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5 Linear Graphs and
Equations
5.1 Coordinates
Firstly, we recap the concept of ( x, y ) coordinates, illustrated in the following examples.
Example 1
On a set of coordinate axes, plot the points A (2, 3), B (0, 4), C (– 2, 3), D (– 1, – 2), E (– 3, 0), F (2, – 4)
Solution
The x -axis and the y -axis cross at the origin, (0, 0). To locate the point A (2, 3), go 2 units horizontally from the origin in the positive x -direction and then 3 units vertically in the positive y -direction, as shown in the diagram.
x
y
1
2
3
4
B (0, 4) A (2, 3)
D (– 1, – 2)
E (– 3, 0)
F (2, – 4)
0 1 2 3
5
C (– 2, 3)
Example 2
Identify the coordinates of the points A, B, C, D, E, F, G and H shown on the following grid:
x
y
1
2
3
4
G
0 1 2 3
5
E
F
D
A
C B
H
Solution
A (3, 1), B (0, 2), C (– 2, 2), D (– 3, 0),
E (– 2, – 4), F (0, – 2), G (2, – 3), H (2, 0)
Example 3
Marc has ten square tiles like this: Marc places all the square tiles in a row. He starts his row like this:
For each square tile he writes down the coordinates of the corner which has a.
2 cm
x
y
1 2 3 4 5 6 7 8 9
1
2
3
- On a set of coordinate axes, with x values from – 5 to 5, y values from – 5 to 5, plot the following points: A (2, 4), B (1, 2), C (– 2, 5), D (– 3, – 3), E (– 2, – 4), F (0, – 3), G (– 4, 0), H (2, – 3) What can you say about A, B and E?
- On a suitable set of coordinate axes, join the points (3, 0), (0, 4) and (– 3, 0). What shape have you made?
- Three corners of a square have coordinates (4, 2), (– 2, 2) and (4, – 4). Plot these points on a grid, and state the coordinates of the other corner.
- Three corners of a rectangle have coordinates (4, 1), (– 2, 1) and (– 2, – 3). Plot these points on a grid and state the coordinates of the other corner.
- Two adjacent corners of a square have coordinates (– 1, 1) and (2, 1). (a) What is the length of a side of the square? (b) What are the possible coordinates of the other two points?
- Daniel has some parallelogram tiles. He puts them on a grid, in a continuing pattern. He numbers each tile. The diagram shows part of the pattern of tiles on the grid.
2
4
6
0 2 4 6 8
1
2
3
y 1
x Daniel marks the top right corner of each tile with a. The coordinates of the corner with a on tile number 3 are (6, 6). (a) What are the coordinates of the corner with a on tile number 4? (b) What are the coordinates of the corner with a on tile number 20? Explain how you worked out your answer. (c) Daniel says: " One tile in the pattern has a in the corner at (25, 25). " Explain why Daniel is wrong.
(d) Daniel marks the bottom right corner of each tile with a. Copy and complete the table to show the coordinates of each corner with a.
(e) Copy and complete the statement: 'Tile number 7 has a in the corner at ( ....... , ....... ).' (f) Copy and complete the statement: 'Tile number ....... has a in the corner at (20, 19).' (KS3/99/Ma/Tier 4-6/P1)
- A robot can move about on a grid. It can move North, South, East or West. It must move one step at a time. The robot starts from the point marked on the grid below. It takes 2 steps. 1st step: West 2nd step: North It gets to the point marked. (a) The robot starts again from the point marked. It takes 2 steps. 1st step: South 2nd step: South Copy the grid below and mark the point it gets to with a.
1 step South 1 step West 1 step East
1 step North
Tile Number Coordinates of the Corner with a 1 (2, 1) 2 3 4
5.2 Straight Line Graphs
We look in this section at how to calculate coordinates and plot straight line graphs. We also look at the gradient and intercept of a straight line and the equation of a straight line. The gradient of a line is a measure of its steepness_._ The intercept of a line is the value where the line crosses the y-axis.
The equation of a straight line is y = m x + c , where m = gradient and c = intercept (where the line crosses the y -axis).
Example 1
Draw the graph with equation y = 2 x + 3.
Solution
First, find the coordinates of some points on the graph. This can be done by calculating y for a range of x values as shown in the table.
The points can then be plotted on a set of axes and a straight line drawn through them.
x
y
1
2
3
4
5
6
7
8
9
–3 –2–1 0 1 2 3 4
y = 2 x + 3
Step
Rise Intercept
y
x
Gradient = RiseStep
x – 2 – 1 0 1 2 3 y – 1 1 3 5 7 9
Example 2
Calculate the gradient of each of the following lines: (a) (b) (c) (d)
Solution
(a)
Rise = 6 Gradient = (^66)
= 1 Step = 6 (b)
Rise = 6 Gradient = (^63)
= 2 Step (^) = 3 (c)
Rise = 6 Gradient = 126
= 1 2 Step = 12 (d)
Rise = − 6 Gradient = − 2 6
= − 3 Step = 2
(b)
Gradient =^84 = 2
Intercept = − 1
So m = 2 and c = − 1. The equation is: y = m x + c y = 2 x + (^) (− 1 ) or y = 2 x − 1
Exercises
- (a) Copy and complete the following table for y = 2 x − 2.
(b) Draw the graph of y = 2 x − 2.
- Draw the graphs with the equations given below, using a new set of axes for each graph. (a) y = x + 3 (b) y = x − 4 (c) y = 4 x − 1 (d) y = 3 x + 1 (e) y = 4 − x (f) y = 8 − 2 x
- Calculate the gradient of each of the following lines, (a) - (g):
y
–5 – 4–3–2–1 (^1 2 34 5) x
1
2
3
4
5
6
7
8
9
Rise = 8
Step = 4
x – 2 – 1 0 1 3 5 y
(a) (b) (c) (d)
- Write down the equations of the lines with gradients and intercepts listed below: (a) Gradient = 4 and intercept = 2. (b) Gradient = 2 and intercept = – 5.
(c) Gradient = 12 and intercept = 1. (d) Gradient = – 1 and intercept = – 5.
- Copy and complete the following table, which gives the equation, gradient and intercept for a number of straight lines.
- (a) Plot the points A, B and C with coordinates: A (2, 4) B (7, 5) C (0, 10) and join them to form a triangle. (b) Calculate the gradient of each side of the triangle.
(e) (f)^ (g)
Equation Gradient Intercept y = 5 x + 7 3 – 2 y = − 3 x + 2 y = − 4 x − 2
- 2 3 1 2 1 y = 4 − x y = 10 − 3 x
- (a) On a set of axes, plot the points with coordinates (– 2, – 2), (2, 0), (4, 1) and (6, 2) and then draw a straight line through these points. (b) Determine the equation of the line.
- (a) On the same axes, draw the lines with equations y = 2 x + 3 and
y = 8 −^12 x. (b) Write down the coordinates of the point where the lines cross.
- The point A has coordinates (4, 2), the point B has coordinates (8, 6) and the point C has coordinates (5, 9). (a) Plot these points on a set of axes and draw straight lines through each point to form a triangle. (b) Determine the equation of each of the lines you have drawn.
- Look at this diagram:
x
y
5
10
0 5 10
C
F A
B E
D
(a) The line through points A and F has the equation y = 11. What is the equation of the line through points A and B?
(b) The line through points A and D has the equation y = x + 3. What is the equation of the line through points F and E? (c) What is the equation of the line through points B and C? (KS3/98/Ma/Tier 4-6/P1)
The s give the graph p = 3 s + 1. The s give the graph p = 2 s + 1. The s give the graph p = s + 1. Selma has 16 pins. (a) Use the correct graph to find the number of squares she can pin up with 4 pins in each square. How many squares can she pin up with 3 pins in each square? (b) The line through the points for p = 3 s + 1 climbs more steeply than the line through the points for p = 2 s + 1 and p = s + 1. Which part of the equation p = 3 s + 1 tells you how steep the line is? (c) On a copy of the grid at the beginning of this question, plot three points to show the graph for 8 pins in each square. (d) What is the equation of this graph? (KS3/95/Ma/Levels 6-8/P2)
Total Number of Pins (p)
Number of Squares (s)
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
44
46
48
50
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
4 PINS 3 PINS
2 PINS
(b) x^ + 6 4 = 3
x + 4 = 3 × 6 (multiplying both sides by 6) x + 4 = 18 x = 18 − 4 (subtracting 4 from both sides) x = 14
(c) 4 ( x + 4 ) = 18 4 x + 16 = 18 (removing brackets) 4 x = 18 − 16 (subtracting 16 from both sides) 4 x = 2 x = 24 (dividing both sides by 4)
x = (^12)
Example 3
Solve the following equations: (a) 4 x + 2 = 3 x + 5 (b) 4 x − 4 = 10 − 3 x
Solution
(a) 4 x + 2 = 3 x + 5 x + 2 = 5 (subtracting 3 x from both sides) x = 5 − 2 (subtracting 2 from both sides) x = 3
(b) 4 x − 4 = 10 − 3 x 7 x − 4 = 10 (adding 3 x to both sides) 7 x = 10 + 4 (adding 4 to both sides) 7 x = 14 x = 147 (dividing both sides by 7) x = 2
Example 4
Use graphs to solve the following equations:
(a) 4 x − 7 = 9 (b) x + 7 = 3 x − 3
Solution
(a) Draw the lines y = 4 x − 7 and y = 9.
The solution is given by the value on the x- axis immediately below the point where y = 4 x − 7 and y = 9 cross.
The solution is x = 4.
(b) Draw the lines (^) y = x + 7 and (^) y = 3 x − 3.
The lines cross where x = 5 , so this is the solution of the equation.
10 9 8 7 6 5 4 3 2 1
(^0 1 2 3 4 5 6 )
y = 9
y = 4 x − 7
Solution x = 4
y
x
1 2 3 4 5 6 7
y
x
y = x + 7 11
12
13
10 9 8 7 6 5 4 3 2 1 0
y = 3 x − 3
Solution x = 5
- Solve the equation 2 x^ −^3 =^9 by drawing the graphs y^ =^2 x −^3 and y = 9.
- Use a graph to solve the equation 4 x − 5 = 3.
- (a) On the same set of axes, draw the lines with equations y = x + 1 and y = 2 x − 3. (b) Use the graph to find the solution of the equation x + 1 = 2 x − 3
- Use a graph to solve the following equations: (a) 2 x = − x + 3 (b) 4 − 2 x = 2 x − 8
- The following graph shows the lines with equations y = 2 x + 1 , y = x + 2 and y = 10 − x. y
x
10 9 8 7 6 5 4 3 2 1
(^0 01 2 3 4 5 6 7 8 9 )
Use the graph to solve the equations: (a) 2 x + 1 = 10 − x (b) x + 2 = 10 − x (c) 2 x + 1 = x + 2
- On the same set of axes, draw the graphs of three straight lines and use them to solve the equations: (a) 2 x − 2 = x + 3 (b) 2 x − 2 = 8 (c) x + 3 = 8
- Solve these equations. Show your working. (a) 4 − 2 y = 10 − 6 y (b) 5 y + 20 = (^3) ( y − (^4) ) (KS3/99/Ma/Tier 6-8/P1)
5.4 Parallel and Perpendicular Lines
In this section we consider the particular relationship between the equations of parallel lines and perpendicular lines. The key to this is the gradient of lines that are parallel or perpendicular to each other.
Example 1
(a) Draw the lines with equations y = x y = x + 4 y = x − 2 (b) What do the three equations have in common?
Solution
(a) The following graph shows the three lines: y
1 2 3 4 5
1
2
3
4
5
6
7
- 6
- 5
- 4
- 3
- 1
- 7
- 8
- 9
- 8 – 7– 6– 5– 4 –3–2–1 0 6 7 8 x
y = x + 4
y = x
y = x − 2
(b) Note that the three lines are parallel, all with gradient 1. All the equations of the lines contain ' x '. This is because the gradient of each line is 1, and so the value of m in the equation y = m x + c is always 1.
