# Lesson Plan: Analyzing and Comparing Linear Relationships, Including Parallel and Perpendicular Lines

## Lesson Objectives

This lesson can be completed in one 50-minute class period. However, for a more in-depth exploration of the concepts, especially the real-world applications and group activities, it may be beneficial to extend it to two class periods.

- Understand the properties of parallel and perpendicular lines
- Determine the slopes of parallel and perpendicular lines
- Write equations of parallel and perpendicular lines
- Apply knowledge of parallel and perpendicular lines to solve geometric problems

## Florida BEST Standards

- MA.8.AR.3.2: Given a table, graph or written description of a linear relationship, determine the slope.
- MA.8.AR.3.4: Given a table, graph or written description of a linear relationship, write an equation in slope-intercept form.
- MA.8.GR.2.1: Given a preimage and image, describe the transformation applied to the preimage that results in the image.

## Prerequisite Skills

- Understanding of linear equations
- Graphing linear functions
- Basic knowledge of slope

## Key Vocabulary

- Rate of change
- Initial value
- Linear function
- Qualitative analysis
- Parallel lines
- Perpendicular lines

## Additional Resources

- Ratios and Proportional Relationships
- Solving Percent Problems
- Graphing and Interpreting Proportional Relationships
- Direct Variations

## Warm-up Activity (10 minutes)

Present the following scenarios to students and ask them to sketch possible graphs and discuss their reasoning:

- Two cars are traveling at the same speed. One car leaves at time t = 0 and the other one leaves several minutes later. What might the distance-vs-time graphs look like?
- What would need to change in scenario 1 so that the second car catches up to the first car?

Show the following graphs. One of them represents two cars traveling at the same speed and the other represents cars traveling at different speeds, with one car catching up with the other.

Discuss student responses, emphasizing that in scenario 1, the graphs would be parallel lines (same slope, different y-intercepts), while in scenario 2, the second car would need to travel at a faster speed (steeper slope) to catch up.

## Teach (25 minutes)

### Definitions

**Rate of change**: The slope of a linear function, representing how much the dependent variable changes for each unit increase in the independent variable**Initial value**: The y-intercept of a linear function, representing the value of the dependent variable when the independent variable is zero**Linear function**: A function that can be represented by a straight line on a coordinate plane**Qualitative analysis**: Describing the overall behavior of a function without specific numerical values**Parallel lines**: Lines in the same plane that never intersect and have the same slope**Perpendicular lines**: Lines that intersect at a right angle and have slopes that are negative reciprocals of each other

Use this slide show to review these and other related terms:

### Instruction

Demonstrate how to compare linear functions in different representations, explain how to interpret slope and y-intercept in context, show how to construct linear functions from real-world situations, introduce the concept of parallel lines having equal slopes, and explain that perpendicular lines have slopes that are negative reciprocals of each other.

As needed, introduce the following videos:

**Linear graphs of parallel lines**:__https://www.media4math.com/library/44928/asset-preview__**Linear graphs of perpendicular lines**:__https://www.media4math.com/library/44929/asset-preview__

### Example 1: Distance vs. Time for Two Cars Traveling at the Same Speed

Consider two cars traveling at the same speed of 60 miles per hour. Car A leaves at time t = 0, and Car B leaves 30 minutes later.

Let t be the time in hours and d be the distance in miles.

Car A: d = 60t

Car B: d = 60(t - 0.5)

Graph these equations:

- Car A: Starts at the origin (0,0) and has a slope of 60.
- Car B: Starts at (0.5, 0) and has the same slope of 60.

The lines are parallel because they have the same slope, representing the same speed.

Use this Desmos activity for demonstration purposes. It mirrors the graph shown above but it also has a slider for changing the speed of the cars, demonstrating that same slopes result in parallel lines:

__https://www.desmos.com/calculator/bisy8qf93s__

### Example 2: Distance vs. Time for Two Cars, One Traveling Faster

Now consider Car A traveling at 60 miles per hour and Car B traveling at 80 miles per hour, with Car B leaving 30 minutes later.

Let t be the time in hours and d be the distance in miles.

Car A: d = 60t

Car B: d = 80(t - 0.5)

Graph these equations:

- Car A: Starts at the origin (0,0) and has a slope of 60.
- Car B: Starts at (0.5, 0) and has a slope of 80.

The lines are not parallel because they have different slopes. Car B's steeper slope indicates a faster speed, allowing it to catch up to Car A. Use this Desmos activity for support:

__https://www.desmos.com/calculator/iizio5t9jy__

### Example 3: Perpendicular Lines in Graphing

Graph the lines y = 2x + 3 and y = -0.5x + 1. Explain why these lines are perpendicular.

Solution: After graphing, we observe that the lines intersect at a right angle. They are perpendicular because their slopes are negative reciprocals of each other (2 and -0.5).

Use this Desmos activity to explore graphs of perpendicular lines. In particular have them explore the role of the y-intercept, which doesn't affect if the lines are perpindicular:

__https://www.desmos.com/calculator/g2ewan7fjr__

### Example 4: Business Application - Smartphone Production

Samsung's largest smartphone factory in Noida, India, can produce approximately 120 million smartphones per year. Let's assume they operate 360 days a year. The company needs to produce 10 million smartphones over the next month (30 days). However, there were machinery malfunctions that cost the company 4 days of work. Find the equations for the original production line and the new production line that accounts for the delay.

Solution:

Original daily production: For simplicity, use M to represent millions. So the smartphone rate per day is 0.33M smartphones/day. | 120 million / 360 days = 333,333 smartphones/day. |

Original production line (direct variation): where y is the number of smartphones and x is the number of days: | y = kx |

Plug in specific values to solve for k: | 10,000,000 = k(30) k = 0.33M |

This is the direct variation. | y = 0.33M•x |

We need to find a line that passes through two points: (0, -1.33M) to account for the 4-day delay (30, 10M) to meet the original target. Use the point-slope form: | y - y₁ = m(x - x₁) y - (-1.3M) = m(x - 0) |

Calculate slope m: | m = (10M - (-1.33M)) / (30 - 0) = 11.3M / 30 = 0.378M |

This is the new equation: | y = 0.378M•x - 1.3M |

Interpret results: The original production rate was 333,333 smartphones/day. The new production rate is 377,778 smartphones per day, an increase of about 13.3%. The y-intercept of -1.3M represents the production deficit due to the 4-day delay.

Graph both lines to visualize how the new production line intersects the original line but has a steeper slope to make up for lost time.

This Desmos activity provides a solution to this problem but uses fractional coefficients to give a more precise result:

__https://www.desmos.com/calculator/n68f2pa4tc__

## Review (10 minutes)

Use this slide show to review of examples of linear equations with slopes that are either parallel or perpendicular to each other:

Group activity: Students analyze and compare different linear functions representing real-world scenarios, including identifying parallel and perpendicular relationships.

### Example 1: Agriculture

A farmer records the following data on wheat yield based on the amount of nitrogen fertilizer applied:

Nitrogen (lbs/acre) | Yield (bushels/acre) |
---|---|

0 | 30 |

50 | 45 |

100 | 60 |

150 | 75 |

Find the linear function that models this relationship and interpret the slope and y-intercept.

Solution:

Let x be the amount of nitrogen and y be the yield. | f(x) = mx + b |

Use two points to find the slope: | (0, 30) and (150, 75) |

Calculate the slope: | (75 - 30) / (150 - 0) = 45 / 150 = 0.3 |

From the data we know y-intercept | 30 |

Write the equation: | y = 0.3x + 30 |

Interpretation:

- Slope: For every 1 lb/acre increase in nitrogen, the yield increases by 0.3 bushels/acre.
- Y-intercept: Without any nitrogen applied, the expected yield is 30 bushels/acre.

This Desmos activity can provide additional support:

__https://www.desmos.com/calculator/b1howyffjg__

This activity also includes an extension: A scenario for a farm with slightly less productive soil. This one results in a parallel line.

### Example 2: Music

A piano tuner observes the following relationship between the length of a piano string and its frequency:

Length (cm) | Frequency (Hz) |
---|---|

100 | 110 |

80 | 137.5 |

60 | 183.3 |

40 | 275 |

Find the linear function that models this relationship and interpret the slope and y-intercept.

Solution:

Let x be the length of the string and y be the frequency. | y = f(x) |

Use two points: | (100, 110) and (40, 275) |

Calculate slope: | (275 - 110) / (40 - 100) = 165 / (-60) = -2.75 |

Calculate y-intercept: | b = 110 + 2.75(100) = 385 |

Linear function: | y = -2.75x + 385 |

Interpretation:

- Slope: For every 1 cm increase in string length, the frequency decreases by 2.75 Hz.
- Y-intercept: A theoretical string of 0 cm would have a frequency of 385 Hz (note: this is not physically meaningful in this context).

Here is a companion Desmos activity, which shows that the graph isn't exactly linear, but a line of best fit is very close to the calculated on above.

__https://www.desmos.com/calculator/cxyznlkkqu__

## Assess (5 minutes)

Administer the 10-question quiz.

## Quiz

- Compare the rates of change of y = 2x + 3 and y = 0.5x - 1.

- A car rental company charges a flat fee of \$30 plus \$0.25 per mile driven. Write a function to represent the total cost.

- Which has a greater initial value: y = 3x + 5 or y = 2x + 7?

- Describe the relationship between x and y in the graph of y = -2x + 4.

- If f(x) = 3x + 2 and g(x) = 3x - 1, how do these functions compare?

- A line passes through (2, 5) and (4, 9). Find its equation and interpret the slope in context.

- Write an equation for a line parallel to y = 2x - 3 that passes through the point (0, 5).

- Find the slope of a line perpendicular to y = 3x + 1.

- Are the lines y = 2x + 1 and y = -0.5x + 3 perpendicular? Explain why or why not.

- If two lines are parallel and one passes through (1, 4) and (3, 8), what is the slope of both lines?

## Answer Key

- y = 2x + 3 has a greater rate of change (2 vs 0.5)
- y = 0.25x + 30, where y is total cost and x is miles driven
- y = 2x + 7 has a greater initial value
- As x increases, y decreases at a constant rate of 2 units per 1 unit of x
- They have the same slope but different y-intercepts; they are parallel.
- y = 2x + 1. The slope represents the rate of change of y with respect to x.
- y = 2x + 5
- -1/3
- Yes, because the slopes are negative reciprocals of each other.
- 2

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