Learn on PengiOpenstax Elementary Algebra 2EChapter 4: Graphs

Lesson 4.4: Understand Slope of a Line

In this lesson from OpenStax Elementary Algebra 2E, students learn to understand and calculate the slope of a line using the rise-over-run formula (m = rise/run). Students explore slope through geoboard models, graphs, and the slope formula to find the slope between two points, and also examine the slope of horizontal and vertical lines. Real-world applications such as roof pitch, highway grades, and wheelchair ramps are used to connect the concept of slope to everyday contexts.

Section 1

πŸ“˜ Understand Slope of a Line

New Concept

Slope, mm, measures a line's steepness and direction. You'll learn to calculate this 'tilt' as the ratio of vertical change (rise) to horizontal change (run), using graphs (m=rise⁑run⁑m = \dfrac{\operatorname{rise}}{\operatorname{run}}) and coordinates (m=y2βˆ’y1x2βˆ’x1m = \dfrac{y_2 - y_1}{x_2 - x_1}).

What’s next

You're ready to start! Up next are interactive examples and practice cards where you'll find the slope of any line from its graph or given points.

Section 2

Slope of a Line

Property

The slope of a line of a line is m=rise⁑run⁑m = \frac{\operatorname{rise}}{\operatorname{run}}. The rise measures the vertical change and the run measures the horizontal change between two points on the line.

Examples

  • A line rises 4 units for every 5 units it runs horizontally. Its slope is m=45m = \frac{4}{5}.
  • A line drops 2 units for every 3 units it runs horizontally. Its rise is βˆ’2-2, so its slope is m=βˆ’23m = \frac{-2}{3} or βˆ’23-\frac{2}{3}.
  • On a geoboard, if you connect two pegs by going up 3 units and right 2 units, the slope of the line is m=32m = \frac{3}{2}.

Explanation

Slope tells you how steep a line is. Think of it as 'rise over run'. For every number of units you move horizontally (run), the slope tells you how many units you move vertically (rise).

Section 3

Find Slope from a Graph

Property

To find the slope of a line from its graph:

  1. Locate two points on the graph whose coordinates are integers.
  2. Starting with the point on the left, sketch a right triangle, going from the first point to the second point.
  3. Count the rise and the run on the legs of the triangle.
  4. Take the ratio of rise to run to find the slope, m=rise⁑run⁑m = \frac{\operatorname{rise}}{\operatorname{run}}.

Examples

  • A line passes through (1,2)(1, 2) and (4,8)(4, 8). The rise is 8βˆ’2=68 - 2 = 6 and the run is 4βˆ’1=34 - 1 = 3. The slope is m=63=2m = \frac{6}{3} = 2.
  • A line passes through (0,6)(0, 6) and (2,2)(2, 2). The rise is 2βˆ’6=βˆ’42 - 6 = -4 and the run is 2βˆ’0=22 - 0 = 2. The slope is m=βˆ’42=βˆ’2m = \frac{-4}{2} = -2.
  • A line on a graph connects points (βˆ’3,1)(-3, 1) and (5,7)(5, 7). The rise is 7βˆ’1=67-1=6 and the run is 5βˆ’(βˆ’3)=85-(-3)=8. The slope is m=68=34m = \frac{6}{8} = \frac{3}{4}.

Explanation

To find slope from a graph, pick two easy-to-read points. Count the vertical distance (rise) and horizontal distance (run) to get from one point to the other. The slope is simply the rise divided by the run.

Section 4

Slope of Horizontal and Vertical Lines

Property

The slope of a horizontal line, y=by = b, is 0. The slope of a vertical line, x=ax = a, is undefined.

Examples

  • The line y=5y = 5 is a horizontal line. Its slope is 0.
  • The line x=βˆ’3x = -3 is a vertical line. Its slope is undefined.
  • A line passing through the points (2,6)(2, 6) and (9,6)(9, 6) has a rise of 6βˆ’6=06 - 6 = 0, so its slope is m=07=0m = \frac{0}{7} = 0.

Explanation

A horizontal line is perfectly flat, so its 'rise' is always zero, making the slope 0. A vertical line is infinitely steep; its 'run' is zero, and since we can't divide by zero, its slope is undefined.

Section 5

The Slope Formula

Property

The slope of the line between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is m=y2βˆ’y1x2βˆ’x1m = \frac{y_2 - y_1}{x_2 - x_1}. This is the slope formula. The slope is the difference in the y-coordinates divided by the difference in the x-coordinates.

Examples

  • For the points (2,5)(2, 5) and (4,11)(4, 11), the slope is m=11βˆ’54βˆ’2=62=3m = \frac{11 - 5}{4 - 2} = \frac{6}{2} = 3.
  • For the points (βˆ’3,6)(-3, 6) and (1,4)(1, 4), the slope is m=4βˆ’61βˆ’(βˆ’3)=βˆ’24=βˆ’12m = \frac{4 - 6}{1 - (-3)} = \frac{-2}{4} = -\frac{1}{2}.
  • For the points (5,βˆ’1)(5, -1) and (βˆ’2,3)(-2, 3), the slope is m=3βˆ’(βˆ’1)βˆ’2βˆ’5=4βˆ’7=βˆ’47m = \frac{3 - (-1)}{-2 - 5} = \frac{4}{-7} = -\frac{4}{7}.

Explanation

The slope formula is a tool to find a line's steepness without a graph. It calculates the rise by subtracting y-values (y2βˆ’y1y_2 - y_1) and the run by subtracting x-values (x2βˆ’x1x_2 - x_1), then divides them.

Section 6

Graph a Line from a Point and Slope

Property

To graph a line given a point and the slope:

  1. Plot the given point.
  2. Use the slope formula m=rise⁑run⁑m = \frac{\operatorname{rise}}{\operatorname{run}} to identify the rise and the run.
  3. Starting at the given point, count out the rise and run to mark the second point.
  4. Connect the points with a line.

Examples

  • To graph a line through (1,2)(1, 2) with slope m=35m = \frac{3}{5}, start at (1,2)(1, 2), move up 3 units and right 5 units to plot a second point at (6,5)(6, 5).
  • To graph a line through (βˆ’2,4)(-2, 4) with slope m=βˆ’3m = -3, treat the slope as βˆ’31\frac{-3}{1}. From (βˆ’2,4)(-2, 4), move down 3 units and right 1 unit to plot a second point at (βˆ’1,1)(-1, 1).
  • To graph a line through the y-intercept (0,βˆ’1)(0, -1) with slope m=23m = \frac{2}{3}, start at (0,βˆ’1)(0, -1), move up 2 units and right 3 units to plot a second point at (3,1)(3, 1).

Explanation

Think of it as 'point and directions'. Start by plotting the given point. Then use the slope's rise and run as steps to find a second point. Connect them to draw your line.

Section 7

Slope Applications

Property

The 'pitch' of a building’s roof is the slope of the roof. The grade of a road or the slope of a drainage pipe are other real-world applications of slope. We use the formula m=rise⁑run⁑m = \frac{\operatorname{rise}}{\operatorname{run}} to calculate these values.

Examples

  • A roof rises 6 feet for every 18 feet of horizontal run. The pitch of the roof is its slope: m=618=13m = \frac{6}{18} = \frac{1}{3}.
  • A road drops 4 feet for every 100 feet of horizontal distance. The grade of the road is m=βˆ’4100=βˆ’125m = \frac{-4}{100} = -\frac{1}{25}.
  • A wheelchair ramp needs to rise 2 feet. If the required slope is 112\frac{1}{12}, the horizontal run must be 24 feet, since 224=112\frac{2}{24} = \frac{1}{12}.

Explanation

Slope is everywhere in the real world. It describes the steepness of hills, the pitch of roofs, and the angle of wheelchair ramps. A positive slope goes up, and a negative slope goes down.

Book overview

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Chapter 4: Graphs

  1. Lesson 1

    Lesson 4.1: Use the Rectangular Coordinate System

  2. Lesson 2

    Lesson 4.2: Graph Linear Equations in Two Variables

  3. Lesson 3

    Lesson 4.3: Graph with Intercepts

  4. Lesson 4Current

    Lesson 4.4: Understand Slope of a Line

  5. Lesson 5

    Lesson 4.5: Use the Slope-Intercept Form of an Equation of a Line

  6. Lesson 6

    Lesson 4.6: Find the Equation of a Line

  7. Lesson 7

    Lesson 4.7: Graphs of Linear Inequalities

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

πŸ“˜ Understand Slope of a Line

New Concept

Slope, mm, measures a line's steepness and direction. You'll learn to calculate this 'tilt' as the ratio of vertical change (rise) to horizontal change (run), using graphs (m=rise⁑run⁑m = \dfrac{\operatorname{rise}}{\operatorname{run}}) and coordinates (m=y2βˆ’y1x2βˆ’x1m = \dfrac{y_2 - y_1}{x_2 - x_1}).

What’s next

You're ready to start! Up next are interactive examples and practice cards where you'll find the slope of any line from its graph or given points.

Section 2

Slope of a Line

Property

The slope of a line of a line is m=rise⁑run⁑m = \frac{\operatorname{rise}}{\operatorname{run}}. The rise measures the vertical change and the run measures the horizontal change between two points on the line.

Examples

  • A line rises 4 units for every 5 units it runs horizontally. Its slope is m=45m = \frac{4}{5}.
  • A line drops 2 units for every 3 units it runs horizontally. Its rise is βˆ’2-2, so its slope is m=βˆ’23m = \frac{-2}{3} or βˆ’23-\frac{2}{3}.
  • On a geoboard, if you connect two pegs by going up 3 units and right 2 units, the slope of the line is m=32m = \frac{3}{2}.

Explanation

Slope tells you how steep a line is. Think of it as 'rise over run'. For every number of units you move horizontally (run), the slope tells you how many units you move vertically (rise).

Section 3

Find Slope from a Graph

Property

To find the slope of a line from its graph:

  1. Locate two points on the graph whose coordinates are integers.
  2. Starting with the point on the left, sketch a right triangle, going from the first point to the second point.
  3. Count the rise and the run on the legs of the triangle.
  4. Take the ratio of rise to run to find the slope, m=rise⁑run⁑m = \frac{\operatorname{rise}}{\operatorname{run}}.

Examples

  • A line passes through (1,2)(1, 2) and (4,8)(4, 8). The rise is 8βˆ’2=68 - 2 = 6 and the run is 4βˆ’1=34 - 1 = 3. The slope is m=63=2m = \frac{6}{3} = 2.
  • A line passes through (0,6)(0, 6) and (2,2)(2, 2). The rise is 2βˆ’6=βˆ’42 - 6 = -4 and the run is 2βˆ’0=22 - 0 = 2. The slope is m=βˆ’42=βˆ’2m = \frac{-4}{2} = -2.
  • A line on a graph connects points (βˆ’3,1)(-3, 1) and (5,7)(5, 7). The rise is 7βˆ’1=67-1=6 and the run is 5βˆ’(βˆ’3)=85-(-3)=8. The slope is m=68=34m = \frac{6}{8} = \frac{3}{4}.

Explanation

To find slope from a graph, pick two easy-to-read points. Count the vertical distance (rise) and horizontal distance (run) to get from one point to the other. The slope is simply the rise divided by the run.

Section 4

Slope of Horizontal and Vertical Lines

Property

The slope of a horizontal line, y=by = b, is 0. The slope of a vertical line, x=ax = a, is undefined.

Examples

  • The line y=5y = 5 is a horizontal line. Its slope is 0.
  • The line x=βˆ’3x = -3 is a vertical line. Its slope is undefined.
  • A line passing through the points (2,6)(2, 6) and (9,6)(9, 6) has a rise of 6βˆ’6=06 - 6 = 0, so its slope is m=07=0m = \frac{0}{7} = 0.

Explanation

A horizontal line is perfectly flat, so its 'rise' is always zero, making the slope 0. A vertical line is infinitely steep; its 'run' is zero, and since we can't divide by zero, its slope is undefined.

Section 5

The Slope Formula

Property

The slope of the line between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is m=y2βˆ’y1x2βˆ’x1m = \frac{y_2 - y_1}{x_2 - x_1}. This is the slope formula. The slope is the difference in the y-coordinates divided by the difference in the x-coordinates.

Examples

  • For the points (2,5)(2, 5) and (4,11)(4, 11), the slope is m=11βˆ’54βˆ’2=62=3m = \frac{11 - 5}{4 - 2} = \frac{6}{2} = 3.
  • For the points (βˆ’3,6)(-3, 6) and (1,4)(1, 4), the slope is m=4βˆ’61βˆ’(βˆ’3)=βˆ’24=βˆ’12m = \frac{4 - 6}{1 - (-3)} = \frac{-2}{4} = -\frac{1}{2}.
  • For the points (5,βˆ’1)(5, -1) and (βˆ’2,3)(-2, 3), the slope is m=3βˆ’(βˆ’1)βˆ’2βˆ’5=4βˆ’7=βˆ’47m = \frac{3 - (-1)}{-2 - 5} = \frac{4}{-7} = -\frac{4}{7}.

Explanation

The slope formula is a tool to find a line's steepness without a graph. It calculates the rise by subtracting y-values (y2βˆ’y1y_2 - y_1) and the run by subtracting x-values (x2βˆ’x1x_2 - x_1), then divides them.

Section 6

Graph a Line from a Point and Slope

Property

To graph a line given a point and the slope:

  1. Plot the given point.
  2. Use the slope formula m=rise⁑run⁑m = \frac{\operatorname{rise}}{\operatorname{run}} to identify the rise and the run.
  3. Starting at the given point, count out the rise and run to mark the second point.
  4. Connect the points with a line.

Examples

  • To graph a line through (1,2)(1, 2) with slope m=35m = \frac{3}{5}, start at (1,2)(1, 2), move up 3 units and right 5 units to plot a second point at (6,5)(6, 5).
  • To graph a line through (βˆ’2,4)(-2, 4) with slope m=βˆ’3m = -3, treat the slope as βˆ’31\frac{-3}{1}. From (βˆ’2,4)(-2, 4), move down 3 units and right 1 unit to plot a second point at (βˆ’1,1)(-1, 1).
  • To graph a line through the y-intercept (0,βˆ’1)(0, -1) with slope m=23m = \frac{2}{3}, start at (0,βˆ’1)(0, -1), move up 2 units and right 3 units to plot a second point at (3,1)(3, 1).

Explanation

Think of it as 'point and directions'. Start by plotting the given point. Then use the slope's rise and run as steps to find a second point. Connect them to draw your line.

Section 7

Slope Applications

Property

The 'pitch' of a building’s roof is the slope of the roof. The grade of a road or the slope of a drainage pipe are other real-world applications of slope. We use the formula m=rise⁑run⁑m = \frac{\operatorname{rise}}{\operatorname{run}} to calculate these values.

Examples

  • A roof rises 6 feet for every 18 feet of horizontal run. The pitch of the roof is its slope: m=618=13m = \frac{6}{18} = \frac{1}{3}.
  • A road drops 4 feet for every 100 feet of horizontal distance. The grade of the road is m=βˆ’4100=βˆ’125m = \frac{-4}{100} = -\frac{1}{25}.
  • A wheelchair ramp needs to rise 2 feet. If the required slope is 112\frac{1}{12}, the horizontal run must be 24 feet, since 224=112\frac{2}{24} = \frac{1}{12}.

Explanation

Slope is everywhere in the real world. It describes the steepness of hills, the pitch of roofs, and the angle of wheelchair ramps. A positive slope goes up, and a negative slope goes down.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 4: Graphs

  1. Lesson 1

    Lesson 4.1: Use the Rectangular Coordinate System

  2. Lesson 2

    Lesson 4.2: Graph Linear Equations in Two Variables

  3. Lesson 3

    Lesson 4.3: Graph with Intercepts

  4. Lesson 4Current

    Lesson 4.4: Understand Slope of a Line

  5. Lesson 5

    Lesson 4.5: Use the Slope-Intercept Form of an Equation of a Line

  6. Lesson 6

    Lesson 4.6: Find the Equation of a Line

  7. Lesson 7

    Lesson 4.7: Graphs of Linear Inequalities