11.14 Gradient

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=−2−6=−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=7−3=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.

Let's just remind ourselves how we calculated the rise and run again.

rise = $y_2-y_1$y2​−y1​

run = $x_2-x_1$x2​−x1​

This means we can generate a new rule for finding the gradient if we are given two points.

 

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.

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.

 

Question 1

Question 2

Question 3

Question 4