The Straight Line

0

10

20

30

40

50

60

70

80

90

0 60 120 180

Time in Minutes

Dis

tanc

e in

Mile

s fr

om L

ondo

n

T

D

Reminder: Gradient = Change in vertical distance Change in horizontal distance

You can start and end anywhere on the line.

2

1

4

2m

2

1

6

3m

2

1

4

2m

2

1

10

5m

rise run

=

0

10

20

30

40

50

60

70

80

90

0 60 120 180

Time in Minutes

Dis

tanc

e in

Mile

s fro

m L

ondo

n

T

D

What happens when the line slopes down?

In this case the gradient is negative.

3

2

12

8m

Reminder: Gradient Formula12

12

xx

yym

x

y

13

303

3612

12

xx

yym

22

4

)1()3(

3712

12

xx

yym

(0, 3)

(3, 6)(-3, 7)

(-1, 3)

What are the gradients of these two lines?

Positive gradient1

Negative gradient-2

Now check the gradients using the formula.

Gradient Exercise

Reminder: Sketching Lines

Given the equation of a line, we can sketch it by making a table and finding points which lie on the line.

We usually find three points.

Example: Sketch the line .23 xy

x 0 2 4

y -2 4 10

(0, -2) (2, 4) (4, 10)We plot the points

-4

-3

-2

-1

0

1

2

3

4

5

6

7

8

9

10

11

12

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

x

y

(0, -2)

(2, 4)

(4, 10)

.23 xy•Plot the points

•Draw line through the points

x

y

xy 2 2

x

y

xy 2 1

x

y

xy 2 3

x

y

xy 2 2

x

y

xy 2 4

x

y

xy 2 5

x

y

xy 2

x

y

xy 2 1

What do you notice?

What is the gradient of each line?

Where does each line cut the y axis?

We will now look at more lines and their equations

x

y

- 6

x

y

- 2

x

y

+ 1xy 3

Here are some more lines

What do you notice?

What is the gradient of each line?

Where does each line cut the y axis?

So far all the lines we have looked have been of the form

y= mx + c

gradienty-

intercept

We will look at this more closely using Autograph

Autograph.lnk

23 xy

cmxy m = 3 c = -2

What do you think the gradient of the line is?23 xy

-4

-3

-2

-1

0

1

2

3

4

5

6

7

8

9

10

11

12

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

x

y

.23 xy

Check this using the graph.

Where do you think it cuts the y axis?

Again check this using the graph.

What about ?3 xy

-4

-3

-2

-1

0

1

2

3

4

5

6

7

8

9

10

11

12

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

x

y

.3 xy

The equation of the line shown is

62 xy(a)

321 xy(b)

321 xy(c)

62 xy(d)

-4

-3

-2

-1

0

1

2

3

4

5

6

7

8

9

10

11

12

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

x

y

-4

-3

-2

-1

0

1

2

3

4

5

6

7

8

9

10

11

12

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

x

y

-4

-3

-2

-1

0

1

2

3

4

5

6

7

8

9

10

11

12

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

x

y

-4

-3

-2

-1

0

1

2

3

4

5

6

7

8

9

10

11

12

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

x

y

-4

-3

-2

-1

0

1

2

3

4

5

6

7

8

9

10

11

12

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

x

y

The graph of y = 3x + 1 is

A B

DC

Sorry that is incorrect!

Try again.

Well Done!

Click to continue.

Well Done!

Click to continue.

You join a video shop for a membership fee of £3 and then charge £2 for each video you hire.

We can draw a graph of Cost against Number of Videos by making a table.

No of Videos (N) 0 1 2 3 4 5 6Cost of Videos(£) (C)

3 5 7 9 11 13 15

Now draw a graph of the table above.

Consider the following problem

0123456789

10111213141516171819202122232425

0 1 2 3 4 5 6 7 8 9 10

Number of Videos

Co

st o

f V

ideo

s

N

C

We can use y = mx + c to find a formula for the cost of hiring any number of videos.

•Instead of x we have N.

•Instead of y we have C.

•What is the gradient of this line?

•What is the y-intercept?

What is the equation of this line?Answer:C = 2N + 3

21

2m

3c

32

32

3

NC

xy

mxy

You always have to pay £3.

For every square that you move to the right you go two squares up because the cost of each video is £2.

Were you correct?

What does the y-intercept tell us?

What does the gradient of 2 tell us?

Another problem: Find the equation of the line below.

• Write down the coordinates of 2 points on the line.

• Use the gradient formula to find m.

• Read off c.

• Write down the equation in the form y = mx + c.

• Write down the equation in terms of s and P.

-4

-3

-2

-1

0

1

2

3

4

5

6

7

8

9

10

11

12

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

s

P

Answer: 32

1 sP

Method:

Special Cases

Lines parallel to the x and y axes.

-4

-3

-2

-1

0

1

2

3

4

5

6

7

8

9

10

11

12

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

x

y

Using the gradient formula with (0, 3) and (6, 3) gives

03

036

3312

12

xx

yym

3

3)(0

y

xy

cmxy

Notice: the y coordinate of every point on this line is 3We say the equation

of the line is y = 3Also c = 3

3

0

c

mUsing y = mx + c

1. Parallel to the x-axis

-4-3-2-10123456789

101112

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10

x

y

Special Cases

Lines parallel to the x and y axes.

Using the gradient formula with (4, 2) and (4, 5) gives

undefined

xx

yym

0

344

2512

12

Notice: the x coordinate of every point on this line is 4

We say the equation of the line is x = 4

This is because you cannot divide by zero!This means you cannot use y = mx + c

2. Parallel to the y-axis

However……..