Module 15
The Straight Line and The
Absolute Value Function
The straight line is the simplest geometric curve. Despite its simplicity, the straight line is a vital concept of mathematics and enters into our daily activities in numerous interesting and useful ways.
This module will teach you how find the equation of a line or write the equation of a graph of a linear function.
This module will also strengthen your understanding of the connection between equations and the linear functions they represent. Also, you will learn how to deal with real world data that tend to be linear.
Many types of calculations give us negative numbers. When we are interested in real quantities such as distance and location, there is a mathematical way to use these numbers without representing them as signed numbers. The concept of absolute value is used to describe such distances and locations.
The last part of this module focuses on the concept of the absolute value of a number, the graphs of absolute value functions, and solutions of equations involving absolute values.
This module has five lessons:
Lesson 1 Point-Slope Equation of a Line Lesson 2 Two-Point Equation of a Line Lesson 3 Intercept Equation of a Line
Lesson 4 Applications of Linear Functions/Equations Lesson 5 Absolute Value Functions
2
After going through this module, you are expected to:
obtain the equation of a line given the following:
i) the slope and a point ii) any two points
iii) intercepts
use linear equations in two variables to solve problems define an absolute value equation y = |x|
construct a table of ordered pairs and draw the graphs y = x y = |x| + b
y = x + b y = |x| – b y = x – b y = |x| + b y = |x| y = |x| – b
Follow the steps suggested in the flowchart below.
What you are expected to learn
How to learn from this module
Take the Pretest
Take the Posttest
Check your answers with the
Answer Key placed at the later part of this Module
Read each lesson and answer Self-check
Multiple Choice: Read the problems carefully and then select the letter of the correct answer. Write your answers on a separate sheet of pad paper.
1. What is the absolute value of the sum of –12 and 4?
A. -8 B. -16 C. 8 D. 16
2. What is the equation of the line passing through (0, -5) and with slope of 1?
A. y = -5x + 1 B. y = x – 5 C. y = -x + 5 D. y = 5x – 1
3. What is the equation of the graph?
A. 3
3 2
x y
B. 3
2
3
x
y
C. 3
2
3
x y
D. 3
3
2
x
y
4. Which of the following graphs is represented by the table of the ordered pairs?
x -1 1 0 -2 2
y -2 -2 -3 -1 -1
What to do before (Pretest)
Y
4
A. B.
C. D.
5. What is the table of ordered pairs of the equation y = |x| + 2?
6. The graph of the equation x - y = 5 crosses the y-axis at
A. -5 B. -1 C. 1 D. 5
7. What is the equation of the line that passes through the points (-2, 2) and (3,1)? A. x 0 -1 2 1 -4
y 2 3 4 3 6
B. x -2 -1 0 1 2 y 0 1 2 3 4
C. x -1 0 1 2 -3 y 1 2 3 4 -1
D. x -3 -2 -1 0 2 y 5 4 1 2 0
Y
X
Y
X
Y
X
Y
A. 5x + 5y = 8 B. 5x – 5y = 8 C. x + 5y = 8 D. 5x – y = 8
8. What is the equation of the graph at the right?
A. 3x2y 1
B. 2x3y1
C. 3x2y1
D. 2x3y1
9. The graph of a line through (-2, -2) and has a slope of -2/3 is:
A. B.
C. D.
10. What is the slope and the y-intercept of the graph at the right?
X Y
X Y
X Y
X Y
X Y
6
11. What is the equation of the line whose x-intercept is 1/3 and whose y-intercept is -1/4?
A. 3x – 4y = 1 B. 3x + 4y = 8 C. 4x – 3y = 1 D. 4x + 3y = 1
12. What is the slope and the coordinates of a point on the graph of the equation y - 2 = 3(x - 5)?
A. m=3; (2,5) B. m=3; (5,2) C. m=3; (-2,-5) D. m=-3; (-2,-5) 13. A 9-liter container of freshly squeezed orange juice costs P35. A 12-liter
container costs P45. How much would a 16-liter container cost?
A. Php42.15 B. Php48.33 C. Php55.00 D. Php65.00
14. Miss Torres offers tutorial services in mathematics. She charges a flat fee of P150, plus P60 per hour. Write an equation of the tutorial services fee in slope-intercept form.
A. y = 6x + 150 C. y = 150x + 60 B. y = 60x + 150 D. y = 150x – 60
15. What is the value of |a - 3b| when a = -2 and b = 4?
A. -14 B. 10 C. -10 D. 14
Lesson 1 Point-Slope Equation of a Line
A. ; int 2 4
1
y
m
B. ; int 4 2
1
y
m
C. m4;yint4
D. m4;yint4
In the previous activity, you learned about the slope of a line. You have learned how to describe the slopes of lines mathematically and how to apply them as rates of change. You also learned how to find the slope-intercept form of an equation of a line, that is,
y = mx + b, with slope m and y-intercept b.
Now you will explore ways to find the equation of the line when the slope and a point on the line are given. Let's have a simple example.
Example 1. Find an equation of the line through (2, 4) with slope 1/3.
Solution: a) Draw a Cartesian plane and locate the point (2,4). Thus,
b) From the point (2,4), find another point on the line. Since the slope of the line is 1/3, move one unit upward and 3 units to the right. At what point did you stop?___________________
X Y
(2,4)
X Y
(2,4)
(x,y)
8 The slope from (2,4) to (x, y) is
2
4
x
y
m
but 3 1 m , so
2 4 3 1 x y
Multiplying both sides by (x-2)
(x-2) 3 1 = 2 4 x y
(x-2) Simplifying
y - 4 =
3 1
(x - 2)
This equation is said to be in the point-slope form since this form of the equation was generated using the coordinates of a known point other than the y-intercept and the slope of the line.
So, the point-slope form of an equation of a line through the point (x1, y1) with slope m is written as
y - y1 = m(x - x1)
This equation tells us that the point (x1,y1) is any point on the line and that the slope of the line is m.
You can now write an equation in point-slope form for the graph of any nonvertical line if you know the slope of the line and the coordinates of one point on the line.
Example 2. Write the point-slope form of an equation for each line:
a) The line passes through (-2, 5) and has a slope of -3/4. b) The horizontal line passes through (3,1).
Solution:
y - y1 = m(x - x1) point-slope form
y 5 =
-4 3
[x - (-2)]
y 5 =
-4 3
(x + 2) equation of the line
b) The slope of a horizontal line is 0 and using the point-slope form of a line, you get
y - y1 = m(x - x1) point-slope form
y - 1 = 0(x - 3) y - 1 = 0
y = 1 equation of the line
A. Write the equation of the line which passes through the point P and with the slope m. Draw the lines.
1. P(-2, 1), m = 3 2. P(0, 0), m = 5/6 3. P(5, -2), m = -3/2 4. P(-3, -4), m = 1/4
B. State the slope and the coordinates of a point through which the line represented by each equation passes.
1) y - 2 = -3(x + 1) 2) y + 1 =
-5 4
(x - 2)
3) 2(x + 5) = y - 1 4) y = -6
Lesson 2 Two-Point Form of the Equation of a Line
10
In the previous lesson you learned how the concepts of slope and a point on the line are used to write the equation of a line. Now, you will see how to graph a linear equation if you know only two points on the line.
So far, you have been using the equation of a line to find ordered pairs. But suppose you know only the coordinates of two points. How could you find the equation of the line that contains them? There is only one line that passes through two given points. Thus, given two points, one can determine a line following these steps:
To write the equation of a line through two given points, use the slope-intercept form of a line, y = mx + b,
Find m from the given coordinates, Then, find b by solving the equation.
Example 1. Write an equation of the line that passes through the points (-1, 3) and (2, -4).
Solution:
a) First, find the slope of the line.
1 2 1 2
x
x
y
y
m
slope formula
37 1 2 7 1 2 3 4 m slope
b) Substitute this value of m into the slope-intercept form y = mx + b,
y = mx + b slope-intercept form
y =
3 7
x + b
c) Now, find the y-intercept, b, by substituting the coordinates of either points for x and y in the equation. Thus, using the point (-1, 3), you get,
3 =
3 7
(-1) + b
3 = 7 + b 3
3
b =
3 2
d) Finally, substitute b =
3 2
in the equation obtained in part (b). thus,
y =
3 7
x + b
y = 3 7 x + 3 2
This is the equation of the line
that passes through the points (-1, 3) and (2, -4).
An alternative way of finding the equation of a line through two given points is to use the point-slope form of the line. That is,
y - y1 = m(x - x1) point-slope form
but m =
1 2 1 2 x x y y slope formula
Consequently, the equation of the line through (x1, y1) and (x2, y2) is
y - y1 =
1 2 1 2 x x y y
(x - x1)
This equation is called the two-point form of the equation of a straight line.
Example 2. Find the equation of the line determined by the points (2, -2) and (4, 5).
Solution: Using the two-point form of a line, you get,
y - y1 =
1 2 1 2 x x y y
(x - x1)
12 y - (-2) =
2 4
) 2 ( 5
(x - 2)
y + 2 =
2 2 5
(x - 2)
y + 2 =
2 7
(x - 2)
y =
2 7
(x - 2) – 2
y =
2 7
x -
2 7
(2) - 2
y =
2 7
x - 9 equation of the line.
Solve for the equation of the line given:
a) (3, 3) and (2, 1)
b) (5, 1) and (1, 4)
c) (0, 0) and (3, -4)
d) passing through the origin with slope 0
Lesson 3 Intercept Form of the Equation of a Line
Recall the x and y-intercepts of a line. Suppose the x-intercept of a line is a and the y-intercept is b, where a 0 and b 0. See Fig. 1 below:
X Y
(0,b)
(a,0)
Fig. 1
Then the line passes through the points (a, 0) and (0, b). What is the slope of the line described in Fig. 1? ____________________________________________
Applying the point-slope form of a line with slope
a b
, you get
y - y1 = m(x - x1) point-slope form
y - b =
a b
(x - 0)
y - b =
a b
x multiply both sides by a
ay - ay = -bx
bx + ay = ab divide both sides by ab
a x
+
b y
= 1
This equation is called the intercept form of a line because the intercepts are given in the denominators.
Example 1. Write the equation of the line whose x-intercept is 5 and whose y-intercept is -2.
Solution: Using the intercept-form of a line with a = 5 and b = -2, you have
a x
+
b y
14
5
x +
2
y
= 1 multiplying both sides by 10
2x - 5y = 10 equation of the line
Example 2. Write the equation 4x + 7y = 28 in the intercept form.
Solution: Divide both sides of 4x + 7y = 28 by 28.
7
x +
4
y
= 1
A. Write the equation of the line with x-intercepts a and y-intercept b in each of the following:
1. a = -3, b = -8 2. a =
5 3
, b = 1
3. a =
9 4
, b =
2 1
4. a = -1, b = -2
B. Express each equation in the intercept form:
a. 6x - 5y = 30 b. 9y = 4x + 36 c. 2x = 5y + 9 d. 3x - 4y = 8
Lesson 4 Applications of Linear Equations/Functions
You have learned how line, with their slopes and y-intercepts, or any two points on it, can be matched with equations. We continue to look at the relationship between these equations and the practical part everyday function they represent.
Real-world data often have functional relationships that can be written as equations. These equations can be used to make predictions, with or without graphs. Let's have some examples.
Example 1. Ana baked 10 cakes in 5 hours. She worked for 2 ½ more hours and had baked a total of 15 cakes. How many cakes had she baked after 10 hours?
Solution:
a) Know what is asked.
The number of cakes she had baked after 10 hours.
b) Make a table to show the number of cakes she baked in certain number of hours.
No. of hours (x) 5 7 ½ 10
No. of cakes she baked (y)
10 15 20
c) Solve the problem.
Show the table as a rule and as an equation.
Rule: The number of cakes she baked was 2 times the number of hours.
Equation: y = 2x
Check: y = 2x
y = 2(10)
y = 20
She had baked 20 cakes after 10 hours.
16
should exercise to the point where your heart rate reaches a target level based on your age. Here's a rule that is suggested:
To find your target heart rate or pulse . . . .. Subtract your age from 220.
a) Write an equation for the relationship. b) Find your target rate using the equation.
c) What is the target heart rate for a 5-year old child?
Solution:
a) Know what is asked.
i) Your target heart rate
ii) The target heart rate of a 5-year old child
b) Make a table to show the target heart rate for some people.
Age(x) 15 30 45
Target heart rate(y) 205 190 175
c) Solve the problem.
Rule: Subtract the age from 220.
Equation: y = 220 - x
To find your age, say your 16 years old,
y = 220 - 16
y = 204 target heart rate of a 16-year old
Similarly, the target heart rate of a 5-year old child is,
y = 220 - 5 y = 215
Following the procedures used in examples 1 and 2 of this lesson, solve the given problems below:
1. An employee makes 7 pans of pizzas per hour. How many pans of pizzas can he make if he works 8 hours?
2. A swimming pool was being filled with water. It was 1.22 m deep after 3 hours and 1.52 m deep after 5 hours.
a. Write an equation for the relationship.
b. How long will it take a pool to a depth of 2.13 m
Lesson 5 Absolute Value Function
In this lesson, you will learn how to express distances between points mathematically and to solve equations involving distances.
Take a look of the number line below:
5 units 3 units
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
What are the numbers on the right of 0 on the number line? ______________ On the left of 0? ________________________________________________
The numbers to the right of 0 on the number line are positive and the numbers to the left of 0 are negative. The distance from 0 to 3 is 3 while the distance from 0 to -5 is what? _____
Distance is always positive (or zero). In mathematics, we use absolute value to describe distance on a number line.
Consider the following examples:
1. | 9 | = 9 2. | -9 | = 9
18
3. |-1.2 | = 1.2 4. |0| = 0
5. |-3| + 4 = 3 + 4 = 7 6. |5| - 2 = 5 - 2 = 3 7. |-5 + 3| = |-2| = 2 8. |-3 - 1| = |-4| = 4
9. |6 - 4| = |2| = 2 10. |-3 + 4| + |-6 - 1| = |1| + |-7| = 1 + 7 = 8
From the examples above, the absolute value of zero or a positive number is what?______ How about the absolute value of a negative number, what can you say about it? ________
Notice that the absolute value of any real number is a non-negative number. Try the following numbers below:
Find the numbers:
1. |-6| 2. |-12|
3. |4| 4. |10|
5. |6|
The answers are:
1. 6 2. 12
3. 4 4. 10
5. 6
We can build a definition for the absolute value of a number, that is
Having known the definition of the absolute value of a real number, you will see what the graph of the function y = |x| and related functions look like.
Example 1. Find a table of values for y = x and sketch the graph.
The table of ordered pairs of y = x Graph
x 0 1 -2 4
y 0 1 -2 4
x if x 0 |x| =
-x if x 0
X Y
The graph of y = x is in Quadrant I and Quadrant III. Now, graph the function y = |x|.
y = | x | Graph
x 0 1 -1 -3
y 0 1 1 3
Notice that the Quadrant III "arm" of the
graph y = x is not flipped over into
Quadrant II.
The function y = |x| is called the absolute value function.
Thus,
To graph the absolute value function, simply find enough ordered pairs of the function, then plot them and see what the graph looks alike.
In the next example, you will explore and try to graph other functions involving absolute value.
Example 2. Graph y = x + b where b = 2 by finding a table of values.
The absolute value function is a relationship that can be represented by the equation y = |x|, where x is any real number.
X Y
20
y = x + 2 Graph
X 0 1 -1 -3
Y 2 3 1 -1
Fig.1
Example 3. Find a table of values for y = |x + 2| and y = |x - 2|. Compare the graphs with the graph of y = x + 2.
Solution: The table of values for y = |x + 2| and y = |x - 2| is:
x -5 -4 -3 -2 -1 0 1 2 3
y = |x+ 2| 3 2 1 0 1 2 3 4 5
y = |x - 2| 7 6 5 4 3 2 1 2 1
Plotting these ordered pairs, you get
y = |x + 2| y = |x - 2|
Fig. 2 Fig. 3
Notice again that the Quadrant III "arm" of the graph of y = x + 2 (see Fig. 1) is now flipped over into Quadrant II (see Fig. 2). Also the graph of y = |x - 2| was moved four units to the right. So they differ only by their vertices.
X Y
y = x +2
X Y
21
Example 3. a) Copy and complete the table to find points for the graphs of y = |x| + 2 and y = |x| - 2
x -3 -2 -1 0 1 2 3
y = |x| + 2 5 3
y = |x| - 2 1 -1
b) Sketch the graphs in different Cartesian planes. c) How do the graphs differ from one another?
Solution:
a) The table of values for y = |x| +2 and y = |x| - 2:
x -3 -2 -1 0 1 2 3
y = |x| + 2 5 4 3 2 3 4 5
y = |x| - 2 1 0 -1 -2 -1 0 1
b)Graphs are:
y = |x| + 2 y = |x| - 2
Fig. 4 Fig. 5
[image:21.612.83.543.71.544.2]c) The graphs are similar. They only differ by their vertices. The vertex of the graph of y = |x| + 2 , which is (0, 2) was moved to (0, -2).
We can now summarize the graphs of the absolute value function y = |x|, and check how these are similar to other equations you have graphed.
The graphs of the absolute function y = |x| when x is any real number are V-shaped.
The graphs of the functions y = |x + b| and y = |x - b| are also V-shaped but their vertices are moved b units to the right/left from the origin.
X Y
22 A. Perform the indicated operation.
1. | -2 | - | 3 4. | -12 | 2. | 6 – 7 | 5. 4 + | -3 | 3. | -3 + 2 | + | -6 – 5|
B. Identify how to translate the graph of y = | x | to obtain the graph of each function.
c. y = | x | - 5 d. y = | x – 4 | e. y = -6 + | x | f. y = | x + 3 | g. y = | x | + 4
Self-check
Multiple Choice: Choose the letter of the correct answer.
1. What is the absolute value of the sum of -2 and -6?
A. -8 B. 4 C. 8 D. 12
2. What is the equation of the line passing through (0, 6) and whose slope is -2?
A. y = -2x – 6 B. y = 2x + 6 C. y = -2x + 6 D. y = 2x – 6
3. What equation is represented by the graph?
A. y = x + 3
B. y = -x + 3
C. y = x – 3
D. y = -x – 3
The point-slope form of the equation of a line is y - y1 = m(x - x1), where m is the
slope of the line and (x1, y1) is a any point on it.
The two-point form of the equation of a line is y - y1 =
1 2
1 2
x x
y y
(x - x1) , where (x1,
y1) and (x2, y2) are any two distinct points on the line.
The intercept equation of a line is a x
+
b y
= 1, where a is the x-intercept and
b is the y-intercept.
The absolute value function is a function defined by y = |x| where x is any real number.
The graphs of the absolute value function are V-shaped figures.
X Y
24
4. Which of the following graphs is represented by y = | x | + 1?
A. B.
C. D.
5. What is the equation of the line passing through ( -3, 5) and (-1, 7)?
A. y = -x – 8 B. y = -x + 8 C. y = x – 8 D. y = x + 8
6. The graph of the equation x + y = 2 crosses the x-axis at
A. 2 B. -2 C. 1 D. -1
7. Which of the following points is contained in the graph of the equation 3x - y = 6?
A. (1, -9) B. (-1, 9) C. (-1, 3) D. (1, -3)
8. The equation of a line through (1, -3) and having a slope of -1 is
A. x + y = -2 B. x + y = 2 C. x + 2y = 1 D. 2x + y = 1
9. The graph of the equation y = 4x - 1 is
A. B.
X Y
X Y
X Y
X Y
X Y
C. D.
10. What is the slope and the y-intercept of the equation 5x - 10y - 1 = 0?
11. Which equation would give the graph at the right?
12. Which of the following graphs is the graph of y =
4
x ?
A. C.
A. m2;yint5 C.
5 1 int ;
2
y
m
B.
5 1 int ;
2
y
m D. m2;yint 5
A. y = |x - 4|
B. y = |x| - 4
C. y = |x + 4|
D. y = |x| + 4
X Y
X Y
X Y
X Y
26
B. D.
13. When you buy an item from a credit car on a mail order catalog you pay a 10% sales tax. If you want the item set immediately by a delivery service, you must pay an additional P100 per order. The equation that shows this relationship is y = 0.10x + 100, where x is the cost of the item before tax and y is the cost of one item sent immediately. What is the cost of an item priced P875 and shipped immediately?
A. Php1875.00 B. Php1062.25 C. Php962.50 D. Php975.00 14. The maximum heart rate during exercise for a 20-year old is 200 beats per
minute, and the maximum heart rate for a 70-year old is 150 beats per minute.(Nordic Track Brochure). What is the equation that shows the relationship between the maximum heart rate of a person and his/her age?
15. Which of the following points contained in the graph of y = |x| + 1?
A. (0,-1) B. (1,0) C. (-1,2) D. (-1,1) A. x + y - 180 = 0 C. x - y - 180 = 0
B. x - y + 180 = 0 D. x + y + 180 = 0
X Y
Answer Key
Pretest
1. c
2. b
3. a
4. b
5. a
6. d
7. a
8. b
9. c
10. c
11. a
12. b
13. b
14. c
15. d
Lesson 1. Self-Check
A. 1. 3x - y = 7 2. 5x - 6y = 0 3. 3x + 2y = 11 4. x - 4y = -13
B. 1. m = -3 ; (-1, 2) 2. m = -4/5; (2, -1)
3. m = 2; (-5, 1) 4. m = 0; (0, -6)
Lesson 2. Self-Check
1. 2x - y = 3 2. 3x + 4y = 6 3. 3x + 4y = 16 4. 3x + 4y = 0 5. y = 0
Lesson 3. Self-Check
1. a. 8x + 3y = -24 b. 5x - 3y = -3 c. 9x + 8y = 4 d. 2x + y = -2
2. a.
5 x + 6 y = 1 b. 9 x + 4 y = 1 c. 2 9 x + 5 9 y = 1 d. 3 8 x + 2 y = 1
Lesson 4. Self-Check
1.
No. of hours (x) 1 2 3 4 5 6 7 8 No. of pizzas (y) 7 14 21 28 35 42 49 56
Rule: The number of pans of pizzas made is 7 times the number of hours. Equation: y = 7x
28 y = 56
Therefore: the employee made 56 pans of pizzas in 8 hours.
2. a. 2x + y = 11 b. 9 hours
Lesson 5. Self-Check
A. 1. -1 B. 1. Vertex (0, -5) 2. 1 2. Vertex (4, 0) 3. 12 3. Vertex (0, -6) 4. 6 4. Vertex (-3, 0) 5. 7 5. Vertex (0, 4)
Posttest
1. c 2. c 3. c 4. a 5. d 6. a 7. d 8. a 9. c 10. b 11. b 12. a 13. b 14. d 15. c
