first, let’s warm up evaluate each equation for x = –1, 0, and 1. 1. y = 3 x 2. y = x –...

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First, Let’s Warm Up Evaluate each equation for x = –1, 0, and 1. 1. y = 3x 2. y = x – 7 3. y = 2x + 5 4. y = 6x – 2 –3, 0, 3 –8, –7, –6 3, 5, 7 –8, –2, 4

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First, Let’s Warm Up Evaluate each equation for x = –1, 0, and 1. 1. y = 3 x 2. y = x – 7 3. y = 2 x + 5 4. y = 6 x – 2. –3, 0, 3. –8, –7, –6. 3, 5, 7. –8, –2, 4. WARM-UP. - PowerPoint PPT Presentation

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First, Let’s Warm UpEvaluate each equation for x = –1, 0, and 1.

1. y = 3x

2. y = x – 7

3. y = 2x + 5

4. y = 6x – 2

–3, 0, 3

–8, –7, –6

3, 5, 7

–8, –2, 4

WARM-UP

A lift on a ski slope rises according to the equation a = 130t + 6250, where a is the altitude in feet and t is the number of minutes that a skier has been on the lift. Five friends are on the lift. What is the altitude of each person if they have been on the ski lift for the times listed in the table? Draw a graph that represents the relationship between the time on the lift and the altitude.

Here is how I answered it!

Additional Example 2 Continued

The altitudes are: Anna, 6770 feet; Tracy, 6640 feet; Kwani, 6510 feet; Tony, 6445 feet; George, 6380 feet. This is a linear equation because when t increases by 1 unit, a increases by 130 units. Note that a skier with 0 time on the lift implies that the bottom of the lift is at an altitude of 6250 feet.

Continued

Let’s try another one in teams!In an amusement park ride, a car travels according to the equation D = 1250t where t is time in minutes and D is the distance in feet the car travels. Below is a chart of the time that three people have been in the cars. Graph the relationship between time and distance. How far has each person traveled?

Rider Time

Ryan 1 min

Greg 2 min

Colette 3 min

Continued

t D =1250t D (t, D)

1 1250(1) 1250 (1, 1250)

2 1250(2) 2500 (2, 2500)

3 1250(3) 3750 (3, 3750)

The distances are: Ryan, 1250 ft; Greg, 2500 ft; and Collette, 3750 ft.

You looked at slope on the coordinate plane in Lesson 5-5 (p. 244).

Remember!

Learn to find the slope of a line and use slope to understand and draw

graphs.

Linear equations have constant slope. For a line on the coordinate plane, slope is the following ratio:

vertical change horizontal change

change in y change in x=

This ratio is often referred to as , or “rise

over run,” where rise indicates the number of units moved up or down and run indicates the number of units moved to the left or right. Slope can be positive, negative, zero, or undefined. A line with positive slope goes up from left to right. A line with negative slope goes down from left to right.

rise run

If you know any two points on a line, or two solutions of a linear equation, you can find the slope of the line without graphing. The slope of a line through the points (x1, y1) and (x2, y2) is as follows:

yy22 –– yy11 xx22 –– xx11

Find the slope of the line that passes through (–2, –3) and (4, 6).

Finding Slope, Given Two Points

Let (x1, y1) be (–2, –3) and (x2, y2) be (4, 6).

6 – (–3)4 – (–2)

Substitute 6 for y2, –3 for y1, 4 for x2, and –2 for x1.

96=

The slope of the line that passes through (–

2, –3) and (4, 6) is . 32

=y2 – y1

x2 – x1

32=

yy22 –– yy11 xx22 –– xx11

Find the slope of the line that passes through (–4, –6) and (2, 3).

Try this one on your own!

Let (x1, y1) be (–4, –6) and (x2, y2) be (2, 3).

3 – (–6)2 – (–4)

Substitute 3 for y2, –6 for y1, 2 for x2, and –4 for x1.

96=

The slope of the line that passes through (–

4, –6) and (2, 3) is . 32

=y2 – y1

x2 – x1

32=

Use the graph of the line to determine its slope.

Additional Example 2: Finding Slope from a Graph

Additional Example 2 Continued

Choose two points on the line: (0, 1) and (3, –4).

Guess by looking at the graph:

riserun = –5

3 = – 5 3

Use the slope formula.

Let (3, –4) be (x1, y1) and (0, 1) be (x2, y2).

1 – (–4) 0 – 3=

y2 – y1

x2 – x1

5–3= 5

3= –

–5

3

Notice that if you switch (x1, y1) and (x2, y2), you get the same slope:

53The slope of the given line is – .

Let (0, 1) be (x1, y1) and (3, –4) be (x2, y2).

Additional Example 2 Continued

–4 – 1 3 – 0=

y2 – y1

x2 – x1

–5 3= 5

3= –

Use the graph of the line to determine its slope.

You Try This One!

Try This: Example 2 Continued

Choose two points on the line: (1, 1) and (0, –1).

Guess by looking at the graph:

riserun = 2

1 = 2

Use the slope formula.

Let (1, 1) be (x1, y1) and (0, –1) be (x2, y2).

=y2 – y1

x2 – x1

–2–1=

–1 – 1 0 – 1

= 2

12

Recall that two parallel lines have the same slope. The slopes of two perpendicular

lines are negative reciprocals of each

other.

Identifying Parallel and Perpendicular Lines by Slope

Tell whether the lines passing through the given points are parallel or perpendicular.

A. line 1: (–6, 4) and (2, –5); line 2: (–1, –4) and (8, 4)

slope of line 1:

slope of line 2:

Line 1 has a slope equal to – and line 2 has a slope

equal to , – and are negative reciprocals of each

other, so the lines are perpendicular.

98

89

89

98

=y2 – y1

x2 – x1

–9 8= –5 – 4

2 – (–6)

4 – (–4)8 – (–1)=

y2 – y1

x2 – x1

8 9=

9 8= –

Identifying Parallel and Perpendicular Lines by Slope

B. line 1: (0, 5) and (6, –2); line 2: (–1, 3) and (5, –4)

Both lines have a slope equal to – , so the lines are parallel.

76

slope of line 1:

slope of line 2:

=y2 – y1

x2 – x1

–7 6= –2 – 5

6 – 0

=y2 – y1

x2 – x1

7 6= –

–7 6= 7

6= – –4 – 35 – (–1)

Try This:

Tell whether the lines passing through the given points are parallel or perpendicular.

A. line 1: (–8, 2) and (0, –7); line 2: (–3, –6) and (6, 2)

slope of line 1:

slope of line 2:

Line 1 has a slope equal to – and line 2 has a slope

equal to , – and are negative reciprocals of each

other, so the lines are perpendicular.

98

89

89

98

=y2 – y1

x2 – x1

–9 8= –7 – 2

0 – (–8)

2 – (–6)6 – (–3)=

y2 – y1

x2 – x1

8 9=

9 8= –

Try This: Example 3B

B. line 1: (1, 1) and (2, 2); line 2: (1, –2) and (2, -1)

Line 1 has a slope equal to 1 and line 2 has a slope equal to –1. 1 and –1 are negative reciprocals of each other, so the lines are perpendicular.

slope of line 1:

slope of line 2:

=y2 – y1

x2 – x1

1 1= 2 – 1

2 – 1

=y2 – y1

x2 – x1

–1 1= –1 – (–2)

2 – (1)

= 1

= –1

Additional Example 4: Graphing a Line Using a Point and the Slope

Graph the line passing through (3, 1) with slope 2.

Plot the point (3, 1). Then move 2 units up and right 1 unit and plot the point (4, 3). Use a straightedge to connect the two points.

The slope is 2, or . So for every 2 units up, you will move right 1 unit, and for every 2 units down, you will move left 1 unit.

21

Additional Example 4 Continued

1

2(3, 1)

Try This: Example 4

Graph the line passing through (1, 1) with slope 2.

Plot the point (1, 1). Then move 2 units up and right 1 unit and plot the point (2, 3). Use a straightedge to connect the two points.

The slope is 2, or . So for every 2 units up, you will move right 1 unit, and for every 2 units down, you will move left 1 unit.

21

Try This: Example 4 Continued

1

2(1, 1)

Lesson Quiz: Part 1

Find the slope of the line passing through each pair of points.

1. (4, 3) and (–1, 1)

2. (–1, 5) and (4, 2)

3. Use the graph of the line to

determine its slope.

25

53–

34–

Lesson Quiz: Part 2

Tell whether the lines passing through the given points are parallel or perpendicular.

4. line 1: (–2, 1), (2, –1); line 2: (0, 0), (–1, –2)

5. line 1: (–3, 1), (–2, 3); line 2: (2, 1), (0, –3)

parallel

perpendicular