Straight Line Equation.

x

y

22

3y x

x

y

y = -2x + 3

The Video Shop.

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

Complete the table below for the cost of hiring different numbers of videos.

No of videos 0 1 2 3 4 5 6Cost of Videos(£) 3 5 7 9 11 13 15

Now draw a graph of the table above.

1

4 3 2

56789

1011121314

0 1 2 3 4 5 6 No of Videos

Cost

Graph of videos hired against cost.

1

4 3 2

56789

1011121314

0 1 2 3 4 5 6 No of Videos

Cost Now consider the structure of the graph.

The graph cuts the y axis at (0,3) because it cost £3 to join the video shop before you hired any videos.

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

Finding A Formula.Look at the table of values for the video hire once again:

V 0 1 2 3 4 5 6 C 3 5 7 9 11 13 15

Find a formula for the cost of videos (C) given the number of videos (V) :

C = 2V + 3

From the graph we saw that :

This was the number of squares you went up for each one you went along.This is called the gradient of the line.

This was the place were the graph cut the y axis. This is called the y axis intercept.

Now repeat the question again for a video shop charging £5 to join and £3 for each video hired.Start by completing the table below.

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

Answer the questions below:

Where does the graph cut the y axis ?

( 0 , 5 )

What is the gradient of the line:

Gradient = 3

The full graph is shown on the next slide:

Gradient of 3.

Y axis intercept (0,5)

1

4 3 2

56789

1011121314

0 1 2 3 4 5 6 No of Videos

CostC = 3V + 5

Is the equation of the line.

More About The Gradient.The gradient (m) of a straight line is defined to be:

distance.horizontalinchange

height.verticalinchangem

Change in vertical height.

Change in horizontal distance.

We are going to use this definition to calculate the gradient of various straight lines:

3

Find the gradients of the straight lines below:

(1)

distance.horizontalinchange

height.verticalinchangem

4

3

m = 4

4

(2)

7

4

m = 7

4

(3)

4

4

m = 4

m = 1

(5)

8

6

m = 68 = 3

4

(6)

9

3

m =39 = 3

Negative GradientConsider the straight lines shown below:

(a) (c)(b) (d) (e)

Can you split the lines into two groups based on their gradients ?

Lines (a) (c) and (d) slope upwards from left to right.

Lines (b) and (e) slope downwards from left to right.

Positive gradient

Negative gradient

Calculate the gradients of the lines below:

(1)

- 4

5

5

4m

(2)

- 8

6

3

4

6

8

m

The Equation Of A Straight Line.To find the equation of any straight line we require to know two things:

(a) The gradient of the line.

(b) The y axis intercept of the line.

m = gradient.

c = y axis intercept.

The equation of a straight line is : y = m x + c

Examples.

Give the gradient and the y axis intercept for each of the following lines.

(1) y = 6x + 5

m = 6 c = 5

(2) y = 4x + 2

m = 4 c = 2

(3) y = x - 3

m = 1 c = - 3

Finding The Equation.Find the equation of the straight lines below:

x

y(1)

What is the gradient ? m = 1

What is the y axis intercept? c = 2

Now use y = m x + c y = x + 2

x

y(2)

2

1m

c = 1

12

1y x

(3)

x

y

2

3m

c = -2

22

3y x

(4)

x

y m = -2

c = 3

y = -2x + 3

x

y(5)

3

4m

c = 6

63

4y x

(6)

x

y

3

2m

c = 2

23

2y x

The Gradient Formula.The Gradient Formula.

Look at the diagram below:

x1 x2

y1

y2

It shows a straight line passing through the points (x1,y1) and (x2,y2).

We must calculate the gradient of the line using the triangle shown:

Change in vertical height:

y2-y1

y2 – y1

Change in horizontal distance:

x2-x1

x2-x1

Gradient of line:12

12

xx

yym

Calculate the gradient of the line through the points below:

(1) A(4,6) and B( 10,12)

Solution:

Write down the gradient formula:

Gradient of line:12

12

xx

yym

Substitute in your values:

410

612m

Calculate and simplify:

16

6m

(2) C(-4,8) and D(6,-10)

Solution:

12

12

xx

yym

6-4-

(-10)-8m

10-

18m

5

9m

Straight Line From Two Points. Find the equation of the straight line passing through (4,6) and (8,12)

Solution:

Find the gradient of the line:

12

12

xx

yym

32

6

48

612m

Substitute gradient into y = m x + c.

y = 3 x + c

Now substitute one of the points into y = m x + c to find c.

Sub’ (4,6) into y = 3x +c :

6 = 3 x 4 +c

c + 12 = 6

c = - 6

Now write down the equation of the straight line:

y = 3x - 6

Find the equation of a straight line passing through C(6,-7) and D(-12,9)

Solution.

Calculate the gradient:

12

12

xx

yym

3

8

6

16

612

7)(9m

Substitute gradient into y = m x + c.

cx3

8y

Now substitute one of the points into y = m x + c to find c.

Sub’ (6,-7) into equation:

c63

87-

c36821- -6948--21c3

c = -23

Equation of the straight line:

233

8 xy