Some lines have increasing slopes, like these:
And some have decreasing slopes, like these:
This applet will let you create lines with positive and negative gradients:
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:
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 are calculated from two known points on the line.
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:
If you have a pair of coordinates, such as $A=\left(3,6\right)$A=(3,6) and $B=\left(7,-2\right)$B=(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.
Horizontal lines have no rise value. The $\text{rise }=0$rise =0.
So the gradient of a horizontal line is $\text{Gradient }=\frac{\text{rise }}{\text{run }}$Gradient =rise run $=$=$\frac{0}{\text{run}}$0run$=$=$0$0.
Vertical lines have no run value. The $\text{run }=0$run =0.
So the gradient of a vertical line is $\text{gradient }=\frac{\text{rise }}{\text{run }}$gradient =rise run $=$=$\frac{\text{rise }}{0}$rise 0. Division by $0$0 results in the value being undefined.
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)
The gradient of a horizontal line is $0$0. The gradient of vertical line is undefined.
What is the gradient of the line shown in the graph, given that Point A $\left(3,3\right)$(3,3) and Point B $\left(6,5\right)$(6,5) both lie on the line?
What is the gradient of the line going through A $\left(-1,1\right)$(−1,1) and B $\left(5,2\right)$(5,2)?
The gradient of interval AB is $3$3. A is the point ($-2$−2, $4$4), and B lies on $x=3$x=3. What is the $y$y-coordinate of point B, denoted by $k$k?