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3.01 Gradient formula

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.

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

rise = $y_2-y_1$y2y1

run = $x_2-x_1$x2x1

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

Gradient formula

 

For any two points $\left(x_1,y_1\right)$(x1,y1) and $\left(x_2,y_2\right)$(x2,y2)

We can find the gradient using the formula:

 $\frac{y_2-y_1}{x_2-x_1}$y2y1x2x1

 

 

 

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.

 

 

Practice questions

Question 1

Consider the interval between $A$A$\left(-3,4\right)$(3,4) and $B$B$\left(3,16\right)$(3,16).

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  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 2

What is the gradient of any line parallel to the $x$x-axis?

Question 3

What is the gradient of the line going through $A$A and $B$B?

Loading Graph...

Question 4

Consider the line $y=-2x+8$y=2x+8.

  1. Find the $y$y-intercept.

    Enter each line of work as an equation.

  2. Find the $x$x-intercept.

    Enter each line of work as an equation.

  3. Draw a graph of the line.

    Loading Graph...

  4. Find the gradient of the line by substituting the intercepts into the gradient formula.

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