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Stage 5.1-2

3.02 The gradient of an interval

Lesson

Increasing and decreasing

Some lines have increasing slopes, like these:

And some have decreasing slopes, like these:

This applet will let you explore lines with positive and negative gradients:

 

Gradient

The slope of a line is a measure of how steep it is. In mathematics we call this the gradient.

A gradient is a single value that describes:

  • if a line is increasing (has positive gradient)
  • if a line is decreasing (has negative gradient)
  • how far up or down the line moves (how the $y$y-value changes) with every step to the right (for every $1$1 unit increase in the $x$x-value)

Take a look at this line, where the horizontal and vertical steps are highlighted:

We call the horizontal measurement the run and the vertical measurement the rise. For this line, a run of $1$1 means a rise of $2$2, so the line has gradient $2$2.

Sometimes it is difficult to measure how far the line goes up or down (how much the $y$y-value changes) in $1$1 horizontal unit, especially if the line doesn't line up with the grid points on the $xy$xy-plane. In this case we calculate the gradient by using a formula:

$\text{gradient }=\frac{\text{rise }}{\text{run }}$gradient =rise run

The rise and run can be calculated from using any two points on the line.

 

Finding the gradient from a graph

You can find the rise and run of a line by drawing a right triangle created by any two points on the line. The line itself forms the hypotenuse.

This line has a gradient of $\frac{\text{rise }}{\text{run }}=\frac{4}{3}$rise run =43

In this case, the gradient is positive because, over the $3$3 unit increase in the $x$x-values, the $y$y-value has increased. If the $y$y-value decreased as the $x$x-value increases, the gradient would be negative.

This applet allows you to see the rise and run between two points on a line of your choosing:

 

Finding the gradient from a pair of coordinates

If you have a pair of coordinates, such as $A$A$\left(3,6\right)$(3,6) and $B$B$\left(7,-2\right)$(7,2), we can find the gradient of the line between these points using the same formula. It is a good idea to draw a quick sketch of the points, which helps us quickly identify what the line will look like:

Already we can tell that the gradient will be negative, since the line moves downward as we go from left to right.

The rise is the difference in the $y$y-values of the points. We take the $y$y-value of the rightmost point and subtract the $y$y-value of the leftmost point to describe the change in vertical distance from $A$A to $B$B:

$\text{rise}=-2-6=-8$rise=26=8

The run is the difference in the $x$x-values of the points. We take the $x$x-value of the rightmost point and subtract the $x$x-value of the leftmost point to describe the change in horizontal distance from $A$A to $B$B:

$\text{run}=7-3=4$run=73=4

Notice that we subtracted the $x$x-values and the $y$y-values in the same order - we check our sketch, and it does seem sensible that between $A$A and $B$B there is a rise of $-8$8 and a run of $4$4. We can now put these values into our formula to find the gradient:

$\text{gradient }$gradient $=$= $\frac{\text{rise }}{\text{run }}$rise run
  $=$= $\frac{-8}{4}$84
  $=$= $-2$2

We have a negative gradient, as we suspected. Now we know that when we travel along this line a step of $1$1 in the $x$x-direction means a step of $2$2 down in the $y$y-direction.

Gradient formula

 

For any line: $\text{gradient }=\frac{\text{rise }}{\text{run }}$gradient =rise run

To calculate the rise from two points, take the difference of the $y$y-values (subtract left from right).
To calculate the run from two points, take the difference of the $x$x-values (subtract left from right).

 

 

 

 

Gradient of horizontal and vertical lines

Horizontal Lines

On horizontal lines, the $y$y-value is always the same for every point on the line. In other words, there is no rise- it's completely flat.

Let's look at the coordinates for $A$A, $B$B and $C$C on this line.

$A$A$\left(-8,4\right)$(8,4)

$B$B$\left(-2,4\right)$(2,4)

$C$C$\left(7,4\right)$(7,4)

All the $y$y-coordinates are the same. Every point on the line has a $y$y-value equal to $4$4, regardless of the $x$x-value.

The equation of this line is $y=4$y=4.

Since gradient is calculated by $\frac{\text{rise }}{\text{run }}$rise run and there is no rise (i.e. $\text{rise }=0$rise =0), the gradient of a horizontal line is always $0$0.

Vertical Lines

On vertical lines, the $x$x-value is always the same for every point on the line.

Now, let's look at the coordinates for $A$A, $B$B and $C$C on this line.

$A$A$\left(-3,8\right)$(3,8)

$B$B$\left(-3,3\right)$(3,3)

$C$C$\left(-3,-3\right)$(3,3)

All the $x$x-coordinates are the same, $x=-3$x=3, regardless of the $y$y-value.

The equation of this line is $x=-3$x=3.

Vertical lines have no "run" (i.e. $\text{run }=0$run =0).

If we substituted this into the $\frac{\text{rise }}{\text{run }}$rise run equation, we'd have a $0$0 as the denominator of the fraction. However, fractions with a denominator of $0$0 are undefined.

So, the gradient of vertical lines is always undefined.

 

 

Question 1

Consider the intervals $AB$AB and $BC$BC.

Loading Graph...

  1. What is the gradient of $AB$AB?

    undefined

    A

    $4$4

    B

    $0$0

    C

    $2$2

    D

    undefined

    A

    $4$4

    B

    $0$0

    C

    $2$2

    D
  2. What is the gradient of $BC$BC?

    $6$6

    A

    undefined

    B

    $0$0

    C

    $5$5

    D

    $6$6

    A

    undefined

    B

    $0$0

    C

    $5$5

    D

Question 2

Consider the interval between $A$A$\left(4,4\right)$(4,4) and $B$B$\left(7,7\right)$(7,7).

Loading Graph...

  1. Find the rise (change in the $y$y value) between point $A$A and $B$B.

  2. Find the run (change in the $x$x value) between point $A$A and $B$B.

  3. Find the gradient of the interval $AB$AB.

Question 3

Consider the line shown with $y$y-intercept at $Y$Y$\left(0,1\right)$(0,1) and $x$x-intercept at $X$X$\left(5,0\right)$(5,0).

Loading Graph...

  1. Find the rise (change in the $y$y value) using the intercepts.

  2. Find the run (change in the $x$x value) using the intercepts.

  3. Find the gradient of the interval $XY$XY.

Question 4

We want to find the gradient of the line passing through the points $A$A$\left(-5,-3\right)$(5,3) and $B$B$\left(-2,12\right)$(2,12).

Loading Graph...
You can use this $xy$xy-plane and the sketchpad to plot the points and the line.
  1. First, find the rise (change in the $y$y value) between points $A$A and $B$B.

  2. Now, find the run (change in the $x$x value) between points $A$A and $B$B.

  3. What is the gradient of the line?

Outcomes

MA5.1-6NA

determines the midpoint, gradient and length of an interval, and graphs linear relationships

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