# 1.04 Review: Slopes of lines

Lesson

## Slope as a rate of change

The slope is the rate of change over an interval or the ratio of vertical change to horizontal change.

$\text{slope }=\frac{\text{vertical change }}{\text{horizontal change }}$slope =vertical change horizontal change

We call the horizontal measurement the run and the vertical measurement the rise.

$slope=\frac{\text{rise }}{\text{run }}$slope=rise run

To find the slope of a line the ordered pairs, $\left(x_1,y_1\right)$(x1,y1) and $\left(x_2,y_2\right)$(x2,y2), indication the interval of change. Now, run corresponds to the $x$x-axis and rise corresponds to the $y$y-axis.

$slope=\frac{y_2-y_1}{x_2-x_1}$slope=y2y1x2x1

#### Worked examples

##### Question 1

Determine the slope of the line given:

Think: Sometimes it is difficult to measure how far the line goes up or down (how much the $y$y value changes) in 1 horizontal unit. In this case we calculate the slope by using a formula.

$\text{slope }=\frac{\text{rise }}{\text{run }}$slope =rise run

Where you take any two points whose coordinates are known or can be easily found, and look for the rise and run between them.

Do: Substitute the values $3=run$3=run and $4=rise$4=rise into the slope formula.

$\text{slope }=\frac{4}{3}$slope =43

Reflect: The slope can be written as an improper fraction. Why is this the case? Why are both the $run$run and the $rise$rise positive and not negative?

### Increasing and decreasing

Since the slope measures the rate of change, we know that change can increase, decrease, or have no change over time.

Below are a few line graphs. What do they all have in common when you look at them from left to right?

The lines are all increasing and part of the graphs lay in quadrant $I$I  and quadrant $III$III. The slopes of the lines are positive.

Below are a few line graphs. What do they all have in common when you look at them from left to right?

The lines are all decreasing and part of the graphs lay in quadrant $II$II  and quadrant $IV$IV. The slopes of the lines are negative.

#### Exploration

Use the applet below to explore positive and negative slopes. Use the sliders to change the value of the $slope$slope, indicated by $m$m

If given a graph of a line, we can find the rise and run by drawing a right triangle created by any two points on the line. The line itself becomes the hypotenuse.

Use the applet below to explore how changing the rise and run impacts the slope and the graph. Use the $m$m slider to change the slope and use the endpoints to adjust the rise and run.

#### Worked examples

##### Question 2

Find the slope of the line that passes through $\left(3,6\right)$(3,6) and $\left(7,-2\right)$(7,2)

Think: It is good mathematical practice to draw a quick sketch of the points.  This helps to quickly identify what the line looks like (is it increasing or decreasing).

The rise is the difference in the $y$y values of the points. The run is the difference in the $x$x values of the points.

Do:

 $\text{slope }$slope $=$= $\frac{\text{rise }}{\text{run }}$rise run ​ $=$= $\frac{y_2-y_1}{x_2-x_1}$y2​−y1​x2​−x1​​ $=$= $\frac{-2-6}{7-3}$−2−67−3​ $=$= $\frac{-8}{4}$−84​ $=$= $-2$−2

Notice, we subtracted the $x$x values and the $y$y values in the same order.

Reflect: Would it answer change if we using $\text{slope }=\frac{y_1-y_2}{x_1-x_2}$slope =y1y2x1x2? Try it.

##### Question 3

The slope of the line is $\frac{-1}{5}$15 . The line passes through the point $M$M at $\left(-12,4\right)$(12,4) and point $E$E. $E$E has a $y$y -coordinate of $y=-3$y=3 . What is the $x$x -coordinate of point $E$E, denoted by $d$d

Think: The rise is the difference in the $y$y values of the points. The run is the difference in the $x$x values of the points.

$\text{slope }=\frac{-1}{5}$slope =15

$M$M $\left(x_1,y_1\right)=\left(-12,4\right)$(x1,y1)=(12,4)

$E$E $\left(x_2,y_2\right)=\left(d,-3\right)$(x2,y2)=(d,3)

Do: Use substitution to replace the variables in the slope formula to find the value of $d$d.

 $\text{slope }$slope $=$= $\frac{\text{rise }}{\text{run }}$rise run ​ $\frac{-1}{5}$−15​ $=$= $\frac{y_2-y_1}{x_2-x_1}$y2​−y1​x2​−x1​​ $\frac{-1}{5}$−15​ $=$= $\frac{-3-4}{d-\left(-12\right)}$−3−4d−(−12)​ $\frac{-1}{5}$−15​ $=$= $\frac{-7}{d+12}$−7d+12​ $\frac{-1}{5}\left(d+12\right)$−15​(d+12) $=$= $\frac{-7}{d+12}\left(d+12\right)$−7d+12​(d+12) $\frac{-1}{5}\left(d+12\right)$−15​(d+12) $=$= $-7$−7 $d+12$d+12 $=$= $-7\left(\frac{-5}{1}\right)$−7(−51​) $d+12$d+12 $=$= $35$35 $d$d $=$= $35-12$35−12 $d$d $=$= $23$23

The answer is $d=23$d=23

Careful!

When finding the slope, the most common error is when students are not consistent with which point is the first point and which point is the second.

$slope=\frac{y_2-y_1}{x_2-x_1}$slope=y2y1x2x1

$slope=\frac{y_1-y_2}{x_1-x_2}$slope=y1y2x1x2

## Slope 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.

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

Since the slope is calculated by $\frac{\text{rise }}{\text{run }}$rise run  or $\frac{y_2-y_1}{x_2-x_1}$y2y1x2x1 and there is no rise (i.e., $\text{rise }=0$rise =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 $A$A, $B$B and $C$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.

Vertical lines have no "run" (ie. $\text{run }=0$run =0). If we substituted this into the slope formula, $\frac{y_2-y_1}{x_2-x_1}$y2y1x2x1, the fraction will have a $0$0 as the denominator. We know that the denominator of a fraction cannot have the value of zero. Therefore, the slope is undefined.

#### Practice questions

##### Question 4

What is the slope 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.

##### Question 5

What is the slope of the line going through A and B?

The slope 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?