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Lesson 1Plotting Points on the Number Line and Finding the Absolute Value of a Number Lesson 2 Plotting Points on the Cartesian Plane Lesson 3 Finding the Distance and the Midpoint of Two Points

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Module 13

Point to Point

Would you like to find the approximate distance between your house and your school? To do this, you need to draw the map of your barangay, town, or city on a piece of paper that has been divided into small, same-sized squares. By taking your house and your school as points on this paper, this module can help you compute the distance between the two.

In this module, we will study the parts of the number line and the Cartesian coordinate plane, locate points in one and two-dimensional spaces, determine the absolute value of a number, and find the distance and the midpoint of two points.

This module contains three lessons, namely:

Lesson 1 Plotting Points on the Number Line and Finding the Absolute

Value of a Number

Lesson 2 Plotting Points on the Cartesian Plane

Lesson 3 Finding the Distance and the Midpoint of Two Points

After studying this module, you are expected to:

a) determine the coordinate of a point on the number line; b) find the absolute value of a number;

c) compute the distance between two points on the number line; d) simplify expressions containing the absolute value sign; e) describe the parts of the Cartesian plane;

What this module is all about

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f) give the

 

x,y -coordinates of a point; g) plot a point given its

 

x,y -coordinates;

h) determine the quadrant where a given point is located; i) find the distance between two points; and

j) determine the coordinates of the midpoint of two given points.

This is your guide for the proper use of the module:

1. Read the items in the module carefully.

2. Follow the directions as you read the materials.

3. Answer all the questions that you encounter. As you go through the module, you will find help to answer these questions. Sometimes, the answers are found at the end of the module for immediate feedback.

4. To be successful in undertaking this module, you must be patient and industrious in doing the suggested tasks.

5. Take your time to study and learn. Happy learning!

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Direction. Read each item and choose the letter of the correct answer.

1. What is the absolute value of -3?

a. -3 b. 3 c. 3 1

d. 3 1

2. On the number line, what is the distance between -2 and 7? a. 14 b. 9 c. 5 d. -3.5

3. What is the value of 42 275? a. 18 b. 6 c. -2 d. -4

What to do before (Pretest)

Start

Take the Pretest

Check your paper and count your correct answers.

Is your score 80% or above?

Yes Scan the items you missed.

No

Study this module

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4. What is the abscissa or the x-coordinate of the point (-3, 3)?

a. -3 b. 3 1

 c. 3 d. 3 1

5. A point has positive abscissa and negative ordinate. In which quadrant does it lie? a. I b. II c. III d. IV

6. A point lies 4 units to the left of the x-axis and 5 units below the y-axis. What are the coordinates of the point?

a. (4, 5) c. (-4, -5) b. (-4, 5) d. (4, -5)

7. The ordinate of a point on the x-axis is always a. negative c. positive b. zero d. undefined

8. What is the distance between the points (1, 3) and (4, 7)? a. 15 c. 7

b. 5 d. 7

9. What is the distance between the points (0, 2) and (-2, 0)? a. 12 c. 8

b. 4 d. 8

10. What are the coordinates of the midpoint of the points (2,5) and (-6,3)? a. (-2,8) c. (-8,-16)

b. (-2,4) d. (8,2)

Read carefully the questions that follow, answer the questions asked, and then do the suggested activities to enhance your competence in mathematics.

What you will do

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Lesson 1 Plotting Points on the Number Line and Finding the

Absolute Value of a Number

Since our elementary years, we have seen the following diagram called the number line.

The number line is a horizontal line having two arrow heads with 0 at the middle. The middle number 0 is called the starting point or the origin. Numbers on the right side of 0 are called positive numbers, while those on the left are called negative numbers.

By definition, to every point on the number line we associate a unique number called its coordinate. We denote points by capital letters. If P is a point, we denote its coordinate by the corresponding small letter p.

Example 1 Identify the coordinate of each point:

a) A b) B c) C

Answer:

a) A has coordinate -4. b) B has coordinate 0. c) C has coordinate 3.

The absolute value of a number is its distance from the origin. Since distance is a positive quantity, the absolute value of a number cannot be less than 0. For a number x, we denote the absolute value of x by x .

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

A B C

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

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Example 2 a) What is the absolute value of -5?

The answer is 5 because -5 is 5 units from 0.

b) What is the absolute value of 4?

The answer is 4 because 4 is 4 units from 0.

c) What is the absolute value of 0?

The answer is 0 because it is 0 unit from 0.

The distance between two points on the number line is given by d ab , where a is the coordinate of the first point and b is the coordinate of the second point.

A B

The distance between the points A and B is d ab.

Example 3 Find the distance between the following pair of points: a) P and Q b) Q and R C) Q and S.

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

P Q R S

a b

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

0 unit

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

5 units

4 units

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Answers: a) Since p =-5 and q = -3, the distance between P and Q is 2

2 3 5 ) 3 (

5       

 

d units.

b) Since q = -3 and r = 1, the distance between Q and R is 4

4 1

3   

 

d units.

c) Since q = -3 and s = 4, the distance between Q and S is 7

7 4

3   

 

d units.

Example 4 Simplify the following expressions:

a) 72 17

 

1 b) 6

 

3  5

10

Answers: a) 72 17

 

1 = 9 171 = 918 = 27 b) 6

 

3 5

10

= 63 510 = 9156

Answer the following.

1. Find the coordinate of the following points on the number line.

a) B b) A c) O

2. Find the distance between the given pair of points. a) A and T b) T and S c) A and I

3. Simplify the following expressions.

a) 5 94 b) 27(3) 4 c) 102 18

 

3 d) 62 21

 

1

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

B A O

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

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e) 351262 f) 46 35

 

1  74

Lesson 2 Plotting Points on the Cartesian Plane

In Lesson 1, we learned that any point on the number line is associated to a number called its coordinate. Since the location of every point on the number line can be described by a single coordinate only, the number line is said to be a 1-dimensional space.

P

A point may not lie on the number line. To describe the location of the point, we construct another number line that is vertical to and passing through 0, the origin of the original number line. The resulting figure is called the Cartesian coordinate plane or simply the Cartesian plane.

The horizontal line in the Cartesian plane is called the x-axis, while the vertical line is called the y-axis. The intersection of the axes is called the origin, also denoted by O.

Every point in the Cartesian plane is associated with two numbers x and y, called its coordinates. The coordinates of a point are written using ordered pair (x, y). Also, if (x, y) are the coordinates of a point P, we write P(x, y). Note that the origin has coordinates (0, 0).

Since two numbers are associated to every point in the Cartesian plane, this plane is said to be a 2-dimensional space.

p

Point P on the number line has coordinate p.

Did you know?

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 

y-axis

5 The Cartesian Plane 4 x

P(x, y)

3

2 y origin 1

x-axis -5 -4 -3 -2 -1 0 1 2 3 4 5

-1

-2 Point P has coordinates (x, y).

-3

-4 -5

Example 5 Use the diagram below to find the coordinates of the following points.

a) T b) E c) A d) M

y-axis

5 4

T 3

2

E 1

x-axis -5 -4 -3 -2 -1 0 1 2 3 4 5

-1 M -2

-3

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Answer:

To find the coordinates of a point on the Cartesian plane, we draw a rectangle that has the given point and the origin as two of its vertices. The x-coordinate is the distance of the vertical line passing through the given point from the y-axis while the y-coordinate is the distance of the horizontal line passing though the given point from the x-axis.

a) T(-4, 3) c) A(4, -4)

b) E(3, 2) d) M(-3,-2)

Every point on the Cartesian plane has a corresponding distance from the x and the y axes. If a point has coordinates (x, y), the absolute value of x represents the distance of the point from the y-axis and the absolute value of y represents the distance of the point from the x-axis.

Example 6 Plot the following points on the Cartesian plane:

a) A(-3, 1) c) C(-2, -1)

b) B(3, 4) d) D(3, -2).

Answers:

Referring to the Cartesian plane that follows,

a) A(-3, 1) means that point A is located 3 units to the left of the y- axis and 1 unit above the x-axis.

b) B(3, 4) means than point B is located 3 units to the right of the y- axis and 4 units above the x-axis.

c) C (-2, -1) means that point C is located 2 units to the left of the y- axis and 1 unit below the x-axis.

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y-axis

5

4

B(3,4)

3

2 A (-3, 1)

1

x-axis -5 -4 -3 -2 -1 0 1 2 3 4 5

-1

C (-2,-1) -2  D(3, -2)

-3

-4 -5

1. The Cartesian coordinate system, which uses the Cartesian plane in plotting points, is also known as the rectangular coordinate system.

2. The Cartesian plane is named after the French mathematician Rene Descartes, one of the pioneer mathematicians who invented and used the system.

3. The x-value of the coordinates (x, y) of a point is also known as the abscissa, while the y-value is also known as the ordinate.

4. The Cartesian plane is divided into four quadrants by the x and the y axes. These quadrants are labeled counterclockwise as QI, QII, QIII, and QIV. As a general rule, we consider the axes not to belong to any of the quadrants.

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y-axis

The Four Quadrants

5

4

QII 3 QI

(

,+)

2

(+,+)

1

x-axis -5 -4 -3 -2 -1 0 1 2 3 4 5

-1 -2

-3

QIII QIV

(

,

)

-4

(+,

)

-5

Example 7 Determine the quadrant where each point lies.

a) (-4, 6) b) (7, 4) c) (0, 5)

Answers:

a) Since (-4, 6) has a negative abscissa and positive ordinate, it lies in QII.

b) Since (7, 4) has a positive abscissa and positive ordinate, it lies in QI.

c) The point (0, 5) does not belong to any quadrant. It lies along the positive y-axis.

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1. Identify the coordinates of the following points.

a) B b) A c) E d) T

2. Name the point that corresponds to the given coordinates.

a) (-4, -3) b) (-3, 1) c) (0, 2) d) (3, -2)

B. Identify the quadrant/s that satisfy the given condition.

1. The x-coordinate is negative and the y-coordinate is positive. 2. The x and y coordinates are both negative.

3. The ordinate is positive. 4. The abscissa is negative.

5. The quadrant is above the x-axis and to the left of the y-axis.

C. Plot the following points on the Cartesian plane.

1. A (-1, -3) 2. B (4, 0) 3. C (3, -4) 4. D (1, 1) 5. E (0, -2)

D. Define the following terms.

1. origin 2. abscissa 3. ordinate 4. quadrant 5. y-axis

Answer Key on page 20

y-axis

axisaxi s

x-axis

0

1 2 3 4 5 -1

-2 -3 -4 -5

1 2 3 4 5

-1

-2

-3

-4

B . A

E T

.

. .

. P

O

.

.

Q

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Lesson 3 Finding the Distance and the Midpoint of Two Points

In Lesson 2, we learned how to plot points on the Cartesian plane. We now focus on how to find the distance between two given points and their midpoint.

Given two points A(x1,y1) and B(x2,y2), the distance d between A and B is given by the formula:

2 2 1 2 2

1 ) ( )

(x x y y

d    .

This formula is commonly known as the distance formula. Note that the distance d between the points A and B is also the length of the line segment determined by A and B.

y-axis

5

4 B(x2,y2)

d 3

A (x1,y1)

2

1

x-axis -5 -4 -3 -2 -1 0 1 2 3 4 5

-1 -2

-3

d = length of the line segment

determined by A and B

-4 -5

Example 8 Find the distance between each pair of points:

a) (-1, 3) and (5, -5) b) (4, 0) and (-1, 5)

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Answer: a) Let x1 1, y1 3, x2 5, and y2 5. By substituting these into the formula, we have 2 2 )) 5 ( 3 ( ) 5 1 (      d 64 36 ) 8 ( ) 6

( 2  2  

d

10

100 

d units.

b) Let x1 4, y1 0, x2 1, and y2 5. By substituting these into the formula, we have 2 2 ) 5 0 ( )) 1 ( 4 (      d 25 25 ) 5 ( ) 5

( 2   2  

d

50

d units.

Example 9 What is the length of the line segment determined by A(-2, 3) and B(4, 1)?

Answer: The length of the line segment determined by the points A and B is given by d. We let x1 2, y1 3, x2 4, and y2 1. By substituting these values into the formula, we have

2 2 ) 1 3 ( ) 4 2 (     d 25 25 ) 5 ( ) 5

( 2   2  

d

50

d units.

The midpoint of a line segment is the point that divides the segment into two equal parts. It is also the middlemost point of the line segment. To find the coordinates of the midpoint M of the line segment determined by the points A(x1,y1) and B(x2,y2), we use the midpoint formula given by:

         2 , 2 2 1 2

1 x y y x

M .

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M y-axis

5

4 B(x2,y2)

3

A (x1,y1)

2

1

x-axis -5 -4 -3 -2 -1 0 1 2 3 4 5

-1 -2

-3

M = the midpoint of A and B

-4 -5

Example 10 Find the coordinates of the midpoint of the line segment determined by the following pair of points:

a) (-4, 3) and (6, 5) b) (3, 4) and (2, -6)

Answer: a) We let x1 4, y1 3, x2 6, and y2 5. By substitution, the coordinates of the midpoint is given by

 

1,4 2 8 , 2 2 2 5 3 , 2 6 4                 M .

b) We let x1 3, y1 4, x2 2, and y2 6. By substitution, the coordinates of the midpoint is given by

                        

 , 1

2 5 2 2 , 2 5 2 ) 6 ( 4 , 2 2 3 M .

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Answer: Since A has coordinates (7, 2), we let x1 7, y1 2. The coordinates of the midpoint M are (-3,4) and so

3 2

2 1x

x

and 4 2

2 1y

y

.

By substituting x1 7 into 3 2

2 1x 

x

, we have

3 2 7 2   x

2( 3) 2

7

2 2  

  

  x

7x2 6 x2 13.

Also, by substituting y1 2 into 4 2

2 1y 

y

, we have

4 2

2 2

 y

2(4) 2

2

2 2

  

  y

8 2y2  y2 6 . Therefore, point B has coordinates (-13, 6).

1. Use the distance formula to find the distance between the given pair of points.

a) (-4,2) and (1,-11) b) (-3,1) and (2,3)

2. Find the length of the line segment determined by the given pair of points.

a) A(-3,0) and B(7,1) b) P(2,-2) and Q(-1,2)

3. Find the coordinates of the midpoint M of the line segment determined by the given pair of points.

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4. Let M(-2,4) be the midpoint of the line segment AB.

a) If A has coordinates (-5,6), find the coordinates of B.

b) If B has coordinates (7,5), find the coordinates of A.

The following concepts are the highlights of the three lessons contained in this module:

 The absolute value of a number x, denoted by x , is its distance from the origin of the number line.

 The Cartesian plane consists of two perpendicular lines: the horizontal line or the x-axis and the vertical line or the y-axis.

The intersection of the two axes is called the origin.

The two axes separate the plane into four regions called quadrants.

A point on the coordinate plane is represented by an ordered pair (x, y). These numbers are not interchangeable.

The x-coordinate is also called the abscissa, while the y-coordinate is also called the ordinate.

 Given two points A(x1,y1) and B(x2,y2), the distance d between A and B is given by d (x1x2)2 (y1y2)2 , the distance formula.

The midpoint of a line segment is the point that divides the segment into two equal parts. To find the coordinates of the midpoint M of the line segment determined by the points A(x1,y1) and B(x2,y2), we use the midpoint formula given by 

  

  

2 , 2

2 1 2

1 x y y x

M .

Let’s summarize

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Direction. Read each item and choose the letter of the correct answer.

1. What is the absolute value of -7?

a. 7 b. -7 c. 7 1

d. 7 1

2. On the number line, what is the distance between 8 and -2? a. 16 b. 10 c. 6 d. 3

3. What is the value of 34(3) 234 ? a. 19 b. 23 c. 25 d. 27

4. What is the ordinate or the y-coordinate of the point (-3, 3)?

a. -3 b. 3 c. 3 1

 d.

3 1

5. A point has negative abscissa and negative ordinate. Which quadrant does it lie?

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

6. A point lies 4 units to the right of the x-axis and 5 units below the y-axis. What are the coordinates of the point?

a. (4, 5) b. (-4, 5) c. (-4, -5) d. (4, -5)

7. The abscissa of a point on the y-axis is always a. negative c. positive b. zero d. undefined.

8. What is the distance between the points (1, 3) and (-5, -5)? a. 14 c. 5

b. 10 d. 8

9. On a map, Lisa’s house has coordinates (2, 0). If her school has coordinates (0, -2), what is the distance between Lisa’s house and her school?

a. 12 c. 8 b. 4 d. 8

10. What are the coordinates of the midpoint of (7, -6) and (-1, -2)? a. (3, -4) c. (8, -8)

b. (3, 4) d. (6, -8)

What to do after (Posttest)

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Answer Key

Pretest page 3 1) b 2) b 3) c 4) a 5) d 6) c 7) b 8) b 9) c 10) b

Lesson 1 Self-Check 1 page 7

1 a) -5 b) 0 c) 3

2 a) d = | -5 - (-2) |= | -5 + 2 | = | -3 | = 3 units

b) d = | -2 – 3 | = | -5 | = 5 units

c) d = | -5 – 0 | = | -5 | = 5 units

3 a) 594 5 5 550

b) 27

 

3 4273 421042

 

10 420416 c) 102 18

 

3  8 183  8  2182129 d) 62 21

 

1  4 211 4 22 42

 

2 448 e) 351262 36 24 3

   

6 2 4 18810 f) 46 35

 

1  74  103513

103

 

6 3101835 Lesson 2 Self-Check 2 page 12

A 1 a) B=(2,2) b) A=(-4,3) c) E=(-2,-2) d) T=(3,-2)

2 a) P b) O c) Q d) T

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 A

 D

 E

 B C

y-axis

5

4

3

2

1

x-axis -5 -4 -3 -2 -1 0 1 2 3 4 5

-1 -2

-3

-4

-5

D 1) origin – the point of intersection of the x-axis and the y-axis.

2) abscissa – the x-coordinate of a point.

3) ordinate – the y-coordinate of a point.

4) quadrant – one of the four divisions of the Cartesian plane.

5) y – axis – the vertical line of the Cartesian plane.

Lesson 3 Self-Check 3 page 17

1 a) d (41)2(2(11))2 d (5)2 (13)2  25169 d 225 15 units

b) d (32)2(13)2 d (5)2 (2)2  254

d 29 units

2 a) d (37)2(01)2

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d (10)2 (1)2  1001

d 101 units

b) d (2(1))2 (22)2 d (3)2(4)2  916 d 25 5 units

3 a)

4 2

2 4 2 8 2 4 0 2 8 0

M   

                  ( ), ( ) , ,

b) 

                      2 5 1 2 5 2 2 2 4 1 2 6 4

M , , ,

4 a)By substituting x15 into 2 2

x x1 2

  

, we have 2

2 x 5 2    

 2( 2)

2 x 5

2 2 

  

  

 5x2 4  x2 1. Also, by substituting y16 into 4

2 2 1y 

y

, we have 4 2 y 6 2  

 2(4)

2 y 6

2 2

      8 y

6 2   y2 2.

Therefore, point B has coordinates (1, -2).

b)By substituting x2 7 into 2 2

x

x1 2 , we have 2

2 7

x1

 2( 2)

2 7 x

2 1  

  

 

 x174  x111. Also, by substituting y2 5 into 4

2 2 1y 

y

, we have 4 2

5 y1

 2(4)

2 5 y

2 1 

  

 

 y158  y13.

Therefore, point A has coordinates (-11, 3).

Posttest page 19

1) a 2) b 3) a 4) b 5) c 6) d 7) b 8) b 9) c 10) a

(23)

BIBLIOGRAPHY

Fuller, G. (1977). College algebra. (4th ed). New York, NY: Van Nostrand Company.

Leithold, L. (1989). College algebra. USA: Addison-Wesley Publishing Company, Inc.

References

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