2.02 Gradient and intercept

The gradient of a line is a measure of its 'steepness'. It is represented by a single number and has no units of measurement.

Increasing and decreasing lines

A line is said to be increasing if it slopes upwards, as we move from left to right. Increasing lines always have a positive gradient. 

A decreasing line slopes downwards as we move from left to right. Decreasing lines always have a negative gradient.

The following applet allows us to see how the value of the gradient changes with the steepness of the line. Notice that increasing lines have a positive gradient and decreasing lines have a negative gradient.

Created with Geogebra

 

Finding the gradient of a line

The gradient of a line is defined as the vertical change in the $y$y-coordinates ('rise') of two points on the line, divided by the horizontal change in the corresponding $x$x-coordinates ('run').
 

 

Worked example

Example 1

Find the gradient of the line below:

Solution

We begin by choosing any two points on the line and use them to create a right-angled triangle, where the line itself forms the hypotenuse of the triangle. 

In this case we have chosen the points $\left(-1,0\right)$(−1,0) and $\left(0,2\right)$(0,2). The 'run' (highlighted red) and the 'rise' (highlighted blue) form the sides of the right-angled triangle.

If we start at the left most point, we see that the run is $1$1 and the rise is $2$2. Both values are positive because we move first to the right $1$1 unit, then up $2$2 units. We calculate the gradient as follows:

$\text{Gradient }$Gradient	$=$=	$\frac{\text{rise }}{\text{run }}$rise run ​
	$=$=	$\frac{2}{1}$21​
	$=$=	$2$2

 

The applet below shows that the gradient of a line is not actually affected by the location of the points used to calculate the gradient. We can also see how the value of the gradient changes for lines of different steepness and whether the line is increasing or decreasing.

Created with Geogebra

 

Finding the gradient between two points

If we are given the coordinates of two points on the coordinate plane, we can find the gradient of the line that would pass through these points.

 

Worked example

Example 2

Find the gradient of the line between the points $\left(3,6\right)$(3,6) and $\left(7,-2\right)$(7,−2).

Solution

It is good practice to first draw a sketch of the two points. A sketch means the location of the points doesn't have to be exact. As long as the points are in the correct quadrant and correctly positioned relative to each other.

We can then add a right-angled triangle that shows the 'rise' and 'run'.  This allows us to see immediately whether the line is increasing or decreasing, and if it has a positive or negative gradient. 

The run is the horizontal distance between the points. We can see that the left-most point has an $x$x-coordinate of $3$3 and the right-most point has an $x$x-coordinate of $7$7, so the horizontal distance between them is $4$4 units. 

The rise is the vertical distance between the points. We can see that one point is $6$6 units above the horizontal axis and the other point is $2$2 units below, so the vertical distance between them is $8$8 units.

If we start at the left-most point, we move $4$4 units to the right and then $8$8 units down to reach the right-most point. This gives us a run of $4$4 and a rise of $-8$−8.

We can now calculate the gradient:

$\text{Gradient }$Gradient	$=$=	$\frac{\text{rise }}{\text{run }}$rise run ​
	$=$=	$\frac{-8}{4}$−84​
	$=$=	$-2$−2

 

Another way to find the gradient, without drawing a sketch, is to consider the changes in the $y$y-coordinates and $x$x-coordinates of the two points. 

$\text{Gradient }$Gradient	$=$=	$\frac{\text{change in }y\text{-coordinates}}{\text{change in }x\text{-coordinates}}$change in y-coordinateschange in x-coordinates​
	$=$=	$\frac{-2-6}{7-3}$−2−67−3​
	$=$=	$\frac{-8}{4}$−84​
	$=$=	$-2$−2

 

Notice that we always subtract the coordinates of the left-most point from the coordinates of the right-most point (the left-most point will have the lowest value for the $x$x-coordinate).

 

Worked example

Example 1

1. Find the gradient of the line represented by the following table of values.

   $x$x	$-3$−3	$0$0	$3$3	$6$6
   $y$y	$-2$−2	$-3$−3	$-4$−4	$-5$−5

2. Is the line increasing or decreasing?

 

Solution

1. We can choose any two points from the table to work out the gradient.

   Using $\left(-3,-2\right)$(−3,−2) and $\left(3,-4\right)$(3,−4), for example, and subtracting the coordinates of the right-most point from the coordinates of the left-most point:

   $\text{Gradient }$Gradient	$=$=	$\frac{\text{change in }y\text{-coordinates}}{\text{change in }x\text{-coordinates}}$change in y-coordinateschange in x-coordinates​
   	$=$=	$\frac{-4-\left(-2\right)}{3-\left(-3\right)}$−4−(−2)3−(−3)​
   	$=$=	$\frac{-2}{6}$−26​
   	$=$=	$-\frac{1}{3}$−13​

   Note: It doesn't matter which two points we choose. As long as our calculations are correct, the gradient will be the same.

2. Because the gradient is negative, the line is decreasing.

 

Gradient of horizontal and vertical lines

A horizontal line has a 'run' but no 'rise', therefore:

$\text{Gradient of horizontal line }$Gradient of horizontal line	$=$=	$\frac{\text{rise }}{\text{run }}$rise run ​
	$=$=	$\frac{0}{\text{run }}$0run ​
	$=$=	$0$0

 

A vertical line has a 'rise' but no 'run', therefore:

$\text{Gradient of vertical line }$Gradient of vertical line	$=$=	$\frac{\text{rise }}{\text{run }}$rise run ​
	$=$=	$\frac{\text{rise }}{0}$rise 0​
	$=$=	$\text{undefined }$undefined

 

Remember that division by zero is mathematically 'undefined'.

 

Practice questions

Question 1

Question 2

Question 3

 

Intercepts

Lines drawn on the $xy$xy-plane, extend forever in both directions.  If we ignore the special case of horizontal and vertical lines, all other lines will either cross both the $x$x-axis and the $y$y-axis or they will pass through the origin, $\left(0,0\right)$(0,0).

Here are some examples:

We use the word intercept to refer to the point where the line crosses or intercepts with an axis.

 

• The $x$x-intercept is the point where the line crosses the $x$x-axis. The coordinates of the $x$x-intercept will always have a $y$y-coordinate of zero.

• The $y$y-intercept is the point where the line crosses the $y$y-axis. The coordinates of the $y$y-intercept will always have an $x$x-coordinate of zero.

Note: Every line must have at least one intercept but cannot have any more than two intercepts.

 

Worked example

Find the $y$y-intercept for the straight line below:

The $y$y-intercept is $-6$−6, and the coordinates of the $y$y-intercept are $\left(0,-6\right)$(0,−6).

 

Practice Questions

question 4

Question 5

question 6