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Grade 9

6.03 Slope from two points

Lesson

Slope formula

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.

Slope from two points

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 rise is the difference in the $y$y values.  The difference in the $y$y values is $-3-4=-7$34=7 
  •  The run is the difference in the $x$x values.  The difference in the $x$x values is $23-\left(-12\right)=23+12$23(12)=23+12 = $35$35.  

 

Careful!

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:

  1. Sketch the two points to see whether the slope should be positive or negative.
  2. Find the rise by finding the difference between the $y$y-coordinates.
  3. Find the run by finding the difference between the $x$x-coordinates.
  4. Divide the rise by the run to calculate the slope.
  5. Check that the sign of the slope is as expected.

 

Remember!

Description of slope:  $Slope=\frac{\text{rise }}{\text{run }}$Slope=rise run

Slope of Vertical Line is undefined

Slope of Horizontal Line = $0$0

 

 

Practice questions

Question 1

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.

 

 

Slopes of horizontal and vertical lines

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:

Remember!

$\text{Slope }=\frac{rise}{run}$Slope =riserun

 

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. 

$A=\left(-4,4\right)$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.

 

Vertical Lines

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.

 

Did you know?

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.

 

Practice questions

Question 2

Examine the graph attached and answer the following questions.

Loading Graph...
A Cartesian coordinate plane, with both x- and y- axes labeled from -10 to 10, has a horizontal line drawn on it. The line intersects the y-axis at point (0, 4).
  1. What is the slope of the line?

    $4$4

    A

    $0$0

    B

    Undefined

    C
  2. What is the $y$y-value of the $y$y-intercept of the line?

  3. Does this line have an $x$x-intercept?

    No

    A

    Yes

    B

Question 3

Examine the graph attached and answer the following questions.

Loading Graph...

In a coordinate plane, a line parallel to the $y$y-axis is called vertical line. It is a straight line which goes from top to bottom and bottom to top. The graph indicates that the line intercepts the x-axis at $1$1. Dont mention the point of intersection when asked about the graph.
  1. What is the slope of the line?

    $0$0

    A

    $1$1

    B

    undefined

    C
  2. Does this line have a $y$y-intercept?

    Yes

    A

    No

    B
  3. What is the $x$x-intercept of the line?

 

Outcomes

9.C3.1

Compare the shapes of graphs of linear and non-linear relations to describe their rates of change, to make connections to growing and shrinking patterns, and to make predictions.

9.C3.2

Represent linear relations using concrete materials, tables of values, graphs, and equations, and make connections between the various representations to demonstrate an understanding of rates of change and initial values.

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