Straight Line Graphs

Straight Line GraphsSections

1) Horizontal, Vertical and Diagonal Lines (Exercises)

2) y = mx + c(Exercises : Naming a Straight Line

Sketching a Straight Line)

3) Plotting a Straight Line - Table Method (Exercises)

4) Plotting a Straight Line – X = 0, Y = 0 Method (Exercises)

5) Supporting Exercises Co-ordinates Negative Numbers Substitution

x

y

1

-5

-4

-3

-2

-1

4

3

2

1

-5 -4 -3 -2 0 2 3 54-1

Naming horizontal and vertical lines

(-4,-2) (0,-2) (-4,-2)

y = -2

(3,4)

(3,1)

(3,-5)x = 3

(x,y)

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1

-5

-4

-3

-2

-1

4

3

2

1

-5 -4 -3 -2 0 2 3 54-1

Now try these lines

(-4,2) (0,2) (-4,2)

y = 2

(-2,4)

(-2,1)

(-2,-5)x = -2

(x,y)

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y

x

-5

-4

-3

-2

-1

4

3

2

1

1-5 -4 -3 -2 0 2 3 54-1

See if you can name lines 1 to 5(x,y)

1

5 3

4

2 Back to Main Page

y

x

y = 1

x = 1 x = 5

y = -4

x = -4

1

-5

-4

-3

-2

-1

4

3

2

1

-5 -4 -3 -2 0 2 3 54-1

Diagonal Lines

(-4,-3) (0,1) (2,3)

(3,3)

(1,1)

(-3,-3)

y = -x

(x,y)

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(2,-2)

(-1,1)

(-3,3)

y = x

y = x + 1y

x

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1 2

1

-5

-4

-3

-2

-1

4

3

2

1

-5 -4 -3 -2 0 2 3 54-1

3

4

Now see if you can identify these diagonal lines

x

y

y = x - 1

y = x + 1

y = - x - 2

y = -x + 2

y = mx + cEvery straight line can be written in this form. To do this the values for m and c must be found.

y = mx + cc is known as the intercept

m is known as the gradient

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y

x 1 2 3 4 5 6 7 8

12345678

–7 –6 –5 –4 –3 –2 –1 -1-2-3-4-5-6

Find the Value of c

This is the point at which the line crosses the y-axis.

Find the Value of m The gradient means the rate at which the line is climbing.

Each time the lines moves 1 place to the right, it climbs up by 2 places.

Finding m and c

y = 2x +3y = mx +c

So c = 3

So m = 2

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y

x 1 2 3 4 5 6 7 8

12345678

–7 –6 –5 –4 –3 –2 –1 -1-2-3-4-5-6

Find the Value of c

This is the point at which the line crosses the y-axis.

Find the Value of m The gradient means the rate at which the line is climbing.

Each time the line moves 1 place to the right, it moves down by 1 place.

Finding m and c

y = 2x +3y = mx +cSo c = 2

So m = -1

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y

x 1 2 3 4 5 6 7 8

12345678

–7 –6 –5 –4 –3 –2 –1 -1-2-3-4-5-6

Line 1

m =

c =

Equation:

Some Lines to Identify

Line 2

m =

c =

Equation:

12

y = x + 2

Line 3

m =

c =

Equation:

1-1

y = x - 1

-21

y = -2x + 1Back to Main Page

y

x 1 2 3 4 5 6 7 8

12345678

–7 –6 –5 –4 –3 –2 –1 -1-2-3-4-5-6

Exercise

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Click for Answers

12

3

5

4

1) y = x - 2

2) y = -x + 3

3) y = 2x + 2

4) y = -2x - 1

5) y = -2x - 1

2

Further ExerciseSketch the following graphs by using y=mx + c

1) y = x + 42) y = x - 23) y = 2x + 14) y = 2x – 35) y = 3x – 2

6) y = 1 – x7) y = 3 – 2x8) y = 3x9) y = x + 2 210) y = - x + 1 2

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The Table MethodWe can use an equation of a line to plot a graph by substituting values of x into it.

Example

y = 2x + 1x = 0 y = 2(0) +1 y = 1x = 1 y = 2(1) +1 y = 3

x = 2 y = 2(2) +1 y = 5

Now you just have to plot the points on to a graph!

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x 0 1 2

y 1 3 5

The Table Method

0 1-1 432-2-3-4-1

-2-3

-4

1

23

4

y = 2x + 1

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x 0 1 2

y 1 3 5

The Table MethodUse the table method to plot the following lines:

1) y = x + 3

2) y = 2x – 3

3) y = 2 – x

4) y = 3 – 2x

Click to reveal plotted lines

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x 0 1 2

y

The Table Method

0 1-1 432-2-3-4-1

-2-3

-4

1

23

4

4

31

2 Back to Main Page

Click for furtherexercises

Further ExerciseUsing the table method, plot the following graphs.

1) y = x + 22) y = x – 33) y = 2x + 44) y = 2x – 35) y = 3x + 16) y = 3x – 2

7) y = 1 – x8) y = 1 – 2x

9) y = 2 – 3x10) y = x + 1

2

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2

This method is used when x and y are on the same side. Example: x + 2y = 4

The x = 0, y = 0 Method

To draw a straight line we only need 2 points to join together.

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These points are where x = 0 (anywhere along the y axis) and y = 0 (anywhere along the x axis).

If we find the 2 points where the graph cuts the axes then we can plot the line.

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y

x 1 2 3 4 5 6 7 8

12345678

-6 -5 -4 -3 -2 -1 -1-2-3-4-5-6

This is where the graph cuts the y – axis (x=0)

This is where the graph cuts the x – axis (y=0)

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By substituting these values into the equation we can find the other half of the co-ordinates.

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ExampleQuestion: Draw the graph of 2x + y = 4

Solutionx = 02(0) + y = 4

y = 4

1st Co-ordinate = (0,4)

y = 0

2x + 0 = 4

2x = 4

x = 2

2nd Co-ordinate = (2,0)

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So the graph will look like this.y

x 1 2 3 4 5 6 7 8

12345678

–7 –6 –5 –4 –3 –2 –1 -1-2-3-4-5-6

2x + y = 4

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ExercisePlot 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

Click to reveal plotted lines

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Answersy

x 1 2 3 4 5 6 7 8

12345678

–7 –6 –5 –4 –3 –2 –1 -1-2-3-4-5-6

1. 3x + 2y = 62. x + 2y = 23. 2x + 3y = 64. x - 3y = 3

Click for furtherexercises

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Exercise

1) x + y = 4

2) 2x + y = 2

3) x + 2y = 2

4) x + 3y = 6

5) 2x + 5y = 10

6) x – y = 3

7) 2x – y = 2

8) 2x – 3y = 6

9) x + 2y = 1

10) 2x – y = 3

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Using the x = 0, y = 0 method plot the following graphs:

What are the Co-ordinates of these points?

-11

-5

-4

-3

-2

5

4

3

2

1

-5 -4 -3 -2 0 2 3 54-1

(x,y)

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Negative Numbers

(1) 2 + 3 (2) 6 - 5 (3) 3 - 7 (4) -2 + 6

(5) -1 - 2 (6) -4 + 5 (7) -2 - 2 (8) 0 – 4

(9) -3 + 6 (10) -4 - 1 (11) 6 - 8 (12) -5 - 2

(13) -8 + 4 (14) -5 - (- 2) (15) 0 - (- 1)

(16) 7 - 12 + 9 (17) -4 - 9 + -2 (18) 14 - (- 2)

(19) -45 + 17 (20) 4 - 5½

Addition and Subtraction

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Negative Numbers

(1) 4 x -3 (2) -7 x -2

(3) -5 x 4 (4) 28 ÷ -7(5) -21 ÷ -3 (6) -20 ÷ 5(7) -2 x 3 x 2 (8) -18 ÷ -3 x 2

(9) -2 x -2 x -2 (10) 2.5 x -10

Multiplication and Division

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Substituting Numbers into Formulae

Exercise

Substitute x = 4 into the following formulae:

1) x – 2

2) 2x

3) 3x + 2

4) 1 – x

5) 3 – 2x

6) 4 - 2x

7) x - 3 2

8) 3 - x 29) 2x – 6

Click forward to reveal answers

2814-3-5

-4-1

12

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Substituting Negative Numbers into FormulaeExercise

Substitute x = -1 into the following formulae:

1) x – 22) 2x3) 3x + 24) 1 – x5) 3 – 2x

6) 4 - 2x7) x - 3

2

8) 3 - x 2

9) 2x – 6

Click forward to reveal answers

-3-2-1 2 5

6-3½

3½-8

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