Mathematics 10 Quarter 2 Module 6: Applying the Distance Formula to Prove Some Geometric Properties

Mathematics Quarter 2 – Module 6: Applying the Distance Formula to Prove Some Geometric Properties

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CO_Q2_Mathematics 10_ Module 6

Mathematics – Grade 10 Alternative Delivery Mode Quarter 2 – Module 6: Applying the Distance Formula to Prove Some Geometric Properties First Edition, 2020

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Published by the Department of Education Secretary: Leonor Magtolis Briones Undersecretary: Diosdado M. San Antonio

Development Team of the Module

Writer’s Name: Cerion T. Camhit Co-Writer’s Name: Laila B. Kiw-isen Reviewer’s Name: Bryan A. Hidalgo Management Team: May B. Eclar Benedicta B. Gamatero Carmel F. Meris Ethielyn E. Taqued Edgar H. Madlaing Marciana M. Aydinan Lydia I. Belingon

Printed in the Philippines by:

Department of Education – Cordillera Administrative Region

Office Address: Wangal, La Trinidad, Benguet Telefax: (074) 422-4074 E-mail Address: car@deped.gov.ph

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Mathematics

Quarter 2 – Module 6: Applying the Distance Formula to Prove Some Geometric Properties

Introductory Message

This Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and learn while at home. Activities, questions, directions, exercises, and discussions are carefully stated for you to understand each lesson.

Each SLM is composed of different parts. Each part shall guide you step-by-step as you discover and understand the lesson prepared for you.

Pre-tests are provided to measure your prior knowledge on lessons in each SLM. This will tell you if you need to proceed on completing this module or if you need to ask your facilitator or your teacher’s assistance for better understanding of the lesson. At the end of each module, you need to answer the post-test to self-check your learning. Answer keys are provided for each activity and test. We trust that you will be honest in using these.

In addition to the material in the main text, Notes to the Teacher are also provided to our facilitators and parents for strategies and reminders on how they can best help you on your home-based learning.

Please use this module with care. Do not put unnecessary marks on any part of this SLM. Use a separate sheet of paper in answering the exercises and tests. And read the instructions carefully before performing each task.

If you have any questions in using this SLM or any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator.

Thank you.

[illustration: hand pressing a touch point, icon beside “What I Need to Know”]

What I Need to Know

This module focuses generally in applying distance formula in proving properties of some geometric figures.

After going through this module, you are expected to:

1. find the distance between two points using the Distance Formula; and
2. apply Distance Formula in proving properties of some geometric figures.

[illustration: hand holding a pencil, icon beside “What I Know”]

What I Know

Direction: Choose the letter of the correct answer and write it on a separate sheet of paper.

1. In the Cartesian plane, what is the distance of the point (2, -3) from the origin?
   A) √5 B) √13 C) 5 D) 13

   
   

2. What is the distance between point A(3, 4) and point B (10, 4)?
   A) 4 B) 5 C) 6 D) 7

   
   

3. Which of the following should be the value of x so that the distance between the points (x, −2) and (12, −2) is 7?
   A) 2 B) 3 C) 4 D) 5

   
   

4. Which of the following equation describes the distance formula?
   A) d = √((x₂ − x₁)² + (y₂ − y₁)²) C) d = √((x₂ − x₁)² − (y₂ − y₁)²)
   B) d = √((x₂ + x₁)² + (y₂ + y₁)²) D) d = √((x₂ + x₁)² − (y₂ + y₁)²)

   
   

5. What is the distance between two points with coordinates (4, 3) and (1, 7)?
   A) 25 B) 16 C) 5 D) 2

   
   

6. What is the distance between point A (4a, 4a) and point C (0, a)?
   A) 3a B) 5a C) 9a D) 12a

   
   

7. Both points D and U are on the fourth quadrant. If the distance between point D and U is 3 units and D is at (2, -1), which of the following are the coordinates of point U?
   A) (2, -4) B) (1, -6) C) (4, -1) D) (2, -5)

   
   

8. What is the perimeter of an equilateral triangle if two of its vertices have coordinates (0, 2) and (0, 5)?
   A) 3 B) 6 C) 9 D) 12

   
   

9. What kind of triangle is formed when its vertices (0, 4), (−4, −2) and (4, −2) are plotted on the Cartesian plane?
   A) equilateral B) isosceles C) right D) scalene

   
   

10. What type of quadrilateral is formed by the given vertices C (0, 0), A (a, 0), R (a, a) and E (0, a)?
    A) kite B) parallelogram C) rectangle D) square

    
    

11. In Quadrilateral LOVE, what is the distance of point O from the origin?
    A) √(a + b) C) √(a² + b²)
    B) √(a − b) D) √(a² − b²)

    
    

    [chart: diagram; Quadrilateral LOVE. Labels: L(0,b), O, E (0,0), V(a,0). Axes labeled x and y. Values are not labeled on the chart beyond those listed.]

    
    

12. What are the coordinates of point L in isosceles trapezoid LIVE?
    A) (d-c, e) C) (d-c, 0)
    B) (c-d, e) D) (e-c, d)

    
    

    [chart: diagram; Isosceles trapezoid LIVE. Labels: L, I(c,e), E, V(d,0). Axes labeled x and y.]

    
    

13. If GIVE is a square, what is the distance between the points V and G?
    A) a C) 2√(a + b)
    B) 2a D) 2√(a − b)

    
    

    [chart: diagram; Square GIVE. Labels: G, I, V (a,b), E. Axes labeled x and y.]

    
    

14. To prove that quadrilateral RICE is a rectangle, which of the following should be proven congruent?
    A) RI ≅ IC, IC ≅ CE C) IC ≅ CE, RC ≅ IE
    B) RE ≅ IC, RI ≅ CE D) RC ≅ IE, RE ≅ IC

    
    

15. To prove that triangle ABC is a scalene triangle which of the following should be proven?
    A) AB ≅ BA B) AB ≅ BC ≅ AC C) AB ≅ BC D) AB ≇ BC ≇ AC

    
    

[illustration: hand holding a puzzle piece, icon beside “What’s In”]

Lesson Distance Formula

What’s In

In your previous lessons, you have learned how to plot points and name the coordinates of the points on the Cartesian plane. Examine the situation below.

[illustration: classroom with 9 seats arranged in 3 rows and 3 columns; a teacher stands at the front; two students are labeled JUAN and JUANA]

https://www.dreamstime.com/illustration/classroom-cartoon.html

The picture above depicts a classroom with 9 seats arranged in 3 rows and 3 columns. During their Mathematics class, the teacher asked Juan and Juana to describe their location.

If you were Juan and Juana,

• How would you describe your location? What mathematical concepts can you use to describe your location?
• How far are you from each other? How will you determine your distance from each other?

Now let us visualize the situation in the Cartesian plane.

[chart: scatter plot on a Cartesian plane; x and y axes numbered 0 to 6; dots mark seats in a 3 by 3 arrangement; labels: Juan, Juana. Values are not labeled on the chart beyond the labels and axis numbers shown.]

Using the illustration above, the location of Juan and Juana can be determined by naming the point of their location in the Cartesian plane. Remember that the coordinates of a point are always written as an ordered pair of the form (x, y). The first number corresponds to the x-coordinate and the second is the y-coordinate. Hence, Juan is at point (1, 1) and Juana is at point (3, 2).

Now, it’s your turn!

Activity 1.

A) Write the coordinates of each given point: a) D _______ b) I _______ c) S _______ d) T _______

B) Plot the following points on the Cartesian plane: e) A (8, 3) f) N (−6, 4) g) C (−5, −7) h) E (4, −9)

[chart: Cartesian plane with x and y axes numbered −10 to 10; points labeled D, I, S, and T are plotted. Coordinates are not printed on the chart.]

transcriber’s note: In the source, Activity 1 is laid out as a two-column table, with the instructions on the left and the blank Cartesian plane on the right. The plane is transcribed above as a chart.

What’s New

Using the situation of Juan and Juana, find the distance between them if the students in the class were seated 1 meter from each other.

The distance between two points in the coordinate plane is the length of the segment that joins the two points. Hence, to find the distance between Juan and Juana, draw a segment joining the points of their location. The figure below demonstrates that the distance between Juan and Juana can be found by forming a right triangle in which their distance is the hypotenuse, while the horizontal and the vertical segments are the legs of the right triangle.

[chart: coordinate grid; figure showing the locations of Juan and Juana joined by a segment, forming a right triangle] Labels: Juan; Juana; c; vertical distance (a); horizontal distance (b) Axes: y, with marks 1 to 5; x, with marks 1 to 5

Recall the Pythagorean Theorem: The square of the hypotenuse of a right triangle is equal to the sum of the squares of its two legs.

We will apply the Pythagorean Theorem to solve for the distance between Juan and Juana. Let c be their distance. The vertical distance and the horizontal distance which are the legs of the triangle are 1 meter and 2 meters, respectively. Hence,

c² = 1² + 2² c² = 1 + 4 c² = 5 c = √5 ≈ 2.24

Therefore, the distance between Juan and Juana is √5 or approximately 2.24 meters.

What is It

Let us derive the distance formula to find the distance between two points in the coordinate plane. Consider the figure below, let d be the distance between point P₁ and point P₂ with coordinates (x₁, y₁) and (x₂, y₂), respectively. It is also shown from the figure that the horizontal distance, b, is x₂ − x₁ and the vertical distance, a, is y₂ − y₁.

[chart: coordinate diagram; points P₁ and P₂ joined by segment d, with horizontal distance b and vertical distance a] Labels: P₂(x₂, y₂); P₁(x₁, y₁); d; a = y₂ − y₁; b = x₂ − x₁; y₂; y₁; x₁; x₂ Axes: y; x

Using the Pythagorean Theorem, we have:

d² = a² + b² d² = (y₂ − y₁)² + (x₂ − x₁)²

Taking the square roots of both sides of the equation:

√d² = √((y₂ − y₁)² + (x₂ − x₁)²) d = √((y₂ − y₁)² + (x₂ − x₁)²)

The Distance Formula

The distance, d, between points A(x₁, y₁) and B(x₂, y₂) may be found using the formula:

d = √((x₂ − x₁)² + (y₂ − y₁)²)

Example 1. Plot and solve the distance between the points E (1, 1) and R (2,4) in the coordinate plane.

Solution:

A) Graph:	B) Distance:
[chart: coordinate plane; points E(1,1) and R(2,4) plotted and joined by a segment] Labels: R(2,4); E(1,1) Axes: y; x	Let P₁ (x₁, y₁) be equal to E (1, 1) and P₂ (x₂, y₂) be equal to R(2,4). Hence, x₁ = 1 and y₁ = 1; and x₂ = 2 and y₂ = 4 Substitute the coordinates of points E and R in the formula: d = √((x₂ − x₁)² + (y₂ − y₁)²) d = √((2 − 1)² + (4 − 1)²) Simplify it further: d = √(1)² + (3)² d = √1 + 9 d = √10 Therefore the distance between points E and R is √10.

Example 2. Show that the figure formed when points L (-4, 4), O (3, 9), V (8, 2), E (1,-3) are connected consecutively is a square, then find its perimeter.

Solution:

1. Plot the points L (-4, 4), O (3, 9), V (8, 2), E (1,-3) on the coordinate plane.

[chart: coordinate plane; points L(-4,4), O(3,9), V(8,2), and E(1,-3) connected consecutively to form a quadrilateral] Labels: x-axis; L(-4,4); O(3,9); V(8,2); E(1,-3)

2. To show that the figure formed is a square, we need to show that all the sides are equal in length and all angles are right angles.

A) Show that the lengths of LO̅, OV̅, VE̅ and EL̅ are congruent.

i) to solve the length of LO, we will use the points L (-4, 4) & O (3, 9)

LO = √((3 − −4)² + (9 − 4)²) = √74

ii) to solve for the length of OV, use O (3, 9) & V (8, 2)

OV = √((8 − 3)² + (2 − 9)²) = √74

iii) to solve for the length of VE, use V (8, 2) & E (1,-3)

VE = √((1 − 8)² + (−3 − 2)²) = √74

iv) to solve for the length of EL, use E (1,-3) & L (-4, 4)

VE = √((−4 − 1)² + (4 − −3)²) = √74

[transcriber’s note: the source prints “VE” on the left side of this line, where “EL” is expected; transcribed as printed]

Thus, LO=OV=VE=EL.

B) Show that all angles L, O, V, & E are right angles. Note that if two lines are perpendicular, then they formed a right angle and their slopes are negative reciprocal of each other. Hence, we need to find the slope of each side.

Recall: Slope of a line The slope of a line, m, given two points (x₁, y₁) and (x₂, y₂) is

m = (y₂ − y₁)/(x₂ − x₁)

i) The slope of LO, use the points L (-4, 4) & O (3, 9)

m_LO = (9 − 4)/(3 − (−4)) = 5/7

ii) The slope of OV, use O (3, 9) & V (8, 2)

m_OV = (2 − 9)/(8 − 3) = −7/5

iii) The slope of VE, use V (8, 2) & E (1,-3)

m_VE = (−3 − 2)/(1 − 8) = 5/7

iv) The slope of EL, use E (1,-3) & L (-4, 4)

m_EL = (4 − (−3))/(−4 − 1) = −7/5

C) Based from the computed slopes of each segment, we can conclude the following:

i. The slopes of LO and OV are negative reciprocals of each other, then angle O is a right angle. ii. The slopes of OV and VE are negative reciprocals of each other, then angle V is a right angle. iii. The slopes of VE and EL are negative reciprocals of each other, then angle E is a right angle. iv. The slopes of EL and LO are negative reciprocals of each other, then angle L is a right angle.

3. Since all the sides of the figure are congruent and all the angles are right angles, then quadrilateral LOVE is a square.

   
   

4. To solve for the perimeter, use the formula for the perimeter of square.

   
   

P = 4s = 4(√74) ≈ 34.409 units

Example 3. Find the coordinates (in terms of a and b) of points G, O, L, and D in the figure at the right. (Remember that the coordinate of points on the same vertical line share the same x – coordinate while points on the same horizontal line share the same y – coordinate.)

[chart: coordinate diagram of a rectangle with vertices G, O, L, and D on the axes] Labels: y-axis; x-axis; G(−a, ?); O(?, b); D(?, ?); L(a, ?)

Solution:

a) finding the coordinates of G: Since G and O lie on the same horizontal line, it implies that they have the same y – coordinate. Thus, the coordinate of G is (−a, b)

b) finding the coordinates of O: Since O and L lie on the same vertical line, it implies that they have the same x – coordinate. Thus, the coordinate of O is (a, b).

c) finding the coordinates of L: Since L lies on the x-axis, it means that its y – coordinate is 0. Thus, the coordinate of L is (a, 0).

d) finding the coordinates of D: Since D lies on the x – axis, it means that its y- coordinate is 0. D also lies on the same vertical line with G which means that they have the same x – coordinate. Thus, the coordinates of D is(−a, 0).

e) Therefore, the coordinates are G(−a, b), O (a, b), L (a, 0) & D(−a, 0).

Example 4. Prove that the two sides of an isosceles triangle are congruent.

[illustration: isosceles triangle ABI with tick marks on sides AB and IB; vertices labeled B (top), A (bottom left), I (bottom right)]

Given: ΔABI with sides AB̅ , IB̅ ,IA̅ Prove: AB̅ ≅ IB̅

To prove:

1. Place ΔABI on the coordinate plane and label the coordinate points as shown below:

[chart: coordinate diagram of isosceles triangle ABI on the coordinate plane] Labels: B(0,b); A(-a,0); I(a,0) Axes: y; x

Prov

[transcriber’s note: the word “Prov” stands alone in the source before “To prove:”; transcribed as printed]

To prove:

2. Find the distance between A(-a,0) and B (0,b). Substitute –a and 0 to x₂ and x₁, respectively, and 0 and b to y₂ and y₁, respectively.

AB = √((−a − 0)² + (0 − b)²) AB = √(a² + b²)

3. Find the distance between I (a,0) and B (0,b). Substitute a and 0 to x₂ and x₁, respectively, and 0 and b to y₂ and y₁, respectively.

IB = √((a − 0)² + (0 − b)²) IB = √(a² + b²)

4. Since AB = √(a² + b²) and IB = √(a² + b²), by substitution we can say that AB = IB.

∴ AB̅ ≅ IB̅. The two sides of an isosceles triangle are congruent.

Here are some suggestions to help you place figures for your proofs.

1. Use the origin of the coordinate plane as vertex or center of the figure.
2. Place at least one side of the figure on a coordinate axis, either the x or y- axis.
3. Keep the figure within the first quadrant if possible.
4. Use coordinates that make computations as simple as possible.

What’s More

Activity 2: The coordinates of points C and R are (2, 5) and (7, 2), respectively. Plot these points on the coordinate plane and find their distance.

[illustration: blank coordinate plane for plotting, with x-axis and y-axis marked 1 to 7]

Activity 3. Use the given to answer each question.

A) Show that the figure formed when the points F (-2, 6), U (-2, -3), N (7, 6) are connected consecutively is an isosceles right triangle and find its area.

B) Supply the missing coordinates of the points of each figure below without introducing new letters.

[chart: four coordinate-plane figures for item B; labels as printed]

1. PARALLELOGRAM: (a,c); B; (b,0); O; axes Y and X
2. SQUARE: (0, r); T; S; (g, 0); axes Y and X
3. EQUILATERAL TRIANGLE: (a, b); W; o; axes Y and X
4. SQUARE: S; (a,b); T; o; D; axes Y and X

C) Prove using coordinate plane that the diagonals of an isosceles trapezoid are congruent.

What I Have Learned

1. The Distance Formula

The distance, d, between points A(x₁, y₁) and B(x₂, y₂) may be found using the formula:

d = √((x₂ − x₁)² + (y₂ − y₁)²)

2. With the use of the coordinate plane, distance formula is very helpful in proving properties of some geometric figures. However, the geometric figure must be placed properly on the coordinate axes so that it will be easier to prove. Here are some of the appropriate ways of placing geometric figures on the coordinate plane.

What I Can Do

Activity 4. Answer the following problems:

1. Lieutenant Santos orders an air strike in the battlefield targeting the enemy at a coordinate (2, 5). If he is positioned at a coordinate (-14, -12), how far is he from the target area? If the danger zone is within the 10 km radius from the strike point, is Lt. Santos safe? (Let 1 unit = 1 km)

   
   

2. Chester and his father stood on their newly-bought rectangular lot whose length and width are 60 and 40 meters, respectively. His father told him that the place they are standing at is one of the four boundary points of their lot. He then told Chester that they are going to put a marker on each of the four boundary points. Help Chester to locate the coordinates of the 3 boundary points using the Cartesian coordinate plane if the coordinates of their location is at (-30, 20). (Let 1 unit = 1 meter)

   
   

[chart: three coordinate-plane figures with labels and notes as printed]

1. BIO: B (0,a); (0,0)O; I (b,0); “△ BIO is a right triangle”; “Angle O is a right angle”
2. Square: S (0,a); O (b,a); L (0,0); I (b,0); “□ is a square”; “SO=OI=SL=LI”
3. ICE: I (a,c); E (0,0); C (b,0); “△ ICE is a scalene triangle”; “IE ≠ IC ≠ EC”

[transcriber’s note: the source prints the inequality signs in the ICE note as a symbol that resembles “≠”; transcribed as ≠]

<!– transcriber’s note: the cover page of this module is not among the pages provided, so no document title is transcribed here –>

Assessment

Direction: Choose the letter of the correct answer. Write it on a separate sheet of paper.

1. In the Cartesian plane, what is the distance of the point (-5, 6) from the origin?

   • A) 61
   • B) 11
   • C) √61
   • D) 11

2. What is the distance between point A(−3, 1) and point B (11, 1)?

   • A) 11
   • B) 12
   • C) 13
   • D) 14

3. Which of the following should be the value of y so that the distance between the points (2, −2) and (2, y) is 7?

   • A) 2
   • B) 3
   • C) 4
   • D) 5

4. Which of the following describes the distance formula?

   • A) d = √((x₂ − x₁)² + (y₂ − y₁)²)
   • B) d = √((x₂ + x₁)² + (y₂ + y₁)²)
   • C) d = √((x₂ − x₁)² − (y₂ − y₁)²)
   • D) d = √((x₂ + x₁)² − (y₂ + y₁)²)

5. What is the distance between two points whose coordinates are (4,−3) and (−4,3)?

   • A) 4
   • B) 6
   • C) 8
   • D) 10

6. What is the distance between point A(3a, 2a) and point D(−a, −6a)?

   • A) 2a√10
   • B) 4a√5
   • C) 5a
   • D) 8a

7. What is the area of a triangle whose vertices are (0,2), (0,0) (5,0)?

   • A) 5 square unit
   • B) 8 square unit
   • C) 10 square unit
   • D) 12 square unit

8. What kind of triangle is formed when the vertices (−3, 5), (−3, 1) and (2, 1) are plotted on the Cartesian plane?

   • A) equilateral
   • B) isosceles
   • C) right
   • D) scalene

9. What type of quadrilateral is formed by the given vertices C(0,0), A(1,2), R(4,2) and E(3,0)?

   • A) kite
   • B) parallelogram
   • C) rectangle
   • D) square

10. In Quadrilateral LOVE, what is the length of the diagonal LV̅.

    • [chart: Quadrilateral LOVE on the Cartesian plane; labels L(0,b), O, E(0,0), V(a,0); diagonal LV drawn]
    • A) √(a² + b²)
    • B) √(a² − b²)
    • C) √(a + b)
    • D) a² + b²

Assessment

11. What is the distance of point I from the origin in Trapezoid LIVE?

    • [chart: Trapezoid LIVE on the Cartesian plane; labels L, I(c,e), E, V(d,0)]
    • A) √(c − e)
    • B) √(c + e)
    • C) √(c² − e²)
    • D) √(c² + e²)

12. GIVE is a square. Find its perimeter.

    • [illustration: square GIVE drawn on the Cartesian axes; labels G, I, E, V (a,b)]
    • A) 4(a + b)
    • B) 4√(a + b)
    • C) 4(a² + b²)
    • D) 4√(a² + b²)

13. To prove that triangle CER is a right triangle, which of the following should be proven?

    • [chart: triangle CER on a coordinate grid; labels C, E, R]
    • A) ER̅ = RC̅ + CE̅
    • B) RC̅ = ER̅ + CE̅
    • C) ER̅² = RC̅² + CE̅²
    • D) RC̅² = ER̅² + CE̅²

14. To prove that the diagonals of rectangle RICE are congruent which of the following should be proven?

    • A) RI̅ ≅ IC̅
    • B) IC̅ ≅ CE̅
    • C) RC̅ ≅ EC̅
    • D) RC̅ ≅ IE̅

15. To prove that triangle ABC is an equilateral triangle which of the following should be proven?

    • A) AB ≅ BA
    • B) AB ≅ BC ≅ AC
    • C) AB ≅ BC
    • D) AB ≇ BC ≅ AC

Additional Activity

Direction: Answer the following problems on a separate sheet of paper.

1. Draw segment MR whose endpoints are M(-1,5) and R(2, -4) on the Cartesian plane.
   a) Find the length of segment MR.
   b) If point S(x, −1) lies on segment MR and the length of segment SR is √10, what is x?

   
   

2. Refer to Figure 1. Solve the distance d in terms of a and b.

   • [illustration: Figure 1; segment d joins point b on the y-axis to point a on the x-axis]

3. The vertices of a quadrilateral are R(0,0), U(a,0), D(a,b), Y(0,b).
   a) Illustrate the quadrilateral in the coordinate plane.
   b) Find the length of each diagonal.
   c) Compare the lengths of the diagonals.

   
   

4. Refer to Figure 2. The x-coordinate of D is the mean of the x-coordinates of the vertices of triangle ABC and its y-coordinate is the mean of the y-coordinates of the vertices of triangle ABC. Find the distance between points A and D.

   • [illustration: Figure 2; triangle ABC with A(0,0), B(a,0), C(b,c)]

Answer Key

<!– transcriber’s note: this page is printed rotated 180 degrees; the table below is transcribed in upright reading order, and some cell placements in the Additional Activity row are uncertain –>

What I Know	Activity 1	Activity 2
1) B 2) D 3) D 4) A 5) C 6) B 7) A 8) C 9) B 10) D 11) C 12) A 13) B 14) B 15) D	A) [chart: points plotted on the Cartesian plane; labels A, C, E, N; values not labeled] B) a) (−4, 7) b) (7, 8) c) (−7, −6) d) (2, −2)	d = √34 [chart: segment joining R(7,2) and C(2,5) on the Cartesian plane]

Assessment	Activity 4	Activity 3
1) C 2) D 3) D 4) A 5) B 6) B 7) A 8) C 9) B 10) A 11) D 12) D 13) D 14) D 15) B	1) a) √545 ≈ 23.35 km b) Yes 2) a) (30, 20) b) (−30, −20) c) (30, −20)	A) 40.5 square units B) 1) (b + a, c) O (0, 0) 2) T (g, r) S (0, 0) 3) W (2a, 0) O (0, 0) 4) S (−a, b) T (−a, 0) D) (a, 0)

Additional Activity
1) a) 3√10 b) x = 1 2) d = √(a² + b²) 3.a) RD̅ = √(a² + b²) 3.b) YU̅ = √(a² + b²) 3.c) equal 4) d = √(((b + a)² + c²) / 9) [?]	[chart: quadrilateral with vertices R, U, D, Y on the coordinate plane; labels R, U, D, Y, a, b]

References

Callanta, Melvin., Canonigo, Allan M., Chua, Arnaldo I., Cruz, Jerry D., Esparrago, Mirla S., Garcia, Elino S., Magnaye, Aries N., Orines, Fernando B., Perez Rowena S., Ternida, Conception S. Mathematics-Grade 10 Learner’s Module, Meralco Avenue, Pasig City: REX Bookstore, Inc. and Department of Education, 2015

Larson, R. Precalculus with Limits. Boston, MA: Houghton Mifflin Co., 2006

Nivera, Gladys C., Ph.D. and Lapinid, Minie Rose C., Ph.D. Grade 10 Mathematics-Patterns and Practicalities. Makati City: Don Bosco Press, 2015

Young, C. Precalculus. Hoboken, N.J.: Wiley, 2010

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