Math 301 Module 6: Geometry and Unit Conversion

Math 301 Module 6: Geometry and Unit Conversion Department of Mathematics College of the Redwoods June 12, 2011 Contents 6 Geometry and Unit Conver...
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Math 301 Module 6: Geometry and Unit Conversion Department of Mathematics College of the Redwoods June 12, 2011

Contents

6 Geometry and Unit Conversion 6a Applications . . . . . . . . . . . Units of Length . . . . . . . Applications — Geometry . . Application — Area . . . . . Volume of a Prism . . . . . . Consecutive Integers . . . . . Area of a Parallelogram . . . Area of a Triangle . . . . . . Area of a Trapezoid. . . . . . Height of a Triangle . . . . . Circumference of a Circle . . Area of a Circle . . . . . . . The Pythagorean Theorem . 6b Graphs . . . . . . . . . . . . . . Plotting Points . . . . . . . . Graphing Linear Equations . Linear Equations . . . . . . .

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Module

6

Geometry and Unit Conversion Module 6 begins with a review of the units of length common to the American system. Next follows a review of unit conversions and applications to geometry. In Skill 6b, plotting points in the Cartesian coordinate system is reviewed. Ren´e Descartes (1596-1650) was a French philosopher and mathematician. As a philosopher, he is famous for the saying “Cogito ergo sum” (“I think, therefore I am”), and his writings led many to consider him the Father of Modern Philosophy. Even today, a number of his writings are standard faire in university philosophy departments. However, it is Descartes’ work in mathematics that form the basis for this chapter, particularly his invention of the Cartesian Coordinate System which bears his name. Descartes’ invention of the coordinate system created an entirely new branch of mathematics called analytic geometry, which established a permanent link between the plane and solid geometry of the ancient Greeks and the algebra and analysis of modern mathematics. As a result of his work, mathematicians were able to describe curves with equations, unheard of before Descartes’ invention of the coordinate system. Rather than describing a circle as the “locus of all points equidistant from a given point,” mathematicians were now able to refer to a circle centered at the point (0, 0) with radius r as the graph of the equation x2 + y 2 = r2 . The bridge created between geometry and analysis as a result of Descartes’ methods laid the groundwork for the discovery of the calculus by Newton and Leibniz. For his efforts, mathematicians often refer to Descartes as the Father of Analytic Geometry. In this chapter we will introduce readers to the Cartesian coordinate system and explain the correspondence between points in the plane and ordered pairs of numbers. Once an understanding of the coordinate system is sufficiently developed, we will develop the concept of the graph of an equation. In particular, we will address the graphs of a class of equations called linear equations.

1

2

6a

MODULE 6. GEOMETRY AND UNIT CONVERSION

Applications

The most common units of length in the American system are inch, foot, yard, and mile. Converting from one unit of length to another is a requisite skill in geometry and real world applications.

Units of Length American Units of Length. Facts relating common units of length. 1 foot (ft) = 12 inches (in)

1 yard (yd) = 3 feet (ft)

1 mile (mi) = 5280 feet (ft)

You Try It! Change 48 inches to feet.

EXAMPLE 1. Change 24 inches to feet. Solution. Multiply by the conversion factor 1 ft/12 in. 24 in = 24 in · 1 1 ft = 24 in · 12 in 1 ft = 24 ! in · 12 ! in 24 · 1 ft = 12 = 2 ft

Answer: 4 feet

Multiplicative Identity Property. Replace 1 with 1 ft/12 in. Cancel common unit. Multiply fractions. Simplify.

Hence, 24 inches equals 2 feet. ! A summary of conversion factors for units of length is given in Table 6.1. Convert feet to inches yards to feet miles to feet

Conversion Factor 12 in/1 ft 3 ft/1 yd 5280 ft/1 mi

Convert inches to feet feet to yards feet to miles

Conversion Factor 1 ft/12 in 1 yd/3 ft 1 mi/5280 ft

Table 6.1: Conversion factors for units of length. Some conversions require more than one application of a conversion factor. You Try It! Change 8 yards to inches.

EXAMPLE 2. Change 4 yards to inches.

3

6A. APPLICATIONS

Solution. We multiply by a chain of conversion factors, the first to change yards to feet, the second to change feet to inches. 3 ft 12 in · 1 yd 1 ft 3! ft 12 in = 4! yd · !· ft 1! yd ! 1! 4 · 3 · 12 = in 1·1 = 144 in

4 yd = 4 yd ·

Multiply by conversion factors. Cancel common units. Multiply fractions. Simplify.

Hence, 4 yards equals 144 inches.

Answer: 288 inches !

Applications — Geometry Solving many real world problems require some geometry. We start our review with the perimeter of a polygon, in particular, a rectangle. Perimeter of a Polygon. In geometry a polygon is a plane figure made up of a closed path of a finite sequence of segments. The segments are called the edges or sides of the polygon and the points where two edges meet are called the vertices of the polygon. The perimeter of any polygon is the sum of the lengths of its sides.

You Try It! EXAMPLE 3. A quadrilateral (four sides) is a rectangle if all four of its angles are right angles. It can be shown that the opposite sides of a rectangle must be equal. Find the perimeter of the rectangle shown below, where the sides of the rectangle are measured in meters.

3m

5m Solution. To find the perimeter of the rectangle, find the sum of the four sides. Because opposite sides have the same length, we have two sides of length 5 meters and two sides of length 3 meters. Hence, Perimeter = 5 + 3 + 5 + 3 = 16.

A rectangle has length 12 meters and width 8 meters. Find its perimeter.

4 Answer: 40 meters

MODULE 6. GEOMETRY AND UNIT CONVERSION

Thus, the perimeter of the rectangle is 16 meters. Note that the perimeter of a rectangle can be found by summing twice the length and width of the sides. This is given by the formula P = 2L + 2W . !

Application — Area Consider the rectangle shown in Figure 6.1. The length of this rectangle is four inches (4 in) and the width is three inches (3 in).

One square inch (1 in2 )

3 in

4 in Figure 6.1: A rectangle with length 4 inches and width 3 inches. To find the area of the figure, we can count the individual units of area (in2 ) that make up the area of the rectangle, twelve square inches (12 in2 ) in all. However, it is much faster to multiply the number of squares in each row by the number of squares in each column: 4 · 3 = 12 square inches. The argument presented above leads to the rule for finding the area of a rectangle. Area of a Rectangle. Let L and W represent the length and width of a rectangle, respectively. L

W

W

L To find the area of the rectangle, calculate the product of the length and width. That is, if A represents the area of the rectangle, then the area of the rectangle is given by the formula A = LW.

5

6A. APPLICATIONS

You Try It! EXAMPLE 4. A rectangle has width 5 feet and length 12 feet. Find the area of the rectangle. Solution. Substitute L = 12 ft and W = 5 ft into the area formula.

A rectangle has width 17 inches and length 33 inches. Find the area of the rectangle.

A = LW = (12 ft)(5 ft) = 60 ft2 Answer: 561 square inches.

Hence, the area of the rectangle is 60 square feet.

!

Volume of a Prism The natural next step is to look at the volume of a solid. The first one we consider is the rectangular solid or prism. You Try It! EXAMPLE 5. Pictured below is a rectangular prism.

The surface area of the prism pictured in this example is given by the following formula: S = 2(W H + LH + LW )

H

W L The volume of the rectangular prism is given by the formula V = LW H, where L is the length, W is the width, and H is the height of the rectangular prism. Find the volume of a rectangular prism having length 12 feet, width 4 feet, and height 6 feet. Solution. Following “Tips for Evaluating Algebraic Expressions,” first replace all occurrences of of L, W , and H in the formula V = LW H

If L = 12, W = 4, and H = 6 feet, respectively, calculate the surface area.

6

MODULE 6. GEOMETRY AND UNIT CONVERSION

with open parentheses. V =

!

"!

"!

"

Next, substitute 12 ft for L, 4 ft for W , and 6 ft for H and simplify. ! "! "! " V = 12 ft 4 ft 6 ft = 288 ft3

Answer: 288 square feet.

Hence, the volume of the rectangular prism is 288 cubic feet. ! We can simplify a number of formulas by combining like terms.

You Try It! A regular hexagon has six equal sides, each with length x. Find its perimeter in terms of x.

EXAMPLE 6. Find a formula for the perimeter P of the (a) rectangle and (b) square pictured below. Simplify your answer as much as possible. L s

W

s

W

s

s

L

Solution. The perimeter of any polygonal figure is the sum of the lengths of its sides. a) To find the perimeter P of the rectangle, sum its four sides. P = L + W + L + W. Combine like terms. P = 2L + 2W. b) To find the perimeter P of the square, sum its four sides. P = s + s + s + s. Combine like terms. P = 4s. Answer: P = 6x ! Sometimes a variable in a formula is given in terms of a another variable. The following example illustrates replacing a variable with an expression containing another (related) variable.

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6A. APPLICATIONS

You Try It! EXAMPLE 7. The length of a rectangle is three feet longer than twice its width. Find the perimeter P of the rectangle in terms of only its width. Solution. From the previous problem, the perimeter of the rectangle is given by P = 2L + 2W, (6.1)

The length L of a rectangle is 5 meters longer than twice its width W . Find the perimeter P of the rectangle in terms of its width W .

where L and W are the length and width of the rectangle, respectively. This equation gives the perimeter in terms of its length and width, but we’re asked to get the perimeter in terms of only its width. However, we’re also given the fact that the length is three feet longer than twice the width. That is, the length is described in terms of the width. Length

is

Three Feet

longer than

Twice the Width

L

=

3

+

2W

Because L = 3 + 2W , we can replace L with 3 + 2W in the perimeter formula. P = 2L + 2W P = 2(3 + 2W ) + 2W Apply the distributive property, then combine like terms. P = 6 + 4W + 2W P = 6 + 6W. This last equation gives the perimeter P in terms of only its width W .

Answer: P = 6W + 10 !

Consecutive Integers In the application of geometry to solve real world problems, the lengths of geometric figures may be expressed by algebraic expressions. Consecutive Integers. Let k represent an integer. The next consecutive integer is the integer k + 1. Thus, if k is an integer, then k + 1 is the next integer, k + 2 is the next integer after that, and so on. You Try It!

8 The three sides of a triangle are consecutive integers and the perimeter is 57 centimeters. Find the measure of each side of the triangle.

MODULE 6. GEOMETRY AND UNIT CONVERSION

EXAMPLE 8. The three sides of a triangle are consecutive integers and the perimeter is 72 inches. Find the measure of each side of the triangle. Solution. We follow the Requirements for Word Problem Solutions. 1. Set up a Variable Dictionary. In this case, a carefully labeled diagram is the best way to indicate what the unknown variable represents.

k+2

k

k+1 In our schematic diagram, we’ve labeled the three sides of the triangle with expressions representing the consecutive integers k, k + 1, and k + 2. 2. Set up an Equation. To find the perimeter P of the triangle, sum the three sides. P = k + (k + 1) + (k + 2) However, we’re given the fact that the perimeter is 72 inches. Thus, 72 = k + (k + 1) + (k + 2) 3. Solve the Equation. On the right, regroup and combine like terms. 72 = 3k + 3 Now, solve for k. 72 − 3 = 3k + 3 − 3 69 = 3k 3k 69 = 3 3 23 = k

Subtract 3 from both sides. Simplify. Divide both sides by 3. Simplify.

4. Answer the Question. We’ve only found one side, but the question asks for the measure of all three sides. However, the remaining two sides can be found by substituting 23 for k into the expressions k + 1 and k + 2. k + 1 = 23 + 1 = 24

and

k + 2 = 23 + 2 = 25

Hence, the three sides measure 23 inches, 24 inches, and 25 inches.

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6A. APPLICATIONS

5. Look Back. Does our solution make sense? Well, the three sides are certainly consecutive integers, and their sum is 23 inches + 24 inches + 25 inches = 72 inches, which was the given perimeter. Therefore, our solution is correct.

Answer: 18, 19, and 20 cm

!

Area of a Parallelogram Area of a Parallelogram. A parallelogram having base b and height h has area A = bh. That is, to find the area of a parallelogram, take the product of its base and height.

You Try It! EXAMPLE 9. Find the area of the parallelogram pictured below.

5/3 ft

The base of a parallelogram measures 14 inches. The height is 8/7 of an inch. What is the area of the parallelogram?

6 ft Solution. The area of the parallelogram is equal to the product of its base and height. That is, A = bh = (6 ft)

Area formula for parallelogram. #

5 ft 3

$

30 2 ft . 3 = 10 ft2 . =

Substitute: 6 ft for b, 5/3 ft for h. Multiply numerators and denominators. Divide.

Thus, the area of the parallelogram is 10 square feet.

Answer: 16 square inches !

Area of a Triangle Area of a Triangle. A triangle having base b and height h has area A = (1/2)bh. That is, to find the area of a triangle, take one-half the product of the base and height.

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MODULE 6. GEOMETRY AND UNIT CONVERSION

You Try It! The base of a triangle measures 15 meters. The height is 12 meters. What is the area of the triangle?

EXAMPLE 10. Find the area of the triangle pictured below.

6 cm 13 cm Solution. To find the area of the triangle, take one-half the product of the base and height. 1 bh 2 1 = (13 cm)(6 cm) 2 78 cm2 = 2 = 39 cm2 .

A=

Answer: 90 square meters

Area of a triangle formula. Substitute: 13 cm for b, 6 cm for h. Multiply numerators; multiply denominators. Simplify.

Therefore, the area of the triangle is 39 square centimeters. !

Area of a Trapezoid. Area of a Trapezoid. A trapezoid with bases b1 and b2 and height h has area 1 A = h (b1 + b2 ) . 2 That is, to find the area, sum the bases, multiply by the height, and take one-half of the result.

You Try It! A trapezoid has bases measuring 6 and 15 feet, respectively. The height of the trapezoid is 5 feet. Find the area of the trapezoid.

EXAMPLE 11. Find the area of the trapezoid pictured below. 2 21 in 3 in 4 41 in

11

6A. APPLICATIONS Solution. The formula for the area of a trapezoid is A=

1 h (b1 + b2 ) 2

Substituting the given bases and height, we get $ # 1 1 1 . A = (3) 4 + 2 2 4 2 Simplify the expression inside the parentheses first. Change mixed fractions to improper fractions, make equivalent fractions with a common denominator, then add. # $ 1 17 5 A = (3) + 2 4 2 # $ 17 5 · 2 1 + = (3) 2 4 2·2 $ # 1 17 10 = (3) + 2 4 4 # $# $ 27 1 3 = 2 1 4 Multiply numerators and denominators. =

81 8

This improper fraction is a perfectly good answer, but let’s change this result to a mixed fraction (81 divided by 8 is 10 with a remainder of 1). Thus, the area of the trapezoid is A = 10

1 square inches. 8 Answer: 52 21 square feet !

Height of a Triangle You Try It! EXAMPLE 12. The area of a triangle is 20 square inches. If the length of the base is 2 21 inches, find the height (altitude) of the triangle. Solution. We follow the Requirements for Word Problem Solutions.

The area of a triangle is 161 square feet. If the base of the triangle measures 40 41 feet, find the height of the triangle.

12

MODULE 6. GEOMETRY AND UNIT CONVERSION 1. Set up a Variable Dictionary. Our variable dictionary will take the form of a well labeled diagram.

h

2 21 in 2. Set up an Equation. The area A of a triangle with base b and height h is A=

1 bh. 2

Substitute A = 20 and b = 2 21 . 20 =

1 2

# $ 1 2 h. 2

3. Solve the Equation. Change the mixed fraction to an improper fraction, then simplify. # $ 1 1 20 = 2 h 2 2 # $ 1 5 20 = h 2 2 # $ 1 5 20 = · h 2 2 5 20 = h 4

Original equation. Mixed to improper: 2

1 5 = . 2 2

Associative property. Multiply:

5 1 5 · = . 2 2 4

Now, multiply both sides by 4/5 and solve. 4 4 (20) = 5 5 16 = h

#

5 h 4

$

Multiply both sides by 4/5. Simplify: and

4 (20) = 16 5

4 5 · = 1. 5 4

4. Answer the Question. The height of the triangle is 16 inches.

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6A. APPLICATIONS

5. Look Back. If the height is 16 inches and the base is 2 21 inches, then the area is # $ 1 1 A= 2 (16) 2 2 1 5 16 = · · 2 2 1 5 · 16 = 2·2 (5) · (2 · 2 · 2 · 2) = (2) · (2) 5 · 2" · 2" · 2 · 2 = "2 · "2 = 20 This is the correct area (20 square inches), so our solution is correct.

Answer: 8 feet !

Circumference of a Circle You Try It! EXAMPLE 13. Find the circumference of a circle given its radius is 12 feet. Solution. The circumference of the circle is given by the formula C = πd, or, because d = 2r, C = 2πr.

Find the radius of a circle having radius 14 inches. Use π ≈ 3.14

Substitute 12 for r. C = 2πr = 2π(12) = 24π Therefore, the circumference of the circle is exactly C = 24π feet. We can approximate the circumference by entering an approximation for π. Let’s use π ≈ 3.14. Note: The symbol ≈ is read “approximately equal to.” C = 24π ≈ 24(3.14) ≈ 75.36 feet It is important to understand that the solution C = 24π feet is the exact circumference, while C ≈ 75.36 feet is only an approximation.

Answer: 87.92 inches !

Area of a Circle You Try It! EXAMPLE 14. Find the area of a circle having a diameter of 12.5 meters. Use 3.14 for π and round the answer for the area to the nearest tenth of a square meter.

Find the area of a circle having radius 12.2 centimeters. Use π ≈ 3.14

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MODULE 6. GEOMETRY AND UNIT CONVERSION

Solution. The diameter is twice the radius. d = 2r Substitute 12.5 for d and solve for r. Substitute 12.5 for d.

12.5 = 2r 2r 12.5 = 2 2 6.25 = r

Divide both sides by 2. Simiplify.

Hence, the radius is 6.25 meters. To find the area, use the formula A = πr2 and substutite: 3.14 for π and 6.25 for r. A = (3.14)(6.25)2

Substitute: 3.14 for π, 6.25 for r.

= (3.14)(39.0625)

Square first: (6.25)2 = 39.0625.

= 122.65625

Multiply: (3.14)(39.0625) = 122.65625.

Hence, the approximate area of the circle is A = 122.65625 square meters. To round to the nearest tenth of a square meter, identify the rounding digit and the test digit. Test digit 122. 6 5 625 Rounding digit

Answer: 467.3576 cm2

Because the test digit is greater than or equal to 5, add 1 to the rounding digit and truncate. Thus, correct to the nearest tenth of a square meter, the area of the circle is approximately A ≈ 122.7 m2 . !

The Pythagorean Theorem We now state one of the most ancient theorems of mathematics, the Pythagorean Theorem. Pythagorean Theorem. Let c represent the length of the hypotenuse of a right triangle, and let a and b represent the lengths of its legs, as pictured in the image that follows.

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6A. APPLICATIONS

c

a

b The relationship involving the legs and hypotenuse of the right triangle, given by a2 + b 2 = c2 , is called the Pythagorean Theorem.

You Try It! EXAMPLE 15. Given the following right triangle, find the exact length of the missing side. 7

x

The hypotenuse and one leg of a right triangle measure 9 and 7 inches, respectively. Find the length of the remaining leg.

12

Solution. Note that the hypotenuse (across from the right angle) has length 12. This quantity should lie on one side of the Pythagorean equation all by itself. The sum of the squares of the legs go on the other side. Hence, x2 + 72 = 122 Solve the equation for x. x2 + 49 = 144 x2 + 49 − 49 = 144 − 49 2

x = 95 √ x = 95

Exponents first: 72 = 49 and 122 = 144. Subtract 49 from both sides. Simplify both sides.

Take the nonnegative square root of 95. √ Hence, the exact length of the missing side is 95.

Answer:

√ 32 inches !

16

6b

MODULE 6. GEOMETRY AND UNIT CONVERSION

Graphs

Plotting Points You Try It! Plot the following ordered pairs of whole numbers: (2, 2), (5, 5), and (7, 4).

EXAMPLE 1. Plot the following ordered pairs of whole numbers: (3, 2), (8, 6), and (2, 7). Solution. Create a Cartesian coordinate system on graph paper, then: • To plot the ordered pair (3, 2), start at the origin, then move 3 units to the right and 2 units up. • To plot the ordered pair (8, 6), start at the origin, then move 8 units to the right and 6 units up. • To plot the ordered pair (2, 7), start at the origin, then move 2 units to the right and 7 units up.

Answer: 8 7 6 5 4 3 2 1 0

The results are shown on the following Cartesian coordinate system.

(5, 5) (7, 4) (2, 2) 012345678

8 7 6 5 4 3 2 1 0

(2, 7) (8, 6)

(3, 2)

0 1 2 3 4 5 6 7 8

!

You Try It! EXAMPLE 2. What are the coordinates of the points P , Q, R, and S in the Cartesian coordinate system that follows?

17

6B. GRAPHS

5 Q

P Origin

-5

5 R

S -5

Solution. Make all measurements from the origin. • To obtain the coordinates of point P , start at the origin, move 3 units to the right, then 3 units up. Hence, the coordinates of the point P are (3, 3). • To obtain the coordinates of point Q, start at the origin, move 4 units to the left, then 3 units up. Hence, the coordinates of the point Q are (−4, 3). • To obtain the coordinates of point R, start at the origin, move 3 units to the left, then 4 units down. Hence, the coordinates of the point R are (−3, −4). • To obtain the coordinates of point S, start at the origin, move 4 units to the right, then 3 units down. Hence, the coordinates of the point S are (4, −3). These results are shown on the following Cartesian coordinate system. 5 Q (−4, 3)

P (3, 3)

-5

5

R (−3, −4)

S (4, −3) -5

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MODULE 6. GEOMETRY AND UNIT CONVERSION

Graphing Linear Equations Consider y = x + 1 an equation in two variables. If we substitute the ordered pair (x, y) = (1, 2) into the equation y = x + 1, that is, if we replace x with 1 and y with 2, we get a true statement. y =x+1 2=1+1

Original equation. Substitute: 1 for x and 2 for y.

2=2

Simplify.

We say that the ordered pair (1, 2) is a solution of the equation y = x + 1. Solution of an Equation in Two Variables. If substituting the ordered pair (x, y) = (a, b) into an equation (replace x with a and y with b) produces a true statement, then the ordered pair (a, b) is called a solution of the equation and is said to “satisfy the equation.”

You Try It! Which of the ordered pairs (1, 7) and (2, 9) are solution of the equation y = 3x + 4?

EXAMPLE 3. Which of the ordered pairs are solutions of the equation y = 2x + 5: (a) (−3, −2), or (b) (5, 15)?

Solution. Substitute the points into the equation to determine which are solutions. a) To determine if (−3, −2) is a solution of y = 2x + 5, substitute −3 for x and −2 for y in the equation y = 2x + 5. y = 2x + 5 −2 = 2(−3) + 5

Original equation. Substitute: −3 for x and −2 for y.

−2 = −1

Add: −6 + 5 = −1.

−2 = −6 + 5

Multiply first: 2(−3) = −6

Because the resulting statement is false, the ordered pair (−3, −2) does not satisfy the equation. The ordered pair (−3, −2) is not a solution of y = 2x + 5. a) To determine if (5, 15) is a solution of y = 2x + 5, substitute 5 for x and 15 for y in the equation y = 2x + 5. y = 2x + 5

Original equation.

15 = 2(5) + 5 15 = 10 + 5

Substitute: 5 for x and 15 for y. Multiply first: 2(5) = 10

15 = 15

Add: 10 + 5 = 15.

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6B. GRAPHS

The resulting statement is true. The ordered pair (5, 15) does satisfy the equation. Hence, (5, 15) is a solution of y = 2x + 5.

Answer: (1, 7) !

Ordered Pairs and the Graph. Because the graph of an equation is the collection of all ordered pairs that satisfy the equation, we have two important results: 1. If an ordered pair satisfies an equation, then the point in the Cartesian plane represented by the ordered pair is on the graph of the equation. 2. If a point is on the graph of an equation, then the ordered pair representation of that point satisfies the equation.

Linear Equations Let’s plot the graph of an equation. You Try It! EXAMPLE 4. Sketch the graph of y = −2x + 1. Solution. Select arbitrary values of x: −4, −3, . . . , 5. Substitute these values into the equation y = −2x + 1, calculate the resulting value of y, then arrange your results in a table.

y y y y y y y y y y

= −2(−4) + 1 = 9 = −2(−3) + 1 = 7 = −2(−2) + 1 = 5 = −2(−1) + 1 = 3 = −2(0) + 1 = 1 = −2(1) + 1 = −1 = −2(2) + 1 = −3 = −2(3) + 1 = −5 = −2(4) + 1 = −7 = −2(5) + 1 = −9

y = −2x + 1 x y (x, y) −4 9 (−4, 9) −3 7 (−3, 7) −2 5 (−2, 5) −1 3 (−1, 3) 0 1 (0, 1) 1 −1 (1, −1) 2 −3 (2, −3) 3 −5 (3, −5) 4 −7 (4, −7) 5 −9 (5, −9)

We’ve plotted the points in the table in Figure 6.2(a). There is enough evidence in Figure 6.2(a) to imagine that if we plotted all of the points that satisfied the equation y = −2x + 1, the result would be the line shown in Figure 6.2(b).

Sketch the graph of y = 2x − 2.

20 Answer:

MODULE 6. GEOMETRY AND UNIT CONVERSION

y

y

10

10

10

−10 −10

y 10

x

10

−10

−10 (a) Ten points that satisfy the equation y = −2x + 1.

x

10

−10

x

−10 (b) Plotting all points that satisfy the equation y = −2x + 1.

Figure 6.2: The graph of the equation y = −2x + 1 is a line. ! The graph of y = 2x + 5 in Figure ?? is a line. The graph of y = −2x + 1 in Figure 6.2(b) is also a line. This would lead one to suspect that the graph of the equation y = mx + b, where m and b are constants, will always be a line. Indeed, this is always the case. Linear Equations. The graph of y = mx + b, where m and b are constants, will always be a line. For this reason, the equation y = mx + b is called a linear equation.

You Try It! Which of the following equations is a linear equation? a) y = 2x3 + 5 b) y = −3x − 5

EXAMPLE 5. Which of the following equations is a linear equation? (a) y = −3x + 4, (b) y = 32 x + 3, and (c) y = 2x2 + 4. Solution. Compare each equation with the general form of a linear equation, y = mx + b. a) Note that y = −3x + 4 has the form y = mx + b, where m = −3 and b = 4. Hence, y = −3x + 4 is a linear equation. Its graph is a line. b) Note that y = 32 x + 3 has the form y = mx + b, where m = 2/3 and b = 3. Hence, y = 23 x + 3 is a linear equation. Its graph is a line.

21

6B. GRAPHS

c) The equation y = 2x2 + 4 does not have the form y = mx + b. The exponent of 2 on the x prevents this equation from being linear. This is a nonlinear equation. Its graph is not a line.

Answer: y = −3x − 5 ! You Try It!

EXAMPLE 6. Sketch the graph of y = 31 x + 3. 1 3x

+ 3 has the form y = mx + b. Therefore, the Solution. The equation y = equation is linear and the graph will be a line. Because two points determine a line, we need only find two points that satisfy the equation y = 13 x + 3, plot them, then draw a line through them with a ruler. We choose x = −6 and x = 6, calculate y, and record the results in a table. y x −6 6

y = 31 (−6) + 3 = −2 + 3 = 1 y = 31 (6) + 3 = 2 + 3 = 5

Sketch the graph of y = 32 x + 1.

= 13 x + 3 y (x, y) 1 (−6, 1) 5 (6, 5)

Plot the points (−6, 1) and (6, 5) and draw a line through them. The result is shown in Figure 6.3. y

Answer:

y 10

10 (6, 5)

10

−10

(−6, 1) 10

−10

x

x −10

−10 Figure 6.3: The graph of y = 13 x + 3 is a line.

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