Week 12 · Slope of a straight line: inclination angle, parallel lines, perpendicular lines and collinear points · 3 lessons
1
The inclination angle and the slope of a line: m = tan θ and m = (y₂ − y₁) ÷ (x₂ − x₁)
Mathematics Prep 3 Term 1, Unit 3, Lesson 6
GoalFind the slope of a line from two points and from its inclination angle.
New words
slope · how steep a line is: rise ÷ run, written minclination angle · the angle θ between the line and the positive x-axis, measured anticlockwiserise and run · change in y and change in x between two pointsvertical line · a line x = c, its slope is not defined
Explanation
The slope m tells how steep a line is. Take two points (x₁, y₁) and (x₂, y₂) on it. The rise is y₂ − y₁, the run is x₂ − x₁, and m = (y₂ − y₁) ÷ (x₂ − x₁). It does not matter which point comes first, as long as you use the same order on top and bottom. The same slope is also m = tan θ, where θ is the inclination angle, measured from the positive x-axis up to the line. A line rising to the right has m > 0 and θ < 90°; a line falling to the right has m < 0 and θ is between 90° and 180°, found as 180° + tan⁻¹ m (when the calculator gives a negative angle). A horizontal line has m = 0 (θ = 0°). A vertical line has no slope.
Worked examples
The line through A(1, 1) and B(4, 7): m = (7 − 1) ÷ (4 − 1) = 6/3 = 2. Its angle: θ = tan⁻¹ 2 ≈ 63.43°.
The line through (−1, 5) and (3, 1): m = (1 − 5) ÷ (3 − (−1)) = −4/4 = −1. The calculator gives tan⁻¹(−1) = −45°, so θ = 180° − 45° = 135°.
A line has inclination 45°: m = tan 45° = 1. A line has inclination 60°: m = tan 60° = √3.
Class activity
Staircase. Measure the height of one step and its depth in cm. The slope of the staircase is height ÷ depth. Find the angle with tan⁻¹.
Practice
1
Find the slope of the line through (2, 3) and (6, 11).
(11 − 3) ÷ (6 − 2) = 8/4 = 2.
2
Find the slope through (−2, 4) and (1, −5).
(−5 − 4) ÷ (1 + 2) = −9/3 = −3.
3
A line has inclination 45°. Write its slope. And the slope of a horizontal line?
tan 45° = 1; horizontal: 0.
4
A road rises 3 m over a horizontal distance of 4 m. Find its slope and its angle to the nearest degree.
m = 3/4, θ = tan⁻¹(0.75) ≈ 37°.
5
Why has the vertical line x = 3 no slope?
Two points on it have the same x, so the run is 0 and we cannot divide by 0.