1 topic 4.3.1 slope of a line. 2 topic 4.3.1 slope of a line california standard: 7.0: students...

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Topic 4.3.1Topic 4.3.1

Slope of a LineSlope of a Line

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Topic4.3.1


California Standard:7.0: Students verify that a point lies on a line, given an equation of the line. Students are able to derive linear equations by using the point-slope formula.

What it means for you:You’ll find the slope of a line given any two points on the line.

Key words:• slope• steepness• horizontal• vertical• rise over run

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Topic4.3.1

By now you’ve had plenty of practice in plotting lines.


Any line can be described by its slope — which is what this Topic is about.

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Topic4.3.1

The Slope of a Line is Its Steepness

The slope (or gradient) of a line is a measure of its steepness.


The vertical change is usually written y, and it’s often called the rise.

The slope of a straight line is the ratio of the vertical change to the horizontal change between any two points lying on the line.

In the same way, the horizontal change is usually written x, and it’s often called the run.

x = run

y = rise

x is pronounced “delta x” and just means “change in x.”

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Topic4.3.1


Slope = , provided x 0 vertical change

horizontal change =

rise run

y x

=

If you know the coordinates of any two points on a line, you can find the slope. The slope, m, of a line passing through points P1 (x1, y1) and P2 (x2, y2) is given by this formula:

= m = , provided x2 – x1 0 y x

y2 – y1 x2 – x1

There is an important difference between positive and negative slopes — a positive slope means the line goes “uphill” , whereas a line with a negative slope goes “downhill” .

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Topic4.3.1

Find the slope of the line that passes through the points (2, 1) and (7, 4) and draw the graph.

Solution follows…

Solution


Example 1

my2 – y1 x2 – x1

= =4 – 1 7 – 2

=3 5

So the slope is .3 5

Solution continues…

You can use the formula to find the slope of the line.

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Topic4.3.1

Find the slope of the line that passes through the points (2, 1) and (7, 4) and draw the graph.

Solution (continued)


You know that the line passes through (2, 1) and (7, 4), so just join those two points up to draw the graph.

Example 1

Notice how the line has a positive slope, meaning it goes “uphill” from left to right.

In fact, since the slope is ,the line goes 3 units up for every 5 units across.

35

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Topic4.3.1

Guided Practice

Solution follows…

In Exercises 1–4, find the slope of the line on the graph below.


1

21. m = ……..

2. m = ……..

3. m = ……..

4. m = ……..

3

4

4

3–

3

2–

9

Topic4.3.1

Guided Practice

Solution follows…

5. Find the slope of the line that passes through the points (1, 5) and (3, 2), and draw the graph on a copy of the coordinate plane opposite.


6. Find the slope of the line that passes through the points (3, 1) and (2, 4), and draw the graph on a copy of the coordinate plane opposite.

3

2m = –

5.

m = –3

6.

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Topic4.3.1


Example 2

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

Solution

m =4 – (–2)

3 – 6

4 + 2

–3= =

6

–3= –2

So the slope is –2.

Solution follows…

Be extra careful if any of the coordinates

are negative.

Be extra careful if any of the coordinates

are negative.

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Topic4.3.1


This time the line has a negative slope, meaning it goes “downhill” from left to right.

Here the slope is –2, which means that the line goes 2 units down for every 1 unit across.

1

2

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Topic4.3.1


Find the slope of the lines through: a) (2, 5) and (–4, 2)

b) (1, –6) and (3, 3)Solution

Example 3

a) m =5 – 2

2 – (–4)=

3

2 + 4=

3

6=

1

2

b) m =3 – (–6)

3 – 1=

3 + 6

2=

9

2

Solution follows…

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Find the slope m of the line through each pair of points below.

7. (–1, 2) and (3, 2)

8. (0, –5) and (–6, 1)

9. (5, –7) and (–3, –7)

10. (4, –1) and (–3, 5)

11. (–1, –3) and (1, –4)

Topic4.3.1

Guided Practice

Solution follows…


m = = = 0y2 – y1

x2 – x1

–7 – (–7)

–3 – 5

m = = = 0y2 – y1

x2 – x1

2 – 2

3 –(–1)

m = = = –y2 – y1

x2 – x1

–4 – (–3)

1 – (–1)

1

2

m = = = – 67

y2 – y1

x2 – x1

5 – (–1)

–3 – 4

m = = = = –1y2 – y1

x2 – x1

1 –(–5)

–6 – 0

6

–6

14

Find the slope m of the line through each pair of points below.

12. (5, 7) and (–11, –12)

13. (–2, –2) and (–3, –17)

14. (18, 2) and (–32, 7)

15. (0, –1) and (1, 0)

16. (0, 0) and (–14, –1)

Topic4.3.1

Guided Practice

Solution follows…


m = = = = 15y2 – y1

x2 – x1

–17 – (–2)

–3 – (–2)

–15

–1

m = = = = 1y2 – y1

x2 – x1

0 – (–1)

1 – 0

1

1

m = = = =y2 – y1

x2 – x1

–1 – 0

–14 – 0

–1

–14

1

14

m = = = = –y2 – y1

x2 – x1

7 – 2

–32 –18

5

–50

1

10

m = = = =y2 – y1

x2 – x1

–12 – 7

–11 – 5

–19

–16 16

19

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Topic4.3.1


If the slope of the line that passes through the points (4, –1) and (6, 2k) is 3, find the value of k.

Solution Even though one pair of coordinates contains a variable, k, you can still use the slope formula in exactly the same way as before.

my2 – y1

x2 – x1 = , which means that m =

2k – (–1)

6 – 4 =2k + 1

2

But the slope is 3, so = 32k + 1

2

2k + 1 = 6

2k = 5

Example 4

k =52

Solution follows…

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17. (7, –2c) and (10, –c)

18. (b, 1) and (3b, –3)

19. (2, 2k) and (–5, –5k)

20. (3q, 1) and (2q, 7)

21. (3d, 7d) and (5d, 9d)

Topic4.3.1

Guided Practice

Solution follows…

Find the slope m of the lines through the points below.


m = = =–c – (–2c)

7b – 4b

c

3

y2 – y1

x2 – x1

m = = = = – –3 – 1

3b – b

–4

2b

2

b

y2 – y1

x2 – x1

m = = = = k–5k –2k

–5 – 2

–7k

–7

y2 – y1

x2 – x1

m = = = –7 – 1

2q – 3q

6q

y2 – y1

x2 – x1

2d

2dm = = = = 1

9d – 7d

5d – 3d

y2 – y1

x2 – x1

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22. (4a, 5k) and (2a, 7k)

23. (9c, 12v) and (12c, 15v)

24. (p, q) and (q, p)

25. (10, 14d) and (d, –7)

26. (2t, –3s) and (18s, 14t)

Topic4.3.1

Guided Practice

Solution follows…

Find the slope m of the lines through the points below.


m = = = = –y2 – y1

x2 – x1

7k – 5k

2a – 4a

2k

–2a

k

a

m = = = =v

c

3v

3c

15v – 12v

12c – 9c

y2 – y1

x2 – x1

m = =p – q

q – p

y2 – y1

x2 – x1

m = = =–7 –14d

d – 10

14d + 7

10 – d

y2 – y1

x2 – x1

m = = =14t –(–3s)

18s – 2t

14t + 3s

18s – 2t

y2 – y1

x2 – x1

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Topic4.3.1

Guided Practice

Solution follows…

In Exercises 27–29 you’re given two points on a line and the line’s slope, m. Find the value of the unknown constant in each Exercise.

27. (–2, 3) and (3k, –4), m =

28. (4, –5t) and (7, –8t), m =

29. (4b, –6) and (7b, –10), m =


25

27

34

m = = =

= (–7)(5) = 2(3k + 2) –35 = 6k + 4 6k = –39 k = –2

5

y2 – y1

x2 – x1

–7

3k + 2

–4 – 3

3k –(–2)

2

13–7

3k + 2

m = = = = –t

–t = t = –

y2 – y1

x2 – x1

–8t –(–5t)

7 – 42

7

2

7

–3t

3

m = = =

= –16 = 9b b = –

–4

3b3

4

16

9

y2 – y1

x2 – x1

–10 –(–6)

7b – 4b–4

3b

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Topic4.3.1

Guided Practice

Solution follows…

In Exercises 30–32 you’re given two points on a line and the line’s slope, m. Find the value of the unknown constant in each Exercise.

30. (–8, –6) and (12, 4), m = – v

31. (7k, –3) and (k, –1), m = 4

32. (1, –17) and (40, 41), m = – t


17478

25

m = = = =

– v = –4v = 5 v = –

y2 – y1

x2 – x1

1

2

2

5

5

4

10

20

4 – (–6)

12 – (–8)

1

2

m = = = = –

4 = 12k = –1 k = –

y2 – y1

x2 – x1

–1 – (–3)

k – 7k

1

3k1

12

2

–6k1

3k

m = = =

– t = t = – = – = –

y2 – y1

x2 – x1

174

78

58

39

41 – (–17)

40 – 1

58 × 78

39 × 174

4524

6786

2

3

58

39

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Topic4.3.1

Independent Practice

Solution follows…

1. Find the slope m of the lines shown below.


(i) m = ……..

(ii) m = ……..

(iii) m = ……..

(iv) m = ……..

(v) m = ……..

2

–2

0

12

19

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Topic4.3.1


Solution follows…

In Exercises 2–5, find the slope of the line that passes through the given points, and draw the graph on a copy of the coordinate plane below.


2. (–2, 1) and (0, 2)

3. (4, 4) and (1, 0)

4. (–5, 2) and (–1, 3)

5. (3, –3) and (7, 3)

12

m =

43

m =

14

m =

32

m =

2.3.4.5.

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Topic4.3.1


Solution follows…

In Exercises 6–10, find the slope of the line through each of the points.


6. (–3, 5) and (2, 1) 7. (0, 4) and (–4, 0)

8. (2, 3) and (4, 3) 9. (6d, 2) and (4d, –1)

10. (2s, 2t) and (s, 3t)

45

m = –

m = 0

ts

m = –

m = 1

32d

m =

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Topic4.3.1

Solution follows…

In Exercises 11–15, you’re given two points on a line and the line’s slope, m. Find the value of the unknown constant in each Exercise.

11. (3t, 7) and (5t, 9), m = –

12. (3k, 1) and (2k, 7), m =

13. (0, 14d) and (10, –6d), m = –1

14. (2t, –3) and (–3t, 5), m =

15. (0, 8d) and (–1, 4d), m = –


12

54

t = –2

k = –18

d = – 112


13

13

d =12

t = –3225

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Topic4.3.1

Round UpRound Up

Slope is a measure of how steep a line is — it’s how many units up or down you go for each unit across.


If you go up or down a lot of units for each unit across, the line will be steep and the slope will be large (either large and positive if it goes up from left to right, or large and negative if it goes down from left to right).

1 topic 4.3.1 slope of a line. 2 topic 4.3.1 slope of a line california standard: 7.0: students...

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