164594628-Straight-Line-Equation-ppt.ppt

Simple, and strong foundation for pre college mathematics passionate pupil.

1. Straight Line Equation.
x
y2
2
3
y x
x
y
y = -2x + 3

2. 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 6
Cost of Videos(£)3 5 7 9 1113 15
Now draw a graph of the table above.

3. 1
4
3
2
5
6
7
8
9
10
11
12
13
14
0123456 No ofVideos
Cost
Graph of videos
hired against cost.

4. 1
4
3
2
5
6
7
8
9
10
11
12
13
14
0123456 No ofVideos
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.

5. 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 111315
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.

6. 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 6
Cost 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:

7. Gradient of 3.
Y axis intercept
(0,5)
1
4
3
2
5
6
7
8
9
10
11
12
13
14
0123456 No ofVideos
Cost
C = 3V + 5
Is the equation of the line.

8. More About The Gradient.
The gradient (m) of a straight line is defined to be:distance.horizontalinchange
height.verticalinchange
m
Change in vertical height.
Change in horizontal distance.
We are going to use this definition to calculate the gradient
of various straight lines:

9. 3
Find the gradients of the straight lines below:
(1) distance.horizontalinchange
height.verticalinchange
m
4
3
m =
4
4
(2)
7
4
m =
7
4
(3)
4
4
m =
4
m = 1

10. (5)
8
6
m =
6
8
=
3
4
(6)
9
3
m =
3
9= 3

11. Negative Gradient
Consider 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

12. Calculate the gradients of the lines below:
(1)
-4
55
4
m
(2)
-8
63
4
6
8


m

13. 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 = 6c = 5
(2) y = 4x + 2
m = 4c = 2
(3) y = x -3
m = 1c = -3

14. 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 + cy = x + 2
x
y
(2)2
1
m
c = 11
2
1
y x

15. (3)
x
y2
3
m
c = -22
2
3
y x
(4)
x
y m = -2
c = 3
y = -2x + 3

16. x
y
(5)3
4
m
c = 66
3
4
y x
(6)
x
y3
2
m
c = 22
3
2
y x

17. The Gradient Formula.
The Gradient Formula.
Look at the diagram below:
x
1 x
2
y
1
y
2
It shows a straight line passing
through the points (x
1,y
1) and
(x
2,y
2).
We must calculate the gradient of
the line using the triangle shown:
Change in vertical height:
y
2-y
1
y
2–y
1
Change in horizontal distance:
x
2-x
1
x
2-x
1
Gradient of line:12
12
xx
yy
m




18. 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
yy
m



Substitute in your values:410
612
m



Calculate and simplify:1
6
6
m 
(2) C(-4,8) and D(6,-10)
Solution:12
12
xx
yy
m


 6-4-
(-10)-8
m 10-
18
m 5
9
m

19. 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
yy
m


 3
2
6
48
612
m 



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

20. Find the equation of a straight line passing through C(6,-7) and D(-12,9)
Solution.
Calculate the gradient:12
12
xx
yy
m


 3
8
6
16
612
7)(9
m 



Substitute gradient into y = m x + c.cx
3
8
y 
Now substitute one of the points into
y = m x + c to find c.
Sub’ (6,-7) into equation:c6
3
8
7-  c36821-  -6948--21c3 
c = -23
Equation of the straight line:23
3
8
xy