Lesson 1: Slope and Distance

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1 Common Core Georgia Performance Standards MCC8.G.8* (Transition Standard ; asterisks denote Transition Standards) MCC9 1.G.GPE.4 MCC9 1.G.GPE.5 Essential Questions 1. How is the Pythagorean Theorem used to find distance?*. How is finding the distance between two points in two dimensions similar to finding the distance between two points in one dimension? How is it different? 3. How are the distance formula and the Pythagorean Theorem related? 4. How can algebra be used to explore properties of geometric shapes in the plane? 5. What are real-world applications of slope and the distance formula? 6. Is there a relationship between the slopes of parallel lines? 7. Is there a relationship between the slopes of perpendicular lines? WORDS TO KNOW absolute value congruent distance formula hypotenuse leg of a triangle parallel parallelogram perpendicular the distance from zero on the number line having the same measure, length, or size a formula that states the distance between points (x 1, y 1 ) and (x, y ) is equal to ( x x ) + ( y y ) the longest side of a triangle; opposite the right angle of any right triangle either of the two shorter sides of any right triangle lines that never intersect and have equal slope a quadrilateral with opposite sides parallel lines that intersect at a right angle (90 ); their slopes are opposite reciprocals U6-

2 Pythagorean Theorem quadrilateral rectangle rhombus right triangle slope square a theorem that relates the length of the hypotenuse of a right triangle (c) to the lengths of its legs (a and b). The theorem is a + b = c. a polygon with four sides a parallelogram with opposite sides that are congruent and consecutive sides that are perpendicular a parallelogram with four congruent sides a triangle with exactly one right (90 ) angle measure of the rate of change of one variable with respect to another y y1 y rise variable; slope = = = x x x run a parallelogram with four congruent sides and four right angles Recommended Resources Math Open Reference. Distance from a point to a line. This site shows how to find the shortest distance from a point to a line, and provides an animation that allows you to manipulate both a point and a line on a coordinate plane to see how the perpendicular distance changes. MathsNet. Interactive Shape. Use this animation to determine the relationship between lines. Purplemath. Straight-Line Equations: Parallel and Perpendicular Lines. This site summarizes how to write equations that are parallel and perpendicular to a given point. University of Puerto Rico. Practice: Coordinates on the Cartesian Plane. Plot coordinates on a Cartesian plane and calculate the distance between the points. U6-3

3 NAME: CONNECTING ALGEBRA AND GEOMETRY THROUGH COORDINATES Lesson 6.1.1: Applying the Pythagorean Theorem* Warm-Up A house painter places the bottom of his 0-foot ladder 1 feet from a house. The top of the ladder rests against the house. x 0 ft 1 ft 1. How far up the house does the ladder reach?. Suppose the same 0-foot ladder is placed so that it reaches 17 feet up the side of the house. Approximately how far out from the house was the bottom of the ladder placed? State your answer both exactly and approximately rounded to the nearest tenth. U6-4

4 Lesson 6.1.1: Applying the Pythagorean Theorem* Common Core Georgia Performance Standard MCC8.G.8* Warm-Up Debrief A house painter places the bottom of his 0-foot ladder 1 feet from a house. The top of the ladder rests against the house. 1. How far up the house does the ladder reach? When a ladder is leaned against a house, a right triangle is formed. The ladder represents the hypotenuse of the right triangle. The side of the house and the ground represent the legs of the right triangle. To represent this situation algebraically, use the Pythagorean Theorem: a + b = c, where a and b are the legs and c is the hypotenuse. a + b = c Original equation 1 + b = 0 Substitute known values b = 400 Simplify. b = 56 Subtract 144 from both sides. b 56 Take the square root of both sides. b = 16 The ladder reaches 16 feet up the side of the house. U6-5

5 . Suppose the same 0-foot ladder is placed so that it reaches 17 feet up the side of the house. Approximately how far out from the house was the bottom of the ladder placed? State your answer both exactly and approximately rounded to the nearest tenth. The ladder still represents the hypotenuse of the right triangle. This time we know that the ladder reaches 17 feet up the side of the house. Again, use the Pythagorean Theorem to find the distance from the house the ladder is placed. a + b = c Original equation 17 + b = 0 Substitute known values b = 400 Simplify. b = 111 Subtract 89 from both sides. b 111 Take the square root of both sides. b 10.5 The ladder was placed 111 feet, or approximately 10.5 feet, from the side of the house. Connection to the Lesson Students will extend their understanding of the Pythagorean Theorem to find the distance between two points on a coordinate system. The process for finding the distance between two points on a coordinate system is similar to finding the length of a ladder placed a certain distance from the side of a house and reaching a certain height along the side of the house. Students will be asked to first calculate the vertical height and horizontal length before using the Pythagorean Theorem to calculate the distance between the two points. U6-6

6 Prerequisite Skills This lesson requires the use of the following skills: using the Pythagorean Theorem to solve for unknown lengths of a right triangle solving radical equations approximating values for irrational numbers Introduction Finding the distance between two points on a coordinate system is similar to finding the distance between two points on a number line. It is different in that finding the distance between two points on a coordinate system makes use of two dimensions, but the distance along a number line is only one dimension. Also, we can t always easily measure or count the number of units between two points on a coordinate system. Instead, we can use the Pythagorean Theorem to find the number. The Pythagorean Theorem relates the length of the hypotenuse of a right triangle (c) to the lengths of its legs (a and b), a + b = c, to calculate the distance between the two points. Remember that distance is always positive; therefore, we must take the absolute value, or the distance from zero, of distances calculated. Key Concepts Reviewing Distance on a Number Line To find the distance between two points, a and b, on a number line, find the absolute value of the difference of a and b. This can be expressed algebraically as a b or b a. For example, to find the distance between 4 and 5, take the absolute value of the difference of 4 and = 9 = 9 or 5 4 = = 9 = 9 The distance between the numbers 4 and 5 is 9 units. It may not be that easy to find the distance on a coordinate system. To find the distance between two points on a coordinate system, we must use the Pythagorean Theorem. Reviewing the Pythagorean Theorem Right triangles are triangles with one right (90 ) angle. The side that is the longest and is always across from the right angle is called the hypotenuse. The two shorter sides are referred to as the legs of the right triangle. U6-7

7 We can use the Pythagorean Theorem to calculate the length of any one of the three sides. For example, to find the length of the hypotenuse of a triangle with legs of 5 and 7 units, we use the Pythagorean Theorem. a + b = c Pythagorean Theorem = c Substitute known values = c Simplify. 74 = c Simplify. 74 c Take the square root of both sides of the equation. c= The length of the hypotenuse of the right triangle with side lengths 5 and 7 is 74, or approximately 8.6 units. The Pythagorean Theorem can help us find the distance between two points on a coordinate system. Finding the Distance Between Two Points Using the Pythagorean Theorem 1. Plot the points on a coordinate system.. Draw lines to form a right triangle, using each point as the end of the hypotenuse. 3. Calculate the length of the vertical side, a, of the right triangle by taking the absolute value of the difference of the y-values ( y y 1 ). 4. Calculate the length of the horizontal side, b, of the right triangle by taking the absolute value of the difference of the x-values ( x x 1 ). 5. Use the Pythagorean Theorem to calculate the length of the hypotenuse, c. Common Errors/Misconceptions incorrectly identifying the x- and y-values substituting the x- and y-values incorrectly to find the lengths of the triangle s sides forgetting to take the square root of c in order to find the distance between the points U6-8

8 Guided Practice Example 1 Use the Pythagorean Theorem to calculate the distance between the points (, 5) and ( 4, 3). 1. Plot the points on a coordinate system (, 5) (-4, -3) U6-9

9 . Draw lines to form a right triangle, using each point as the end of the hypotenuse (, 5) c a (-4, -3) b Calculate the length of the vertical side, a, of the right triangle. Let (x 1, y 1 ) = (, 5) and (x, y ) = ( 4, 3). y y 1 = 3 5 = 8 = 8 The length of side a is 8 units. 4. Calculate the length of the horizontal side, b, of the right triangle. x x 1 = 4 = 6 = 6 The length of side b is 6 units. U6-10

10 5. Use the Pythagorean Theorem to calculate the length of the hypotenuse, c. a + b = c Pythagorean Theorem = c Substitute values for a and b = c Simplify each term. 100 = c Simplify. 100 c Take the square root of both sides of the equation. 10 = c The distance between the points (, 5) and ( 4, 3) is 10 units. U6-11

11 Example Tyler and Arsha have mapped out locations for a game of manhunt. Tyler s position is represented by the point (, 1). Arsha s position is represented by the point ( 7, 9). Each unit is equivalent to 100 feet. What is the approximate distance between Tyler and Arsha? 1. Plot the points on a coordinate system. Arsha (-7, 9) Tyler (-, 1) U6-1

12 . Draw lines to form a right triangle, using each point as the end of the hypotenuse. Arsha (-7, 9) a c Tyler b 1 (-, 1) Calculate the length of the vertical side, a, of the right triangle. Let (x 1, y 1 ) = (, 1) and (x, y ) = ( 7, 9). y y 1 = 9 1 = 8 = 8 The length of side a is 8 units. 4. Calculate the length of the horizontal side, b, of the right triangle. x x 1 = 7 = 5 = 5 The length of side b is 5 units. U6-13

13 5. Use the Pythagorean Theorem to calculate the length of the hypotenuse, c. a + b = c Pythagorean Theorem = c Substitute values for a and b = c Simplify each term. 89 = c Simplify. 89 c Take the square root of both sides of the equation. c= The distance between Tyler and Arsha is approximately 9.4 units or 940 feet. U6-14

14 Example 3 Kevin is standing miles due north of the school. James is standing 4 miles due west of the school. What is the distance between Kevin and James? 1. Plot the points on a coordinate system. Let (0, 0) represent the location of the school. Kevin is standing miles due north, so his location is units above the origin, or at the point (0, ). James is standing 4 units due west, so his location is 4 units to the left of the origin, or at the point ( 4, 0) Kevin (0, ) James (-4, 0) School (0, 0) U6-15

15 . Draw lines to form a right triangle, using each point as the end of the hypotenuse Kevin (0, ) James (-4, 0) School (0, 0) 3. Calculate the length of the vertical side, a, of the right triangle. Let (x 1, y 1 ) = (0, ) and (x, y ) = ( 4, 0). y y 1 = 0 = = The length of side a is units. 4. Calculate the length of the horizontal side, b, of the right triangle. x x 1 = = 4 0 = 4 = 4 The length of side b is 4 units. U6-16

16 5. Use the Pythagorean Theorem to calculate the length of the hypotenuse, c. a + b = c Pythagorean Theorem + 4 = c Substitute values for a and b = c Simplify each term. 0 = c Simplify. 0 c Take the square root of both sides of the equation. c= The distance between Kevin and James is approximately 4.5 miles. U6-17

17 NAME: CONNECTING ALGEBRA AND GEOMETRY THROUGH COORDINATES Problem-Based Task 6.1.1: Pride Points Serengeti National Park, located in Tanzania, is often referred to as one of the best wildlife reserves in Africa because of the large populations of animals. Park officials created a map to represent an area of the park. Lion prides are located at the points ( 4, 1), (1, 5), (6, ), and (3, 7). Water holes are located at the points (1, 1) and (8, 4). The pride that is farthest from the water hole located at point (1, 1) is the closest pride to the water hole located at (8, 4). How much farther is this pride from the water hole at (1, 1) than the water hole at (8, 4)? Each unit represents 1,000 feet. Explain your reasoning. U6-18

18 NAME: CONNECTING ALGEBRA AND GEOMETRY THROUGH COORDINATES Problem-Based Task 6.1.1: Pride Points Coaching a. What is the distance between the pride located at ( 4, 1) and the water hole located at the point (1, 1)? b. What is the distance between the pride located at (1, 5) and the water hole located at the point (1, 1)? c. What is the distance between the pride located at (6, ) and the water hole located at the point (1, 1)? d. What is the distance between the pride located at (3, 7) and the water hole located at the point (1, 1)? e. Which calculated distance is the longest? f. Which pride is the farthest from the water hole located at ( 1, 1)? g. What is the distance between this pride and the second water hole located at (8, 4)? h. Compare this pride s distance from each of the water holes. What is the difference? U6-19

19 Problem-Based Task 6.1.1: Pride Points Coaching Sample Responses a. What is the distance between the pride located at ( 4, 1) and the water hole located at the point (1, 1)? Let (x 1, y 1 ) = ( 4, 1) and (x, y ) = (1, 1). Use the Pythagorean Theorem to find the lengths of a and b. a = y y 1 = 1 1 = 0 = 0 b = x x 1 = 1 4 = 5 = 5 a + b = c Pythagorean Theorem = c Substitute values for a and b = c Simplify each term. 5 = c Simplify. 5 c Take the square root of both sides of the equation. c = 5 units Each unit represents 1,000 feet. This pride is 5,000 feet from the water hole. b. What is the distance between the pride located at (1, 5) and the water hole located at the point (1, 1)? Let (x 1, y 1 ) = (1, 5) and (x, y ) = (1, 1). Use the Pythagorean Theorem to find the lengths of a and b. a = y y 1 = 1 5 = 6 = 6 b = x x 1 = 1 1 = 0 = 0 a + b = c Pythagorean Theorem = c Substitute values for a and b = c Simplify each term. U6-0

20 36 = c Simplify. 36 c Take the square root of both sides of the equation. c = 6 units This pride is 6,000 feet from the water hole. c. What is the distance between the pride located at (6, ) and the water hole located at the point (1, 1)? Let (x 1, y 1 ) = (6, ) and (x, y ) = (1, 1). Use the Pythagorean Theorem to find the lengths of a and b. a = y y 1 = 1 = 3 = 3 b = x x 1 = 1 6 = 5 = 5 a + b = c Pythagorean Theorem = c Substitute values for a and b = c Simplify each term. 34 = c Simplify. 34 c Take the square root of both sides of the equation. c= units This pride is approximately 5,800 feet from the water hole. d. What is the distance between the pride located at (3, 7) and the water hole located at the point (1, 1)? Let (x 1, y 1 ) = (3, 7) and (x, y ) = (1, 1). Use the Pythagorean Theorem to find the lengths of a and b. a = y y 1 = 1 7 = 6 = 6 b = x x 1 = 1 3 = = a + b = c Pythagorean Theorem 6 + = c Substitute values for a and b. U6-1

21 = c Simplify each term. 40 = c Simplify. 40 c Take the square root of both sides of the equation. c= units This pride is approximately 6,300 feet from the water hole. e. Which calculated distance is the longest? The longest distance is 6,300 feet. f. Which pride is the farthest from the water hole located at ( 1, 1)? The pride located at (3, 7) is the farthest from the water hole at ( 1, 1). g. What is the distance between this pride and the second water hole located at (8, 4)? Let (x 1, y 1 ) = (3, 7) and (x, y ) = (8, 4). Use the Pythagorean Theorem to find the lengths of a and b. a = y y 1 = 4 7 = 3 = 3 b = x x 1 = 8 3 = 5 = 5 a + b = c Pythagorean Theorem = c Substitute values for a and b = c Simplify each term. 34 = c Simplify. 34 c Take the square root of both sides of the equation. c= units This pride is 5,800 feet from the second water hole. U6-

22 h. Compare this pride s distance from each of the water holes. What is the difference? Subtract the two distances found = 500 There is a 500-foot difference between this pride and each of the water holes. Recommended Closure Activity Select one or more of the essential questions for a class discussion or as a journal entry prompt. U6-3

23 NAME: CONNECTING ALGEBRA AND GEOMETRY THROUGH COORDINATES Practice 6.1.1: Applying the Pythagorean Theorem* Use the Pythagorean Theorem to find the distance between the indicated points. Round to the nearest tenth. 1. (, 5) and (4, 10). (4, ) and (5, 0) 3. ( 1, 1) and ( 3, 4) 4. ( 4, 5) and (1, 5) Questions 5 10 refer to the map below. The locations of the points are as follows: Farmer s Market: (0, 6) Marina: (, 4) Park: (6, 7) Bank: ( 5, 3) School: (9, 3) Post Office: (, ) Each unit on the map is equivalent to 1,000 feet. Round your answers to the nearest foot. continued U6-4

24 NAME: CONNECTING ALGEBRA AND GEOMETRY THROUGH COORDINATES 5. What is the distance from the post office to the bank? 6. What is the distance from the marina to the school? 7. What is the distance from the park to the farmer s market? 8. What is the distance from the farmer s market to the bank? 9. What is the distance from the bank to the park? 10. What is the distance from the marina to the farmer s market? U6-5

25 NAME: CONNECTING ALGEBRA AND GEOMETRY THROUGH COORDINATES Lesson 6.1.: Using Coordinates to Prove Geometric Theorems with Slope and Distance Warm-Up 6.1. The graph below shows the average number of miles an airplane travels over the course of its flight Distance (miles) Time (minutes) 1. What is the average rate of change per hour?. How many miles would the airplane travel in 9 hours? U6-6

26 Lesson 6.1.: Using Coordinates to Prove Geometric Theorems with Slope and Distance Common Core Georgia Performance Standards MCC9 1.G.GPE.4 MCC9 1.G.GPE.5 Warm-Up 6.1. Debrief The graph below shows the average number of miles an airplane travels over the course of its flight Distance (miles) Time (minutes) 1. What is the average rate of change per hour? To calculate the average rate of change per hour, first locate any two points on the graph. Let s use (60, 600) and (180, 1800). U6-7

27 Rate of change can be found using the slope formula: Let (x 1, y 1 ) = (60, 600) and (x, y ) = (180, 1800). y y Rate of change= x x = = 10 Original formula y y x x Substitute values for x 1, x, y 1, and y. Simplify. The airplane is traveling 10 miles per minute. The problem asks for the rate of change per hour. Convert minutes to hours. 10 miles 60 minutes 600 miles = 1 minute 1hour 1hour The airplane travels an average of 600 miles per hour... How many miles would the airplane travel in 9 hours? Use the information found in question 1 to determine the number of miles the airplane would travel. 600 miles 9 hours 5400 miles = 1 hour 1 1 The airplane would travel an average of 5,400 miles in 9 hours. Connection to the Lesson In this lesson, students will use introductory ideas about slope to expand upon their understanding of slope. Students will need to be comfortable with calculating slopes of given lines as they prove simple geometric theorems algebraically. U6-8

28 Prerequisite Skills This lesson requires the use of the following skills: calculating slope writing equations of lines using the Pythagorean Theorem Introduction It is not uncommon for people to think of geometric figures, such as triangles and quadrilaterals, to be separate from algebra; however, we can understand and prove many geometric concepts by using algebra. In this lesson, you will see how the distance formula originated with the Pythagorean Theorem, as well as how distance between points and the slope of lines can help us to determine specific geometric shapes. Key Concepts Calculating the Distance Between Two Points To find the distance between two points on a coordinate plane, you have used the Pythagorean Theorem. After creating a right triangle using each point as the end of the hypotenuse, you calculated the vertical height (a) and the horizontal height (b). These lengths were then substituted into the Pythagorean Theorem (a + b = c ) and solved for c. The result was the distance between the two points. This is similar to the distance formula, which states the distance between points (x 1, y 1 ) and (x, y ) is equal to ( x x ) + ( y y ) Using the Pythagorean Theorem: Find the length of a: y y 1. Find the length of b: x x 1. Substitute these values into the Pythagorean Theorem.. U6-9

29 c = a + b c = y y + x x c= y y + x x Using the distance formula: distance = ( x x ) + ( y y ) We will see in the Guided Practice an example that proves the calculations will result in the same distance. Calculating Slope To find the slope, or steepness of a line, calculate the change in y divided by the change in x using the formula m= y y. x x Parallel and Perpendicular Lines Parallel lines are lines that never intersect and have equal slope. To prove that two lines are parallel, you must show that the slopes of both lines are equal. Perpendicular lines are lines that intersect at a right angle (90 ). The slopes of perpendicular lines are always opposite reciprocals. To prove that two lines are perpendicular, you must show that the slopes of both lines are opposite reciprocals. When the slopes are multiplied, the result will always be 1. Horizontal and vertical lines are always perpendicular to each other. Common Errors/Misconceptions incorrectly using the x- and y-coordinates in the distance formula subtracting negative coordinates incorrectly incorrectly calculating the slope of a line incorrectly determining the slope of a line that is perpendicular to a given line assuming lines are parallel or perpendicular based on appearance only making determinations about the type of polygon without making all the necessary calculations U6-30

30 Guided Practice 6.1. Example 1 Calculate the distance between the points (4, 9) and (, 6) using both the Pythagorean Theorem and the distance formula. 1. To use the Pythagorean Theorem, first plot the points on a coordinate system. 10 (4, 9) (-, 6) U6-31

31 . Draw lines to form a right triangle, using each point as the end of the hypotenuse. 10 (4, 9) 9 8 c a 7 (-, 6) 6 b Calculate the length of the vertical side, a, of the right triangle. Let (x 1, y 1 ) = (4, 9) and (x, y ) = (, 6). y y 1 = 6 9 = 3 = 3 The length of side a is 3 units. 4. Calculate the length of the horizontal side, b, of the right triangle. x x 1 = 4 = 6 = 6 The length of side b is 6 units. U6-3

32 5. Use the Pythagorean Theorem to calculate the length of the hypotenuse, c. a + b = c Pythagorean Theorem = c Substitute values for a and b = c Simplify each term. 45 = c Simplify. 45 c Take the square root of both sides of the equation. c= The distance between the points (4, 9) and (, 6) is approximately 6.7 units. 45, or 6. Now use the distance formula to calculate the distance between the same points, (4, 9) and (, 6). Let (x 1, y 1 ) = (4, 9) and (x, y ) = (, 6). ( x x ) + ( y y ) Distance formula ( 4) + (6 9) Substitute (4, 9) and (, 6). ( 6) + ( 3) Simplify Evaluate squares. 45 Simplify. The distance between the points (4, 9) and (, 6) is or approximately 6.7 units. Both calculations will produce the same results each time. 45 units, U6-33

33 Example Determine if the line through the points ( 8, 5) and ( 5, 3) is parallel to the line through the points (1, 3) and (4, 1). 1. Plot the lines on a coordinate plane (-8, 5) 5 4 (1, 3) (-5, 3) 3 (4, 1) U6-34

34 . Calculate the slope of the line through the points ( 8, 5) and ( 5, 3). Let (x 1, y 1 ) = ( 8, 5) and (x, y ) = ( 5, 3). Substitute these values into the slope formula. m= y y x x (3) (5) = ( 5) ( 8) Original formula Substitute (x 1, y 1 ) and (x, y ). = 3 Simplify. The slope of the line through the points ( 8, 5) and ( 5, 3) is Calculate the slope of the line through the points (1, 3) and (4, 1). Let (x 1, y 1 ) = (1, 3) and (x, y ) = (4, 1). Substitute these values into the slope formula. m= y y x x (1) (3) = (4) (1) Original formula Substitute (x 1, y 1 ) and (x, y ). = 3 Simplify. The slope of the line through the points (1, 3) and (4, 1) is Determine if the lines are parallel. Parallel lines have equal slope. The slope of each line is ; therefore, the lines are parallel. 3 U6-35

35 Example 3 Determine if the line through the points (0, 8) and (4, 9) is perpendicular to the line through the points ( 9, 10) and ( 8, 6). 1. Plot the lines on a coordinate plane. (-9, 10) 10 9 (4, 9) 8 (0, 8) 7 (-8, 6) U6-36

36 . Calculate the slope of the line through the points (0, 8) and (4, 9). Let (x 1, y 1 ) = (0, 8) and (x, y ) = (4, 9). Substitute these values into the slope formula. m= y y x x (9) (8) = (4) (0) Original formula Substitute (x 1, y 1 ) and (x, y ). 1 4 Simplify. The slope of the line through the points (0, 8) and (4, 9) is Calculate the slope of the line through the points ( 9, 10) and ( 8, 6). Let (x 1, y 1 ) = ( 9, 10) and (x, y ) = ( 8, 6). Substitute these values into the slope formula. m= y y x x (6) (10) = ( 8) ( 9) Original formula Substitute (x 1, y 1 ) and (x, y ). 4 = = 4 Simplify. 1 The slope of the line through the points ( 9, 10) and ( 8, 6) is 4. U6-37

37 4. Determine if the lines are perpendicular. Perpendicular lines have slopes that are opposite reciprocals. The slope of the first line is 1 4. The reciprocal of 1 4 is 4 or 4. 1 The opposite of 4 is 4. The slope of the second line is 4. The product of the two slopes is 1. The slopes of the lines are opposite reciprocals; therefore, the lines are perpendicular. U6-38

38 Example 4 A right triangle is defined as a triangle with sides that are perpendicular. Triangle ABC has vertices A ( 4, 8), B ( 1, ), and C (7, 6). Determine if this triangle is a right triangle. When disproving a figure, you only need to show one condition is not met. 1. Plot the triangle on a coordinate plane. A (-4, 8) B (-1, ) C (7, 6) U6-39

39 . Calculate the slope of each side using the general slope formula, m= y y. x x1 () (8) slope of AB= = 6 = ( 1) ( 4) 3 slopeof BC (6) () = = 4 (7) ( 1) 8 = 1 (6) (8) slope of AC= = = (7) ( 4) Observe the slopes of each side. The slope of AB is and the slope of BC is 1. These slopes are opposite reciprocals of each other and are perpendicular. 4. Make connections. Right triangles have two sides that are perpendicular. Triangle ABC has two sides that are perpendicular; therefore, it is a right triangle. U6-40

40 Example 5 A square is a quadrilateral with two pairs of parallel opposite sides, consecutive sides that are perpendicular, and all sides congruent, meaning they have the same length. Quadrilateral ABCD has vertices A ( 1, ), B (1, 5), C (4, 3), and D (, 0). Determine if this quadrilateral is a square. 1. Plot the quadrilateral on a coordinate plane. A (-1, ) D -1 (, 0) B (1, 5) C (4, 3) U6-41

41 . First show the figure has two pairs of parallel opposite sides. Calculate the slope of each side using the general slope formula, m= y y. x x1 slopeof AB (5) () = (1) ( 1) = 3 (3) (5) slope of BC= = = (4) (1) 3 3 slopeof CD (0) (3) = = 3 () (4) = 3 (0) () slope of AD= = = () ( 1) Observe the slopes of each side. The side opposite AB is CD. The slopes of these sides are the same. The side opposite BC is AD. The slopes of these sides are the same. The quadrilateral has two pairs of parallel opposite sides. AB and BC are consecutive sides. The slopes of the sides are opposite reciprocals. BC and CD are consecutive sides. The slopes of the sides are opposite reciprocals. CD and AD are consecutive sides. The slopes of the sides are opposite reciprocals. AB and AD are consecutive sides. The slopes of the sides are opposite reciprocals. Consecutive sides are perpendicular. U6-4

42 4. Show that the quadrilateral has four congruent sides. Find the length of each side using the distance formula, d= ( x x ) + ( y y ). lengthof AB= (1 ( 1)) + (5 ) = () + (3) = 4+ 9= 13 lengthof BC= (4 1) + (3 5) = (3) + ( ) = 9+ 4= 13 lengthof CD= ( 4) + (0 3) = ( ) + ( 3) = 4+ 9= 13 lengthof AD= ( ( 1)) + (0 ) = (3) + ( ) = 9+ 4= 13 The lengths of all 4 sides are congruent. 5. Make connections. A square is a quadrilateral with two pairs of parallel opposite sides, consecutive sides that are perpendicular, and all sides congruent. Quadrilateral ABCD has two pairs of parallel opposite sides, the consecutive sides are perpendicular, and all the sides are congruent. It is a square. U6-43

43 NAME: CONNECTING ALGEBRA AND GEOMETRY THROUGH COORDINATES Problem-Based Task 6.1.: Field of Dreams A square is a quadrilateral with two pairs of parallel opposite sides, consecutive sides that are perpendicular, and all sides congruent. The bases and home plate of a baseball diamond form a square. The local recreation department has created a map of its newest baseball field. First base is represented by the point (7, 3), second base is represented by (4, 6), third base is represented by ( 5, 3), and home plate is represented by (, 6). Is this field actually a square? What is the distance from second base to home plate? What is the distance from third base to first base? U6-44

44 NAME: CONNECTING ALGEBRA AND GEOMETRY THROUGH COORDINATES Problem-Based Task 6.1.: Field of Dreams Coaching a. What does the map of the field look like? b. Which sides of the baseball field are opposites? c. What is the slope of the line connecting first base and second base? d. What is the slope of the line connecting home plate and third base? e. What is the slope of the line connecting home plate and first base? f. What is the slope of the line connecting third base and second base? g. Are opposite sides parallel? h. Which sides of the baseball field are consecutive? i. How is the slope of the line connecting first and second base related to the line connecting first base and home plate? j. How is the slope of the line connecting second base and third base related to the line connecting third base and home? continued U6-45

45 NAME: CONNECTING ALGEBRA AND GEOMETRY THROUGH COORDINATES k. How is the slope of the line connecting third base and home plate related to the line connecting home plate and first base? l. How is the slope of the line connecting first base and second base related to the line connecting second base and third base? m. What is the formula used for finding the distance between two points? n. What is the distance between first base and second base? o. What is the distance between second base and third base? p. What is the distance between third base and home plate? q. What is the distance between home plate and first base? r. Is the baseball diamond a square? s. What is the distance between second base and home plate? t. What is the distance between third base and first base? u. What do you notice about these two distances? U6-46

46 Problem-Based Task 6.1.: Field of Dreams Coaching Sample Responses a. What does the map of the field look like? Plot each point on a coordinate plane. Label each of the bases and home plate. Third Base Home Plate Second Base First Base b. Which sides of the baseball field are opposites? The line connecting first base and second base is opposite the line connecting home plate and third base. The line connecting second base and third base is opposite the line connecting home plate and first base. U6-47

47 c. What is the slope of the line connecting first base and second base? The slope of the line can be found using the slope formula. y y x x ( 3) (6) (7) (4) Slope formula 9 = 3 Simplify. 3 Substitute (4, 6) and (7, 3). The slope of the line connecting first base and home plate is 3. d. What is the slope of the line connecting home plate and third base? y y x x (3) ( 6) ( 5) ( ) Slope formula Substitute (, 6) and ( 5, 3). 9 3 = 3 Simplify. The slope of the line connecting home plate and third base is 3. e. What is the slope of the line connecting home plate and first base? y y x x ( 3) ( 6) (7) ( ) Slope formula Substitute (, 6) and (7, 3). Simplify. The slope of the line connecting home plate and first base is 1 3. U6-48

48 f. What is the slope of the line connecting third base and second base? y y x x (6) (3) (4) ( 5) Slope formula Substitute ( 5, 3) and (4, 6). Simplify. The slope of the line connecting third base and second base is 1 3. g. Are opposite sides parallel? Opposite sides have the same slope. Opposite sides are parallel. h. Which sides of the baseball field are consecutive? The line connecting first base and second base is consecutive to the line connecting second base and third base. The line connecting second base and third base is consecutive to the line connecting third base and home plate. The line connecting third base and home plate is consecutive to the line connecting home plate and first base. The line connecting home plate and first base is consecutive to the line connecting first base and second base. i. How is the slope of the line connecting first base and second base related to the line connecting first base and home plate? The slopes are opposite reciprocals; therefore, they are perpendicular. j. How is the slope of the line connecting second base and third base related to the line connecting third base and home? The slopes are opposite reciprocals; therefore, they are perpendicular. U6-49

49 k. How is the slope of the line connecting third base and home plate related to the line connecting home plate and first base? The slopes are opposite reciprocals; therefore, they are perpendicular. l. How is the slope of the line connecting first base and second base related to the line connecting second base and third base? The slopes are opposite reciprocals; therefore, they are perpendicular. m. What is the formula used for finding the distance between two points? The formula for finding the distance between two points is the distance formula. ( x x ) + ( y y ) n. What is the distance between first base and second base? Use the distance formula to calculate the distance between each base. ( x x ) + ( y y ) ((4) (7)) + ((6) ( 3)) ( 3) + (9) Distance formula Substitute (7, 3) and (4, 6). Simplify Evaluate squares. 90 Simplify. The distance between first base and second base is 90 units. o. What is the distance between second base and third base? ( x x ) + ( y y ) (( 5) (4)) + ((3) (6)) ( 9) + ( 3) Distance formula Substitute (4, 6) and ( 5, 3). Simplify Evaluate squares. 90 Simplify. The distance between second base and third base is 90 units. U6-50

50 p. What is the distance between third base and home plate? ( x x ) + ( y y ) (( ) ( 5)) + (( 6) (3)) (3) + ( 9) Distance formula Substitute ( 5, 3) and (, 6). Simplify Evaluate squares. 90 Simplify. The distance between third base and home plate is 90 units. q. What is the distance between home plate and first base? ( x x ) + ( y y ) ((7) ( )) + (( 3) ( 6)) (9) (3) Distance formula Substitute (, 6) and (7, 3). Simplify Evaluate squares. 90 Simplify. The distance between home plate and first base is 90 units. r. Is the baseball diamond a square? How do you know? The baseball diamond is a square because all sides have the same length of 90 units. Opposite sides have the same slope, so they are parallel. Consecutive sides have opposite reciprocal slopes, so they are perpendicular. U6-51

51 s. What is the distance between second base and home plate? ( x x ) + ( y y ) (( ) (4)) + (( 6) (6)) ( 6) + ( 1) Distance formula Substitute (4, 6) and (, 6). Simplify Evaluate squares. 180 Simplify. The distance between second base and home plate is 180 units. t. What is the distance between third base and first base? ( x x ) + ( y y ) ((7) ( 5)) + (( 3) (3)) (1) + ( 6) Distance formula Substitute ( 5, 3) and (7, 3). Simplify Evaluate squares. 180 Simplify. The distance between third base and first base is 180 units. u. What do you notice about these two distances? The distance from first base to third base is equal to the distance from second base to home plate. Recommended Closure Activity Select one or more of the essential questions for a class discussion or as a journal entry prompt. U6-5

52 NAME: CONNECTING ALGEBRA AND GEOMETRY THROUGH COORDINATES Practice 6.1.: Using Coordinates to Prove Geometric Theorems with Slope and Distance Use the distance formula to calculate the distance between the points indicated. 1. ( 1, 1) and (4, 11). ( 3, ) and (6, 1) A right triangle has a 90 angle made up of two perpendicular sides. Graph each triangle and then determine if each one is a right triangle. Use the slope formula and/or distance formula to justify your answer. 3. A ( 6, ), B (, ), and C (, 1) 4. A ( 3, ), B (1, 4), and C (3, 0) A parallelogram is a quadrilateral with opposite sides parallel. Graph each quadrilateral and then determine if each one is a parallelogram. Use the slope formula and/or distance formula to justify your answer. 5. A (, 1), B ( 1, 3), C (4, 3), and D (3, 1) 6. A (, ), B (1, 8), C (4, 5), and D (3, ) A rectangle is a parallelogram with opposite sides that are congruent and consecutive sides that are perpendicular. Graph each quadrilateral and then determine if each one is a rectangle. Use the slope formula and/or distance formula to justify your answer. 7. A (, 1), B ( 3, 1), C (1, 3), and D (, 1) 8. A (1, 3), B (0, 0), C (4, 4), and D (5, 1) A square is a parallelogram with four congruent sides and four right angles. Graph each quadrilateral and then determine if each one is a square. Use the slope formula and/or distance formula to justify your answer. 9. A (, 1), B (1, 3), C (4, 1), and D (1, 1) 10. A ( 1, 4), B (0, 6), C (, 5), and D (1, 3) U6-53

53 NAME: CONNECTING ALGEBRA AND GEOMETRY THROUGH COORDINATES Lesson 6.1.3: Working with Parallel and Perpendicular Lines Warm-Up Civil engineers are planning a new shopping center downtown. They would like the entrance to the 1 center to run perpendicular to the street represented by the equation y= x+. 5 Water Street Write an equation that would represent the entrance to the shopping center. Explain your reasoning.. A second entrance will run parallel to the first entrance. Write an equation that would represent the second entrance to the shopping center. U6-54

54 Lesson 6.1.3: Working with Parallel and Perpendicular Lines Common Core Georgia Performance Standard MCC9 1.G.GPE.5 Warm-Up Debrief Civil engineers are planning a new shopping center downtown. They would like the entrance to the 1 center to run perpendicular to the street represented by the equation y= x+. 5 Water Street Write an equation that would represent the entrance to the shopping center. Explain your reasoning. 1 The equation that represents Water Street is y= x+. The entrance will be perpendicular to Water Street. 5 The slopes of perpendicular lines are opposite reciprocals. U6-55

55 The slope of the equation representing Water Street is 1 5. The reciprocal of 1 5 is 5 or 5. 1 The opposite of 5 is 5. The location of the entrance was not specified, so any equation with a slope of 5 will be 1 perpendicular to the equation y= x+. 5 Possible answers: y = 5x + or y = 5x 8. A second entrance will run parallel to the first entrance. Write an equation that would represent the second entrance to the shopping center. The slopes of parallel lines are equal. The slope of the equation representing the first entrance is 5. The location of the second entrance was not specified, so any equation with a slope of 5, other than the original equation, will be parallel to the first entrance. Possible answers: y = 5x + 7 or y = 5x 3 Connection to the Lesson Students will extend their understanding of parallel and perpendicular lines in the upcoming lesson. Students will be asked to take a scenario like this a step further and find a line parallel and/or perpendicular to a given line through a specified point. U6-56

56 Prerequisite Skills This lesson requires the use of the following skills: writing linear equations graphing linear equations Introduction The slopes of parallel lines are always equal, whereas the slopes of perpendicular lines are always opposite reciprocals. It is important to be able to determine whether lines are parallel or perpendicular, but the creation of parallel and perpendicular lines is also important. In this lesson, you will write the equations of lines that are parallel and perpendicular to a given line through a given point. Key Concepts You can write the equation of a line through a given point that is parallel to a given line if you know the equation of the given line. It is necessary to identify the slope of the given equation before trying to write the equation of the line that is parallel or perpendicular. Writing the given equation in slope-intercept form allows you to quickly identify the slope, m, of the equation. If the given equation is not in slope-intercept form, take a few moments to rewrite it. Writing Equations Parallel to a Given Line Through a Given Point 1. Rewrite the given equation in slope-intercept form if necessary.. Identify the slope of the given line. 3. Write the general point-slope form of a linear equation: y y 1 = m(x x 1 ). 4. Substitute the slope of the given line for m in the general equation. 5. Substitute x and y from the given point into the general equation for x 1 and y Simplify the equation. 7. Rewrite the equation in slope-intercept form if necessary. Writing the equation of a line perpendicular to a given line through a given point is similar to writing equations of parallel lines. The slopes of perpendicular lines are opposite reciprocals. U6-57

57 Writing Equations Perpendicular to a Given Line Through a Given Point 1. Rewrite the given equation in slope-intercept form if necessary.. Identify the slope of the given line. 3. Find the opposite reciprocal of the slope of the given line. 4. Write the general point-slope form of a linear equation: y y 1 = m(x x 1 ). 5. Substitute the opposite reciprocal of the given line for m in the general equation. 6. Substitute x and y from the given point into the general equation for x 1 and y Simplify the equation. 8. Rewrite the equation in slope-intercept form if necessary. The shortest distance between two points is a line. The shortest distance between a given point and a given line is the line segment that is perpendicular to the given line through the given point. Finding the Shortest Distance Between a Given Point and a Given Line 1. Follow the steps outlined previously to find the equation of the line that is perpendicular to the given line through the given point.. Find the intersection between the two lines by setting the given equation and the equation of the perpendicular line equal to each other. 3. Solve for x. 4. Substitute the x-value into the equation of the given line to find the y-value. 5. Find the distance between the given point and the point of intersection of the given line and the perpendicular line using the distance formula, ( x x ) + ( y y ). Common Errors/Misconceptions attempting to identify the slope of the given line without transforming the equation into slope-intercept form incorrectly identifying the slope of the given line incorrectly finding the slope of the line parallel to the given line incorrectly identifying the slope of the line perpendicular to the given line improperly substituting the x- and y-values into the general point-slope equation U6-58

58 Guided Practice Example 1 Write the slope-intercept form of an equation for the line that passes through the point (5, ) and is parallel to the graph of 8x y = Rewrite the given equation in slope-intercept form. 8x y = 6 Given equation y = 6 8x Subtract 8x from both sides. y = 3 + 4x Divide both sides by. y = 4x 3 Write the equation in slope-intercept form.. Identify the slope of the given line. The slope of the line y = 4x 3 is Substitute the slope of the given line for m in the point-slope form of a linear equation. y y 1 = m(x x 1 ) y y 1 = 4(x x 1 ) Point-slope form Substitute m from the given equation. 4. Substitute x and y from the given point into the equation for x 1 and y 1. y y 1 = 4(x x 1 ) Equation y ( ) = 4(x 5) Substitute (5, ) for x 1 and y 1. U6-59

59 5. Simplify the equation. y ( ) = 4(x 5) Equation with substituted values for x 1 and y 1 y ( ) = 4x 0 Distribute 4 over (x 5). y + = 4x 0 y = 4x Simplify. Subtract from both sides. The equation of the line through the point (5, ) that is parallel to the equation 8x y = 6 is y = 4x. This can be seen on the following graph x - y = (5, -) y = 4x U6-60

60 Example Write the slope-intercept form of an equation for the line that passes through the point ( 1, 6) and is perpendicular to the graph of 10x + 5y = Rewrite the given equation in slope-intercept form. 10x + 5y = 0 Given equation 5y = x Add 10x to both sides. y = 4 + x Divide both sides by 5. y = x + 4 Write the equation in slope-intercept form.. Identify the slope of the given line. The slope of the line y = x + 4 is. 3. Find the opposite reciprocal of the slope of the given line. The slope of the given line is. The opposite of is. The reciprocal of is Substitute the opposite reciprocal for m in the point-slope form of a linear equation. y y 1 = m(x x 1 ) Point-slope form 1 y y = x x ( ) 1 1 Substitute m from the given equation. 5. Substitute x and y from the given point into the equation for x 1 and y 1. 1 y y = x x ( ) 1 1 Equation 1 y 6= x ( ( 1)) Substitute ( 1, 6) for x and y. 1 1 U6-61

61 6. Simplify the equation. 1 y 6= x ( ( 1)) Equation with substituted values for x and y y 6= x Distribute 1 over (x ( 1)) y= x+ Add 6 to both sides. The equation of the line through the point ( 1, 6) that is perpendicular 1 11 to the graph of 10x + 5y = 0 is y= x+. This can be seen on the following graph y = - x + (-1, 6) x + 5y = U6-6

62 Example 3 Find the point on the line y = 4x + 1 that is closest to the point (, 8). 1. Find the line perpendicular to the given line, y = 4x + 1, that passes through the point (, 8).. Identify the slope of the given line. The slope of the line y = 4x + 1 is Find the opposite reciprocal of the slope of the given line. The opposite of 4 is 4. The reciprocal of 4 is Substitute the opposite reciprocal for m in the point-slope form of a linear equation. y y 1 = m(x x 1 ) Point-slope form 1 y y = x x 4 ( ) 1 1 Substitute m from the given equation. 5. Substitute x and y from the given point into the equation for x 1 and y 1. 1 y y = x x 4 ( ) 1 1 Equation 1 y 8= x 4 ( ( )) Substitute (, 8) for x and y. 1 1 U6-63

63 6. Simplify the equation. 1 y 8= x 4 ( ( )) Equation with substituted values for x and y y 8= x Distribute 1 over (x ( )) y= x+ 4 Add 8 to both sides. The equation of the line through the point (, 8) that is perpendicular 1 15 to the graph of y = 4x + 1 is y= x+ 4. This can be seen on the following graph. 1 y = - x (-, 8) y = 4x U6-64

64 7. Find the intersection between the two lines by setting the given equation equal to the equation of the perpendicular line, then solve for x x+ 1= x x= x x 4 6 x 17 Set both equations equal to each other. Subtract 1 from both sides. 1 Add x to both sides. 4 Divide both sides by U6-65

65 8. Substitute the value of x back into the given equation to find the value of y. y = 4x + 1 Given equation y = Substitute 6 for x y 17 Simplify. The point on the line closest to (, 8) is the point 6 17, y = - x (-, 8) 10 y = 4x (, 11 ) U6-66

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