Straight Line Graphs

y
10
Consider two points on this graph.
8
(4, 4) 6
4
(6, 4)
m y2 y1 x2 x1
2
– 10 – 8 – 6 – 4 – 2 –2 –4
2 4 6 8 10 x
44 64
0
–6
–8
The equation of the line is y 4
y=15x+40
Now think about the total cost for a range of different hire times:
If we put these values into a table,we can plot a graph.
A point lying on the line
– 10
All horizontal lines have an equation of the form y c
Consider two points on this graph.
y
10
8
(4,8)
6
m y2 y1 x2 x1
4 2
– 10 – 8 – 6 – 4 – 2 –2
2 4 6 8 10 x
Gradient is - 4 and the y axis intercept is 2 .
3
3
Calculate the gradient of each line.
Calculate the gradient of the line that passes through both points in each case.
86 44
(undefined)
–4
–6
(4, 6)
–8
The equation of the line is x 4
– 10
All vertical lines have an equation of the form x a
All straight lines have an equation of the form y mx c
m = gradient
y axis intercept
Undefined and zero gradient
Gradient is a measure of slope. If a line has zero gradient it has zero slope. A line with zero slope is horizontal.
A point not lying on the line
Points lying on the straight line
(x, y) : symbolic representation of a point
y-coordinate
x-coordinate
y=15x+40 (Equation of a straight line)
Back to Main Page
If we find the 2 points where the graph cuts the axes then we can plot the line.
These points are where x = 0 (anywhere along the y axis) and y = 0 (anywhere along the x axis).
8
2x + y = 4
7
6
5
4
3
2
1
–7 –6 –5 –4 –3 –2 –1 -1 -2 -3 -4 -5 -6
x 1 2 3 4 5 6 7 8
Back to Main Page
Exercise
Plot the following graphs using the x=0, y=0 method. 1) x + y = 5 2) x + 2y = 2 3) 2x + 3y = 6 4) x + 3y = 3
–7 –6 –5 –4 –3 –2 –1 -1 -2 -3 -4 -5 -6
x 1 2 3 4 5 6 7 8
Back to Main Page
Exercise
Using the x = 0, y = 0 method plot the following graphs:
1) x + y = 4 2) 2x + y = 2 3) x + 2y = 2 4) x + 3y = 6 5) 2x + 5y = 10
Click to reveal plotted lines
Back to Main Page
Answers
1. 3x + 2y = 6 2. x + 2y = 2 3. 2x + 3y = 6 4. x - 3y = 3
Click for further exercises
y
8 7 6 5 4 3 2 1
x 1 2 3 4 5 6 7 8
Back to Main Page
Example
By substituting these values into the equation we can find the other half of the co-ordinates.
Question: Draw the graph of 2x + y = 4
1) y = x + 2 2) y = x – 3 3) y = 2x + 4 4) y = 2x – 3 5) y = 3x + 1 6) y = 3x – 2
7) y = 1 – x 8) y = 1 – 2x
2
9) y = 2 – 3x 10) y = x + 1
2
Back tBiblioteka Main Page(a) A (1, 2) and B (3, 8) (b) A (2, -1) and B (4, 2)
(c) A (0, 6) and B(3, 9) (d) A(-1, -4) and B(-3, 2)
Exercise
y
8
7
3
Click for Answers 1) y = x - 2
6
2) y = -x + 3
Back to Main Page
The Table Method
Use the table method to plot the following lines:
1) y = x + 3 2) y = 2x – 3
x012
3) y = 2 – x
y
4) y = 3 – 2x
Click to reveal plotted lines
Back to Main Page
The Table Method
4 3 2 1
-4 -3 -2 -1 0 1 2 3 4 -1
-2
-3
3
1
Click for further
-4
exercises
4
2
Back to Main Page
Further Exercise
Using the table method, plot the following graphs.
For every position P the gradient of AP is
m yc yc
x0
x
x y c mx y mx c
Straight Lines: standard form
y = mx + c
To see what the constant, m, tells us, consider how y changes when x changes:
Solution x=0
y=0
2(0) + y = 4
2x + 0 = 4
y=4
2x = 4
x=2
1st Co-ordinates = (0,4)
2nd Co-ordinates = (2,0)
Back to Main Page
y So the graph will look like this.
y135
x = 2 y = 2(2) +1 y = 5
Now you just have to plot the points on to a graph!
Back to Main Page
The Table Method
x012
y 135
y = 2x + 1
4 3 2 1
-4 -3 -2 -1 0 1 2 3 4 -1 -2 -3 -4
The Table Method
We can use an equation of a line to plot a graph by substituting values of x into it.
Example
y = 2x + 1
x012
x = 0 y = 2(0) +1 y = 1 x = 1 y = 2(1) +1 y = 3
6) x – y = 3 7) 2x – y = 2 8) 2x – 3y = 6 9) x + 2y = 1 10) 2x – y = 3
Back to Main Page
B(x2, y2 )
y2 y1
A(x1, y1)
x2 x1
Gradient m = Vertical Height Horizontal Distance