The gradient of a line describes how steep it is.
The diagram below shows two lines, one with a positive gradient and the other with a negative gradient.
The gradient of a line between two points, A and B, is calculated using
| gradient of AB | = |
| = | |
| = |
Worked Examples
Find the gradient of the line shown in the diagram.
Draw a triangle under the line below to show the horizontal and vertical distances.
Here the vertical distance is 10 and the horizontal distance is 6.
| Gradient | = |
| = | |
| = or |
Find the gradient of the line joining the point A with coordinates (2, 4) and the point B with coordinates
| Gradient | = |
| = | |
| = | |
| = 3 |
Find the gradient of the line that joins the points with coordinates (–2, 4) and (4, 1).
The diagram shows the line. It will have a negative gradient because of the way it slopes.
| Gradient | = |
| = | |
| = | |
| = |
So the gradient is .
Parallel lines have the same gradient.
The coordinates of the points A, B, C and D are listed below.
A (2, 4) B (8, 7) C (–1, –5) D (5, –2)
Show that the line segments AB and CD are parallel.
Show meGradient of AB = = =
Gradient of CD = = =
The line segments have the same gradients and so must be parallel.
Are the line segments AC and BD parallel?
Show meGradient of AC = = = 3
Gradient of BD = = = 3
The line segments have the same gradient and so are parallel.
The results in Worked Example 4 mean that the quadrilateral ABDC is, in fact, a parallelogram.
Exercises
