We have looked at finding the slope through creating right triangles and by using the formula:
$\text{Slope }=\frac{rise}{run}$Slope =riserun
Finding the slope using the idea of $\frac{\text{rise }}{\text{run }}$rise run is a really critical skill for our further studies in linear relationships and graphing.
Consider the segment joining point A$\left(1,1\right)$(1,1) and point B$\left(5,5\right)$(5,5). We can read off a variety of information from this diagram.
Remembering that we move from left to right, we can see that the rise is $4$4, and the run is $4$4. So we could use the rule: $\frac{\text{rise }}{\text{run }}$rise run ,
and calculate the slope to be $\frac{4}{4}=1$44=1.
We could also look for how far the line rises, for every $1$1 horizontal unit increase. We can see that this is also $1$1.
If we don't have such a nice grid, where the rise and run are so easy to read off, then this information is a bit more complicated to find.
What if the points A and B were instead something more obscure like A$\left(-12,4\right)$(−12,4) and B$\left(23,-3\right)$(23,−3)?
In this case, good mathematical practice is certainly to draw yourself a quick sketch of where the points are on the plane.
From this sketch you can identify if the slope of the line will be positive or negative. Can you see how it will be negative?
To find the rise and run we could draw a right triangle on our sketch and carry on as we did before, but there is another way to think about it.
The most common error here is when students are not consistent with which point is the first point and which point is the second.
Now that we have the rise and the run we can calculate the slope:
$\frac{\text{rise }}{\text{run }}=\frac{-7}{35}$rise run =−735 = $\frac{-1}{5}$−15 (negative as we suspected from our sketch!)
Procedure for finding the slope from two points:
Description of slope: $Slope=\frac{\text{rise }}{\text{run }}$Slope=rise run
Slope of Vertical Line is undefined
Slope of Horizontal Line = $0$0
Find the slope of the line that passes through Point A $\left(6,5\right)$(6,5) and Point B $\left(13,2\right)$(13,2), using $a=\frac{rise}{run}$a=riserun.
We know that the slope of a line is a measure of its steepness. Straight lines on the $xy$xy-plane can literally be in any direction and pass through any two points.
This means that straight lines can be:
$\text{Slope }=\frac{rise}{run}$Slope =riserun
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.
$A=\left(-4,4\right)$A=(−4,4)
$B=\left(2,4\right)$B=(2,4)
$C=\left(4,4\right)$C=(4,4)
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 slope is calculated by $\frac{\text{rise }}{\text{run }}$rise run and there is no rise (ie. $\text{rise }=0$rise =0), the slope of a horizontal line is always $0$0.
On vertical lines, the $x$x value is always the same for every point on the line.
Let's look at the coordinates for A, B and C on this line.
$A=\left(5,-4\right)$A=(5,−4)
$B=\left(5,-2\right)$B=(5,−2)
$C=\left(5,4\right)$C=(5,4)
All the $x$x-coordinates are the same, $x=5$x=5, regardless of the $y$y value.
The equation of this line is $x=5$x=5.
Vertical lines have no "run" (ie. $\text{run }=0$run =0). l 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 slope of vertical lines is always undefined.
Linear equations can be written in the form $y=ax+b$y=ax+b, where $a$a is the slope. $y=mx+b$y=mx+b and $y=mx+c$y=mx+c are alternatives.
Notice how the equations of horizontal and vertical lines look a bit different.
Examine the graph attached and answer the following questions.
What is the slope of the line?
$4$4
$0$0
Undefined
What is the $y$y-value of the $y$y-intercept of the line?
Does this line have an $x$x-intercept?
No
Yes
Examine the graph attached and answer the following questions.
What is the slope of the line?
$0$0
$1$1
undefined
Does this line have a $y$y-intercept?
Yes
No
What is the $x$x-intercept of the line?