Module - Linear Equation...

32
LAST REVISED November., 2008 Copyright This publication © The Northern Alberta Institute of Technology 2002. All Rights Reserved. Algebra Module A45 Linear Equation Theory - 1

Transcript of Module - Linear Equation...

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LAST REVISED November., 2008

Copyright This publication © The Northern Alberta Institute of Technology 2002. All Rights Reserved.

Algebra Module A45

Linear Equation Theory - 1

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Module A45 − Linear Equation Theory - 1

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Linear Equation Theory - 1 Statement of Prerequisite Skills Complete all previous TLM modules before beginning this module.

Required Supporting Materials Access to the World Wide Web. Internet Explorer 5.5 or greater. Macromedia Flash Player.

Rationale Why is it important for you to learn this material? This module begins to introduce the student to the relationship between an algebraic equation and the properties of the graph of that equation. A straight line is a simple enough concept to grasp. The properties of a straight line such as slope, midpoint, and length are concepts that have many technical applications and will be applied in any future study of plane analytic geometry.

Learning Outcome When you complete this module you will be able to… Analyze the characteristics of a straight line.

Learning Objectives 1. Apply basic terms necessary for graphing linear equations. 2. Sketch the graph of a linear function using both the table method and the xy intercept

method. 3. Determine the slope of an oblique line given its graph. 4. Determine the slope of an oblique line given two points on the line. 5. Determine the distance between any two points on a line. 6. Determine the midpoint of a line segment.

Connection Activity Consider all the straight lines you encounter in a day; the edge of a wall; the rise in a road, the slope of a roofline. What properties did the engineers who constructed these straight lines have to understand?

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Module A45 − Linear Equation Theory - 1

OBJECTIVE ONE When you complete this objective you will be able to… Apply basic terms necessary for graphing linear equations.

Exploration Activity Cartesian Coordinate System In earlier modules, equations have been solved and their solutions have been expressed algebraically. These solutions may be expressed pictorially using a graph. To display these solutions graphically we use the Cartesian coordinate system that is constructed by the intersection of two real number lines (called axes) perpendicularly at their “zeros” on a flat surface (called a plane). The point of intersection of these lines is called the origin and the four regions that these lines divide the plane into are called Quadrants. The system is depicted below. The system is also referred to as the rectangular coordinate system.

x

y

5 10 15

-5

-5

5

-10

-10

10

-15

-15

15

Vertical Axis

Quadrant IQuadrant II

Quadrant III Quadrant IV

Horizontal AxisO rigin

2

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Module A45 − Linear Equation Theory - 1

Plotting Points To locate a point we need a pair of numbers such as (4, 3). The first number, called the abscissa, is a distance in the direction of the horizontal axis (right or left). The second number, called the ordinate, is a distance in the direction of the vertical axis (up or down). The ordered pairs (4, 3) and (3, 4) represent different points. The numbers that locate a point are called coordinates of the point. Remember, it always takes two numbers to locate one point. The horizontal axis is called the x axis. The vertical axis is called the y axis.

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Determine whether a given ordered pair satisfies a given linear equation. Consider the equation x + y = 5. This equation states an algebraic relationship between a pair of numbers (x and y). Their sum is 5. Many pairs of numbers satisfy this relationship. To determine if a given ordered pair satisfies a given relationship, use the same procedure as used in checking the solution to an equation.

EXAMPLE 1 Does (1, 4) satisfy x + y = 5? Substituting 1 for x and 4 for y we get:

LHS RHS x + y = 1 + 4 = 5

5

Therefore LHS = RHS and we say that (1, 4) satisfies x + y = 5. Comment: Saying that an ordered pair satisfies an equation also means that the ordered pair makes the equation a true statement.

(5,−4)

y

−8

−6

−4

−2

2 4 6 8

2

4

6

8

−4 −6 −2 −8

x

Quadrant Quadrant I

Quadrant III Quadrant IV

Horizontal Axis

Origin (0,0)

Vertical Axis

(4, 3)

(−6,−5)

(+, +) (−, +)

(−, −) (+, −)

(−3,2)

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Module A45 − Linear Equation Theory - 1

Experiential Activity One 1. The x-coordinate of the point A(x,y) is called the _________. 2. If p(x,y) represents a point that is 5 units below the x-axis, what is the value of y? 3. In the following figure, what is the ordered pair corresponding to G?

x

y

5 10 15

-5

-5

5

-10

-10

10

-15

-15

15

G(x, y)

4. Which ordered pair satisfies y = 2(x – 3)? Show Me.

a. (−4, −1) b. (2, −2) c. (−4, 1) d. (0, 0)

5. Which quadrant contains (−2, 1)?

a. I b. II c. III d. IV

6. Quadrants I and IV are separated from quadrants II and III by the _____________ axis. 7. The value of the ordinate of point A(3, 4) is

a. 6 b. 4 c. 2 d. none of these

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Module A45 − Linear Equation Theory - 1

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Experiential Activity One Answers 1. Abscissa

2. −5

3. (8, −6)

4. b

5. b

6. y

7. b

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Module A45 − Linear Equation Theory - 1

OBJECTIVE TWO When you complete this objective you will be able to… Sketch the graph of a linear function using both the table method and the xy intercept method.

Exploration Activity Method A: Sketching a Graph Using a Table of Values Given y = −x + 2. Draw the graph of the solution to y = −x + 2. We must construct a table and choose values of x. For each x value chosen we then calculate the corresponding y value by substituting the x value into the given equation. Usually we choose x values between −5 and +5. It is sufficient to choose 3 or 4 values for x. y = −x + 2 In the following table the x's are chosen arbitrarily, and they are used to calculate the y's.

let x = −3 0 1 2 3 then y = 5 2 1 0 −1

For instance we substituted x = −3 into y = −x + 2 to obtain y = 5. Similarly we calculated the remaining y values. It only requires 2 points to define a straight line, the other points serve as a check. Plot the points indicated in the above table and you will get the following graph:

x2 4 6

-2

-2

2

-4

-4

4

-6

-6

6

y

(0, 2)

(3, -1)

y = -x + 2

6

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Module A45 − Linear Equation Theory - 1

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Method B. Sketching a Graph using the X and Y intercepts

X Intercept By definition, the x intercept is where a graph crosses the x axis. To determine the x intercept, set the value of y to zero in the given equation and solve for x.

Y Intercept By definition, the y intercept is where a graph crosses the y axis. To determine the y intercept, set the value of x to zero in the given equation and solve for y.

x intercept y intercept

Set y = 0 and solve for x: y = −x + 2 0 = −x + 2 therefore x = 2 (2, 0)

Set x = 0 and solve for y: y = −x + 2 y = −0 + 2 therefore y = 2 (0, 2)

Place the ordered pairs (2, 0) and (0, 2) on the coordinate axis and join them. In the graph shown, both methods produce the same line. See the graph on the previous page. Note: The xy intercept method will not work if the line passes through (0, 0). For an example let us sketch the graph of x – 2y = 0.

EXAMPLE 1 Sketch: x − 2y = 0

for x intercept: for y intercept:

set y = 0 and solve for x: set x = 0 and solve for y: x − 2y = 0 x − 2y = 0 x − 2(0) = 0 0 − 2y = 0 therefore x = 0 therefore y = 0 (0, 0) (0, 0)

This example produces only one point, (0,0). Since one point is not enough to define a straight line another ordered pair must be determined using the table method. Pick a value for x, say x = 2, and substitute x = 2, into x − 2y = 0 to get: 2 − 2y = 0 −2y = −2 y = 1 Thus we have the point (2, 1). Plot the points (0, 0) and (2, 1) and join them to get the following graph:

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Module A45 − Linear Equation Theory - 1

x

y

2 4 6

-2

-2

2

-4

-4

4

-6

-6

6

(2, 1)(0, 0)

x - 2y = 0

NOTE: x = k is the equation of a VERTICAL line passing through k on the x-axis. y = k is the equation of a HORIZONTAL line passing through k on the y-axis.

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Module A45 − Linear Equation Theory - 1

Experiential Activity Two Using either the table method or the x y intercept method, sketch the graph of each of the following: 1. y = −3x + 2 2. y = 3x + 1 3. 2x − 3y = 4 Show Me. 4. x = −2 5. y = 5 6. x – y + l = 0 7. −2y + 4x = 8

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Module A45 − Linear Equation Theory - 1

Experiential Activity Two Answers 1.

x

y

2 4 6

-2

-2

2

-4

-4

4

-6

-6

6

2.

x

y

2 4 6

-2

-2

2

-4

-4

4

-6

-6

6

10

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Module A45 − Linear Equation Theory - 1

3.

x

y

2 4 6

-2

-2

2

-4

-4

4

-6

-6

6

4.

x

y

2 4 6

-2

-2

2

-4

-4

4

-6

-6

6

11

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Module A45 − Linear Equation Theory - 1

5.

x

y

2 4 6

-2

-2

2

-4

-4

4

-6

-6

6

6.

x

y

2 4 6

-2

-2

2

-4

-4

4

-6

-6

6

12

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Module A45 − Linear Equation Theory - 1

7.

x

y

2 4 6

-2

-2

2

-4

-4

4

-6

-6

6

13

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Module A45 − Linear Equation Theory - 1

OBJECTIVE THREE When you complete this objective you will be able to… will be able to… Determine the slope of an oblique line given its graph. Determine the slope of an oblique line given its graph.

Exploration Activity Exploration Activity The slope of a line is defined to be the ratio of the change in y to the change in x between any two points on the line. This is called rise over run, where the rise is y2 − y1 and the run is x2 − x1. In formula form we have:

The slope of a line is defined to be the ratio of the change in y to the change in x between any two points on the line. This is called rise over run, where the rise is y2 − y1 and the run is x2 − x1. In formula form we have:

2 1

2 1

y ym slopex x−

= =−

y EXAMPLE 1

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Given: the graph of a line determine the slope of the line.

NOTE: P1(xl, y1) from the graph is (1, 1) and P2(x2, y2) from the graph is (3, 4). Substituting these values into the slope formula we have:

2 1

2 1

4 13 132

y ym slopex x−

= =−

−=

=

= 32

so for a rise of 3 units we have a run of 2 units.

NOTE: 1. If the slope works out to be an integer, like 7, this means the rise is 7 for a run of 1. 2. Also, express the run as a positive number. For example, if a line has a slope of −7/3, then its rise is −7 and its run is 3.

(3, 4)

−4

−3

−2

−1

1 2 3 4

1

2

3

4

−2 −3 −1 −4

x

(1, 1)

P2 (x2, y2)

P1 (x1, y1)

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Module A45 − Linear Equation Theory - 1

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Module A45 − Linear Equation Theory - 1

Experiential Activity Three From the graphs given at the end of this module determine the slopes of the following: 1. Figure 1 5. Figure 9 Show Me. 2. Figure 2 6. Figure 10 3. Figure 3 7. Figure 11 4. Figure 4 8. Figure 12

Experiential Activity Three Answers 1. −1

2. 1

3. −1

4. 56

5. 25

6. −1

7. 1

8. 3/2

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Module A45 − Linear Equation Theory - 1

OBJECTIVE FOUR When you complete this objective you will be able to… Determine the slope of an oblique line given two points on the line.

Exploration Activity Using the slope formula from objective 3 the problem becomes one of identifying (x2, y2) and (x1, y1), and then substituting these values into the formula.

EXAMPLE 1 Find the slope of the straight line joining the points (−8, 22) and (15, 43). SOLUTION: Let (x2, y2) be the point (−8, 22), then (x1, y1) is (15, 43). The slope is:

2 1

2 1

22 438 15

2123

y ym slopex x−

= =−

−=− −

=

i.e. the line rises 21 units for every 23 units it moves horizontally. Now we will let (x2, y2) be (15, 43) and then (x1, y1) = (−8, 22). This will show the student it does not matter which point you choose as (x2, y2).

( )43 22

15 82123

m −=

− −

=

Same as the slope above.

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Module A45 − Linear Equation Theory - 1

Experiential Activity Four Determine the slope of the lines that pass through the given points 1. (3, 6) and (−10, 5)

2. (−6, 0) and (30, 69)

3. (150, −1) and (43, 65)

4. ( 12

, −3) and (4, 78

− ) Show Me.

5. (8.26, −1) and (0, −100)

NOTE: Slope questions from the Computer will prompt you for 2 answers: rise and run. If the slope is negative put the minus sign with the rise, always enter the run as a positive quantity. Also, reduce fractions to lowest terms.

Experiential Activity Four Answers

1. 1

13

2. 2312

3. 66

107−

4. 1728

5. 99

8.26

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Module A45 − Linear Equation Theory - 1

OBJECTIVE FIVE When you complete this objective you will be able to… Determine the distance between any two points on a line.

Exploration Activity Distance Formula

x

y

2 4 6

-2

-2

2

-4

-4

4

-6

-6

6

a

bc

P2 (6, 5)

P1 (3, 2) P3 (x2, y1)

To get the formula for the distance between P1 and P2; i.e. the length of Pl P2, we employ Pythagorean Theorem. From the above diagram we see that: P2 = (x2, y2) = (6, 5) Pl = (x1, y1) = (3, 2) and P3 = (x2, yl ) = (6, 2) Also c = distance P1, P2 a = distance P1P3 = (x2 – x1) b = distance P2P3 = (y2 – y1)

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Module A45 − Linear Equation Theory - 1

Using this information and Pythagorean Theorem we have:

( )

2 2 2

2 2

222 1 2 1( )

c a b

c a b

c x x y y

= +

= +

= − + −

c represents the distance P1, P2, so the distance formula is:

( )222 1 2 1( )d x x y y= − + −

where d is the commonly accepted letter used to represent the distance formula. This formula is very powerful. It gives the distance between any 2 points on a line. NOTE: If you haven’t used the Pythagorean Theorem before, you will learn about it in trigonometry. For now, you just need to know the distance formula. We simply use Pythagorean Theorem here to show how the distance formula was developed. Specifically for the distance between points Pl and P2 we have:

Pl = (3, 2) P2 = (6, 5) x1 = 3, y1 = 2, x2 = 6, y2 = 5

( )22

2 2

(6 3) 5 2

3 3

9 9

184.2426

d = − + −

= +

= +

==

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Module A45 − Linear Equation Theory - 1

Experiential Activity Five Five 1. Determine the length of the line segment whose endpoints are (2, −3) and (2, 4). 1. Determine the length of the line segment whose endpoints are (2, −3) and (2, 4). 2. Determine the length of the line segment, whose endpoints are (−3, 5) and (−7, −1). 2. Determine the length of the line segment, whose endpoints are (−3, 5) and (−7, −1). 3. The coordinates of the endpoints of a diameter of a circle are (−3, 5) and (2, −3).

Determine the length of a radius of the circle. 3. The coordinates of the endpoints of a diameter of a circle are (−3, 5) and (2, −3).

Determine the length of a radius of the circle.

2

diameterradius =

radius diameter 4. Determine the length of a diagonal of a square with vertices (−3, 2), (2, 1), (1, −4)

and (−4, −3). Show Me.4. Determine the length of a diagonal of a square with vertices (−3, 2), (2, 1), (1, −4)

and (−4, −3). Show Me. 5. The endpoints of a radius of a circle are (2, 3) and (−1, 1). Determine the length of

the diameter of the circle.

Experiential Activity Five Answers 1. 7 2. 2 13 7.2111=

3. 89 4.71702

=

4. 2 13 7.2111= 5. 2 13 7.2111=

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Module A45 − Linear Equation Theory - 1

OBJECTIVE SIX When you complete this objective you will be able to… Determine the midpoint of a line segment.

Exploration Activity MIDPOINT FORMULA: The midpoint of the line joining (x2, y2) and (x1, y1) is found by using the formula:

1 2 1

2 22x x yx y+ +

= =y

where x and y are the coordinates of the midpoint.

EXAMPLE 1 Find the midpoint of the line joining the points (6, −8) and (−40, 3). SOLUTION: Let (x2, y2) = (−40, 3), then (x1, yl) = (6, −8)

6 40 34so, 172 2

8 3 5 2.502 2

x

y

− −= = = −

− + −= = = −

EXAMPLE 2 Determine the coordinates of the centre of the circle that has the points (6, 10) and (−8, −12) as the endpoints of a diameter. SOLUTION: Let (x2, y2) = (6, 10) and so (x1, yl) = (−8, −12)

8 6 12

12 10 12

x

y

− += = −

− += = −

therefore the center of the circle is at (−1, −1).

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Module A45 − Linear Equation Theory - 1

Experiential Activity Six 1. Find the midpoint of the straight line joining the points (−100, 56) and (−5, 300).

2. A line segment has endpoints (4, 3) and (−3, 5). Determine the coordinates of the

midpoint of the segment.

3. Determine the coordinates of the centre of a circle if a diameter has endpoints (5, 1) and (−3, 7). Show Me.

Experiential Activity Six Answers 1. x = −52.5

y = 178

2. 1 ,42

⎛ ⎞⎜ ⎟⎝ ⎠

3. (1, 4)

Practical Application Activity Complete the linear equation theory - 1 assignment in TLM.

Summary This module presented the student with a method of plotting points, sketching straight lines, finding slopes, mid-points, and distances. It should be evident to the student that a straight line is a simple enough concept to grasp, but now we have started to analyze some of its characteristics as mentioned above. The theory is continued further in the next module where we see methods for determining the equations of straight lines.

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Appendix 1.

x

y

5 10 15

-5

-5

5

-10

-10

10

-15

-15

15

P1

P2

2.

x

y

5 10 15

-5

-5

5

-10

-10

10

-15

-15

15

P1

P2

3.

x

y

5 10 15

-5

-5

5

-10

-10

10

-15

-15

15

P1

P2

4.

x

y

5 10 15

-5

-5

5

-10

-10

10

-15

-15

15

P1

P2

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5.

x

y

5 10 15

-5

-5

5

-10

-10

10

-15

-15

15

P1 P2

6.

x

y

5 10 15

-5

-5

5

-10

-10

10

-15

-15

15

P1

P2

7.

x

y

5 10 15

-5

-5

5

-10

-10

10

-15

-15

15

P1

P2

8.

x

y

5 10 15

-5

-5

5

-10

-10

10

-15

-15

15

P1 P2

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9.

x

y

5 10 15

-5

-5

5

-10

-10

10

-15

-15

15

P1

P2

10.

x

y

5 10 15

-5

-5

5

-10

-10

10

-15

-15

15

P1

P2

11.

x

y

5 10 15

-5

-5

5

-10

-10

10

-15

-15

15

P1

P2

12.

x

y

5 10 15

-5

-5

5

-10

-10

10

-15

-15

15

P1

P2

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13.

x

y

5 10 15

-5

-5

5

-10

-10

10

-15

-15

15

P1

P2

14.

x

y

5 10 15

-5

-5

5

-10

-10

10

-15

-15

15

P1 P2

15.

x

y

5 10 15

-5

-5

5

-10

-10

10

-15

-15

15

P1

P2

16.

x

y

5 10 15

-5

-5

5

-10

-10

10

-15

-15

15

P1

P2

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17.

x

y

5 10 15

-5

-5

5

-10

-10

10

-15

-15

15

P1 P2

18.

x

y

5 10 15

-5

-5

5

-10

-10

10

-15

-15

15

P1

P2

19.

x

y

5 10 15

-5

-5

5

-10

-10

10

-15

-15

15

P1

P2

20.

x

y

5 10 15

-5

-5

5

-10

-10

10

-15

-15

15

P1

P2

Page 32: Module - Linear Equation Theorytlment.nait.ca/tlm40FileStore/Math/Asset/25000/index_htm.25038/A45... · This equation states an algebraic relationship between a ... Linear Equation