise).
Slope = Rise over Run.
Practice questions
- 1Find the slope of the line through (2, 3) and (6, 11).
- 2Find the slope of the line through (1, 5) and (4, 2).
- 3Find the slope of a line making 45° with the positive x-axis.
- 4A line makes 120° with the positive x-axis. Find its slope.
- 5Find the slope of the line through (3, −1) and (3, 6).
- 6A road rises 5 m over a horizontal distance of 40 m. Find its slope.
- 7Find x if the slope of the line through (2, 5) and (x, 9) is 2.
- 8What angle does the line through (0, 0) and (−1, 1) make with the positive x-axis?
Show answers (try them yourself first)
- 12
- 2-1
- 31
- 4tan 120° = −√3
- 5x-coordinates are equal, so slope is undefined (vertical line).
- 61/8
- 7(9 − 5)/(x − 2) = 2, so x = 4
- 8slope −1, θ = 135°
Parallel, perpendicular and collinear
Railway lines never meet because they have the same direction. Two streets crossing at right angles have slopes that are negative reciprocals.
Two non-vertical lines with slopes m₁ and m₂ are parallel if m₁ = m₂ and perpendicular if m₁ m₂ = −1. Three points are collinear if the slopes of AB and BC are equal.
Formula: parallel: m₁ = m₂; perpendicular: m₁m₂ = −1
Rule 1: Parallel or perpendicular — Find both slopes and compare.
- Easy: Slopes 2 and 2: parallel.
- Medium: Slopes 3/4 and −4/3: product −1, perpendicular.
- Exam-type: Line through (0, 0), (2, 4) has slope 2; line through (1, 1), (5, 3) has slope 1/2; product 1, not perpendicular.
Rule 2: Collinear points — Find slope of AB and BC.
- Easy: A(1, 2), B(3, 6), C(5, 10): slopes 2 and 2, collinear.
- Medium: A(0, 0), B(2, 3), C(4, 5): slopes 3/2 and 1, not collinear.
- Exam-type: Find k so that (1, 2), (3, 6), (5, k) are collinear: slope 2, k = 10.
Perpendicular slope: flip the fraction and change the sign.
Using m₁m₂ = 1 for perpendicular lines.
Parallel: Same. Perpendicular: flip and change sign.
Spot the difference
| Parallel | Perpendicular |
|---|---|
| m₁ = m₂ | m₁ × m₂ = −1 |
| Same steepness and direction | Meet at 90° |
Practice questions
- 1Slope of a line is 5. What is the slope of a parallel line?
- 2Slope of a line is 2/3. What is the slope of a perpendicular line?
- 3Are the lines with slopes 4 and −1/4 perpendicular?
- 4Are A(1, 1), B(2, 3), C(3, 5) collinear?
- 5Find k so that A(0, 1), B(2, 5), C(4, k) are collinear.
- 6Show that the line through (0, 0), (3, 4) is perpendicular to the line through (0, 0), (−4, 3).
- 7Show that A(0, 0), B(4, 0), C(4, 3) form a right-angled triangle.
- 8Find the slope of the altitude from A to BC for A(1, 1), B(4, 2), C(2, 6).
Show answers (try them yourself first)
- 15
- 2−3/2
- 3Yes: 4 × (−1/4) = −1
- 4Slopes 2 and 2: yes
- 5slope AB = 2, so k = 9
- 6slopes 4/3 and −3/4; product −1
- 7AB is horizontal (slope 0), BC is vertical, so the angle at B is 90°
- 8slope BC = -2 = −2, so altitude slope = 1/2
Angle between two lines
Scissors have an opening angle. The angle between two lines depends on how different their slopes are.
If two lines have slopes m₁ and m₂, the acute angle θ between them satisfies tan θ = |(m₂ − m₁)/(1 + m₁m₂)|. If 1 + m₁m₂ = 0, the lines are perpendicular (θ = 90°).
Formula: tan θ = |(m₂ − m₁)/(1 + m₁m₂)|
Rule 1: Find the angle — Put m₁ and m₂ in the formula.
- Easy: m₁ = 1, m₂ = 0: tan θ = 1, θ = 45°.
- Medium: m₁ = 1/√3, m₂ = √3: tan θ = |(√3 − 1/√3)/2| = 1/√3, θ = 30°.
- Exam-type: m₁ = 2, m₂ = 3: tan θ = 1/7, θ ≈ 8.13°.
Rule 2: Find a slope — Use the given angle.
- Easy: Lines with slopes m and 0 make 45°: |m| = 1.
- Medium: Line with slope 1/2 and another make 45°: |(m − 1/2)/(1 + m/2)| = 1 gives m = 3 or −1/3.
- Exam-type: Angle 90° requires 1 + m₁m₂ = 0.
The formula gives the acute angle between the lines; the obtuse angle is 180° − θ.
Dropping the modulus and getting a negative angle.
Different over One-plus-product.
Practice questions
- 1Find the angle between lines of slope 0 and 1.
- 2Find the angle between lines of slope 1 and √3.
- 3Find tan θ for lines with slopes 3 and 2.
- 4Find the angle between lines of slope 1/2 and 3.
- 5Two lines of slopes 2 and −1/2: what is the angle?
- 6The angle between the x-axis and a line is 60°. Find the slope.
- 7Find the slope m of a line that makes 45° with the line of slope 2.
- 8Find the angle between lines of slope √3 and 0.
Show answers (try them yourself first)
- 145°
- 2tan θ = (√3 − 1)/(1 + √3) = 0.268, so θ = 15°
- 31/7
- 4tan θ = |(3 − 1/2)/(1 + 3/2)| = 1, θ = 45°
- 51 + m₁m₂ = 0, so 90°
- 6tan 60° = √3
- 7|(m − 2)/(1 + 2m)| = 1 gives m = −3 or m = 1/3
- 860°