Transcript Straight Line Graphs

Straight Line Graphs
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S4 Credit
The gradient Vertical ÷ Horizontal
The gradient from Coordinates
Drawing Straight line graphs
General Equation y = mx + c
Equation from 2 Points
Modelling Situations
Best – fitting straight line
Exam Type Questions
Starter Questions
S4 Credit
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Q1.
Calculate 7 – 5 x 2
2
1
2 1
5
2
Q2.
Calculate
Q3.
Is this triangle right angled ?
Explain
9
8
5
30-Apr-20
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The Gradient of a Line
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S4 Credit
Learning Intention
1. To show how to get the
equations for horizontal
and vertical lines.
Success Criteria
1. Understand that vertical
lines have equations of the
form x = a
2. Understand that horizontal
lines have equations of the
form y = a
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Equations for Vertical and Horizontal lines
y
x = -6
10
9
8
7
6
5
4
3
2
1
-10 -9
-8 -7 -6 -5
-4 -3
-2 -1
0
-1
-2
-3
-4
-5
-6
-7
-8
-9
-10
30-Apr-20
1
2
3
4
5
6
7
8
9
Mark the points
on your grid below
and then join them
together
Vertical lines
have equations
of the form
10 x
x= 8
Horizontal lines
have equations
of the form
y= 2
y = -9
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Dept
The Gradient
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Now Try MIA
Exercise 2.1
Ch7 (page 136)
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Starter Questions
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S4 Credit
Q1.
-2
Write this number in full to 1 sig. figs 2.5  10
Q2.
Calculate the volume of the triangular prism.
20cm
12cm
8cm
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The Gradient of a Line
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S4 Credit
Learning Intention
1. To explain how to calculate
the gradient using a right
angle triangle
Success Criteria
1. Gradient is :
change in vertical height
divided by
change in horizontal distance
2. Calculate simple gradients.
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The Gradient Difference in
y -coordinates
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S4 Credit
The gradient is the measure of steepness of a line
Change in vertical height
Change in horizontal distance
V

H
The steeper a line the bigger theDifference
gradient in
x -coordinates
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The Gradient
S4 Credit
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6
V
m

H
V
m

H
4
2
-6
-4
-2
0
2
4
6
-2
-4
30-Apr-20
-6
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3
4
3
2
V
3
m

H 5
1
V
2
m
 3
H 6
The Gradient
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S4 Credit
Now Try MIA
Exercise 3.1
Ch7 (page 139)
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Starter Questions
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S4 Credit
Q1.
A house 4 years ago is valued at £50 000.
Calculate it’s value if it has increased by 5%.
Q2.
Calculate 3.36 x 70 to 2 significant figures.
Q3.
Write down the 3 ways of factorising.
30-Apr-20
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The Gradient of a Line
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S4 Credit
Learning Intention
1. To explain positive and
negative gradients using
coordinates.
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Success Criteria
1. Know gradient formula.
2. Calculate gradients given
two coordinates.
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The gradient
using coordinates
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S4 Credit
y-axis
y2
(x2 , y2 )
y2  y1
(x1 , y1 )
y2  y1
m
x2  x1
x2  x1
y1
O
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m = gradient
x1
x-axis
x2
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13
The gradient
using coordinates
S4 Credit
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Find the gradient of the line.
y-axis
(5,10)
(2, 4)
O
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x-axis
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m = gradient
y2  y1
m
x2  x1
10  4
m
52
6
m 2
3
14
The gradient
using coordinates
S4 Credit
Find the gradient of the two lines.
y2  y1
m
x2  x1
82
m
3  (1)
6
m
 3
2
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y-axis
(-3,8)
(3,8)
(1,2)
(-1,2)
O
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x-axis
y2  y1
m
x2  x1
82
m
3 1
6
m 3
2
15
The gradient
using coordinates
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S4 Credit
The gradient formula is :
y2  y1
gradient = m 
x2  x1
It is a measure of how steep a line is
A line sloping up from left to right is a positive gradient
A line sloping down from left to right is a negative gradient
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The gradient
using coordinates
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S4 Credit
Now try MIA
Ex 4.2
Ch7 (page 143)
30-Apr-20
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Straight Line Graphs
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S4 Credit
Learning Intention
1. To draw graphs by using a
coordinate table
Success Criteria
1. Understand the keypoints of
drawing a straight line graph
2. Be able to plot a straight
line graph
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Drawing Straight Line Graphs y = ax + b
y
y=x
10
y = 3x+1
9
8
x -2 0 2
7
x
0 3
-4
y
0 3
-4
6
y -5 1 7
5
4
3
2
1
-10 -9
y=x-3
x 0 4
8
y -3 1 5
-8 -7
-6 -5
-4 -3 -2
-1
0
1
2
3
4
5
6
7
-1
-2
-3
-4
-5
-6
-7
-8
-9
-10
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8
9
10
x
y = 2x
x 0
3
-4
y
6
-8
0
Straight Line Graphs
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S4 Credit
Key Points
1. Make a table
2. Calculate and plot 3 coordinates
3. Draw a line through points
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Straight Line Graphs
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S4 Credit
Now try MIA
Ex5.1 Q5 Only
Ch7 (page 145)
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Starter Questions
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S4 Credit
Q1.
Write out in full to 2 sig. figs. 4.53  105
Q2.
A superstore make 20% profit on each can of
soup they sell. If they buy in a can for 50p.
What is the selling price.
Q3.
Explain why 4x2 – 16 = 4(x + 2)( x – 2)
30-Apr-20
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The Straight Line Equation
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S4 Credit
Learning Intention
1. To explain the connection
between the straight line
equation and the gradient.
30-Apr-20
Success Criteria
1. Understand the term
standard form.
2. Identity the gradient m
from the standard form.
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Straight line equation and the gradient connection
y
10
y = -x - 5
x
0
1
9
y = 2x + 1
8
7
3
6
y -5 -6 -8
5
4
x 0
1
3
y
3
7
1
3
2
1
-10 -9
-8 -7
-6 -5
-4 -3 -2
-1
0
1
2
3
4
5
6
7
8
9
10
x
-1
y2  y1
m
x2  x1
m = -1
-2
-3
-4
-5
-6
-7
-8
-9
-10
30-Apr-20
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y2  y1
m
x2  x1
m=2
Straight Line Equation
S4 Credit
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All straight lines have the equation of the form
y = mx + c
Let’s investigate properties
(You need GeoGebra to run link)
30-Apr-20
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Straight Line Equation
y
S4 Credit
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All straight lines have
the equation of the form
lines are
parallel if
they have
the same
gradient
Where line
Gradient
meets y-axis
y = mx + c
10
9
8
7
6
5
4
3
2
1
0
1
2
3
4
5
6
7
8
9
10
x
Find the equations of the following lines
y = x
30-Apr-20
y = x+4
y = 4x+2
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y = -0.5x+2
The Gradient of a Line
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S4 Credit
Now try MIA
Ex 6.1
Ch7 (page 146)
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Starter Questions
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S4 Credit
Q1.
Write out in full to 1 significant fig. 1.58  10-5
Q2.
A computer store buys in a laptop for £500.
They want to make a 40% profit.
How much do they sell it for.
Q3.
A line is parallel to y = 2x.
Write down its equation
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lines are
parallel ifStraight
same gradient
S4 Credit
Line Equation
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All straight lines have the equation of the form
y = mx + c
Slope left to
right upwards
positive
gradient
Gradient
Slope left to
right downwards
negative
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gradient
y - intercept
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y intercept is
were line cuts
y axis
Equation of a Straight Line
y = mx+c
S4 Credit
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To find the equation of a straight line we need to know
Two points that lie on the line
( x1, y1) and ( x2, y2)
Or
The gradient and a point on the line
m and (a,b)
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Equation of a Straight Line
y = mx+c
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S4 Credit
Find the equation of the straight line passing through
the points (4, 4) and (8,24).
Solution
y2  y1 24  4 20
m


5
x2  x1
84
4
Using the point (4,4) and the gradient m = 5
sub into straight line equation
y = mx + c

4 = 5 x4 + c
Equation : y = 5x - 16

c = 4 - 20 = -16
Equation of a Straight Line
y = mx+c
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S4 Credit
Find the equation of the straight line passing through
the points (3, -5) and (6,4).
Solution
m
y2  y1 4  (5) 9

 3
x2  x1
6 3
3
Using the point (6,4) and the gradient m = 3
sub into straight line equation
y = mx + c

4 = 3 x6 + c
Equation : y = 3x - 14

c = 4 - 18 = -14
Straight Line Graphs
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S4 Credit
Now try MIA
Ex 6.2
Ch7 (page 147 )
30-Apr-20
Created by Mr.Lafferty Maths Dept
Starter Questions
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S4 Credit
Q1.
The points ( 1, 4) and (3, 11) lie on a line.
Find the gradient of the line.
Q2.
Complete the table given :
x
y
Q3.
-3
0
3
Are the two lines parallel. Explain your answer
y=x+2
30-Apr-20
y = 3x + 1
and y = 2x + 2
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Modelling Using
Straight Line Equation
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S4 Credit
Learning Intention
1. How to model real life
situations using straight line
theory
Success Criteria
1. Be able to work out gradient
and y intercept using a
graph.
2. Form an equation for any
straight line graph.
30-Apr-20
Created by Mr.Lafferty Maths Dept
Modelling Real – life
Pick any
2 points
The cost for hiring a plumber per hour is shown below
y 2  y1
100  30

x 2  x1
70
 10
10
(a)
Calculate the gradient.
(b)
What is the value of C
when T = 0
30
(c)
Write down an equation
connecting C and T.
(7,100)
C
100
90
Cost £
m
80
70
60
50
40
(0,30) 2030
10
C=
(d)
T +
0
1
2
3
4
6
7
8
9
Time (hours)
Find the cost for a plumber for 10 hours?
T = 10
5
C = 10x10 + 30 = £130
10
T
Modelling Real – life
Pick any
2 points
The graph shows how a the volume of water tank drains over time.
y 2  y1
x 2  x1

40  80
80
-5

5
(a)
Calculate the gradient.
(b)
What is the value of V
when T = 0
80
(c)
Write down an equation
connecting V and T.
V
100
(0,80) 90
Volume
m
80
70
60
(8,40)
50
40
30
20
10
V=
(d)
0
T +
1
2
3
4
5
6
7
8
9
10
T
Time (mins)
How long before the tank is empty?
V=0
0 = -5xT + 80
T = (-80) ÷(–5) = 16mins
Straight Line Equation
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S4 Credit
Now try Ex 7.1
MIA (page 149)
30-Apr-20
Created by Mr.Lafferty Maths Dept
Starter Questions
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S4 Credit
Q1.
The points ( 5, 7) and (7, 21) lie on the same line.
Find the gradient of the line.
Q2.
Complete the table given :
x
y
Q3.
-3
3
Are the two lines parallel. Explain answer
y = -2x + 1
30-Apr-20
0
y = 2x + 10
and y = 2x + 1
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Best Fit
Straight Line Equation
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S4 Credit
Learning Intention
1. How to model real life
situations using straight line
theory
Success Criteria
1. Be able to work out gradient
and y intercept using a
graph.
2. Form an equation for any
straight line graph.
30-Apr-20
Created by Mr.Lafferty Maths Dept
Best-Fitting Straight line
Data was collected on pupils weight and height. Data is plotted below.
(b)
y  y1
m 2
x 2  x1
180  0

 2.57
2.57
70  0
c=0
h=
(d)
w +
Height (cms)
(a)
Pick any
Is there correlation between height and weight.
2 points
Yes, a positive correlation
h
200
Draw in the best-fit line and
180
x (70,180)
160
find an equation relating
140
height and weight.
120
100
80
60
40
20
(0,0) 0x 10
20 30 40 50 60 70 80 90 100
Weight (Kgs)
What height is a pupils who weights 40 kgs?
w = 40
h = 2.57x40 = 102.8 cms
w
Best-Fitting Straight line
A survey was carried out on the value of cars depending on their age
The data is plotted below.
(b)
y  y1
m 2
x 2  x1
8  18


80
-1.25
1.25
c = 18
v=
(d)
value £ ‘ 000
(a)
Pick any
Is there correlation between value and age.2 points
Yes, a negative correlation
v
20
Draw in the best-fit line and
18
x(0,18)
16
find an equation relating
14
value and year.
12
y+
10
x (8,8)
8
6
4
2
0
1
2
3
4
5
6
7
8
9
10
Years
What is the cost of a car after 4years?
y=4
v = -1.25x 4 + 18 = 13
Value = £13 000
y
Best Fit
Straight Line Equation
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S4 Credit
Now try Ex 8.1
MIA (page 151)
30-Apr-20
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Exam Type
Straight Line Questions
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S4 Credit
a)
b)
Example
The graph shows an electrician’s charging system.
He has a call-out charge plus an hourly rate.
Work out the equation of the line.
Calculate the cost of a 10 hour job.
Exam Type
Straight Line Questions
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S4 Credit
a) Pick two convenient points on the line, say (0, 30) and (5, 80).
Thinking of the form y = mx + c
The y-intercept gives c = 30
The equation is:
y = 10x + 30
80  30
m
= 10
50
The label on the y-axis is C and the label on the x-axis is T
So the equation is C = 10 T + 30
So the equation is C = 10 T + 30
b) Find C when T = 10
C = 10 x 10 + 30 = 130
The cost of a 10 hr job is £130.
Exam Type
Straight Line Questions
S4 Credit
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1. The tank of a car contains 5 litres of petrol.
The graph below shows how the volume of petrol in
this tank changes as a further 45 litres of petrol is
pumped in at a steady rate for 60 seconds.
Find the equation of the straight line in terms of V and t.
Exam Type
Straight Line Questions
S4 Credit
2. In the diagram below.
A is the point (-1, -7) and B is the point (4, 3).
(a)
Find the gradient of the
line AB.
(b)
AB cuts the y-axis at
the point (0, -5).
Write down the equation
of the line AB.
(c)
The point (3k, k) lies on
the line AB.
Find the value of k.
Exam Type
Straight Line Questions
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S4 Credit
3. A water pipe runs between two buildings. These are
represented by the points A and B in the diagram below.
(a)
Using the information in the diagram, show that the
equation of the line AB is 3y – x = 6.
Exam Type
Straight Line Questions
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S4 Credit
(b) An emergency outlet pipe has to be built across the
main pipe. The line representing this outlet pipe has
equation
4y + 5x = 46.
Calculate the coordinates of the point on the diagram
at which the outlet pipe will cut across the main water
pipe.
Exam Type
Straight Line Questions
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S4 Credit