Geometry Analytic Geometry

Transcription

1 Slide 1 / 202

2 Slide 2 / 202 Geometry Analytic Geometry

3 Slide 3 / 202 Table of Contents Origin of Analytic Geometry The Distance Formula The Midpoint Formula Partitions of a Line Segment Slopes of Parallel & Perpendicular Lines Equations of Parallel & Perpendicular Lines Triangle Coordinate Proofs Equation of a Circle & Completing the Square click on the topic to go to that section PARCC Sample Questions General Problems

4 Slide 4 / 202 Throughout this unit, the Standards for Mathematical Practice are used. MP1: Making sense of problems & persevere in solving them. MP2: Reason abstractly & quantitatively. MP3: Construct viable arguments and critique the reasoning of others. MP4: Model with mathematics. MP5: Use appropriate tools strategically. MP6: Attend to precision. MP7: Look for & make use of structure. MP8: Look for & express regularity in repeated reasoning. Additional questions are included on the slides using the "Math Practice" Pull-tabs (e.g. a blank one is shown to the right on this slide) with a reference to the standards used. If questions already exist on a slide, then the specific MPs that the questions address are listed in the Pull-tab.

5 Slide 4 (Answer) / 202 Throughout this unit, the Standards for Mathematical Practice are used. MP1: Making sense of problems & persevere in solving them. MP2: Reason abstractly & quantitatively. MP3: Construct viable arguments and critique the reasoning of others. MP4: Model with mathematics. MP5: Use appropriate tools strategically. MP6: Attend to precision. MP7: Look for & make use of structure. MP8: Look for & express regularity in repeated reasoning. Math Practice Additional questions are included on the slides using the "Math Practice" Pull-tabs (e.g. a blank one is [This shown object is a to pull the tab] right on this slide) with a reference to the standards used. If questions already exist on a slide, then the specific MPs that the questions address are listed in the Pull-tab.

6 Slide 5 / 202 Origin of Analytic Geometry Return to Table of Contents

7 Slide 6 / 202 The Origin of Analytic Geometry Analytic Geometry is a powerful combination of geometry and algebra. Many jobs that are looking for employees now, and will be in the future, rely on the process or results of analytic geometry. This includes jobs in medicine, veterinary science, biology, chemistry, physics, mathematics, engineering, financial analysis, economics, technology, biotechnology, etc.

8 Slide 7 / 202 The Origin of Analytic Geometry Euclidean Geometry Was developed in Greece about 2500 years ago. Was lost to Europe for more than a thousand years. Was maintained and refined during that time in the Islamic world. Its rediscovery was a critical part of the European Renaissance. books-manuscripts/euclid-millietdechales-claude-francoismilliet details.aspx

9 Slide 8 / 202 The Origin of Analytic Geometry Algebra Started by Diophantus in Alexandria about 1700 years ago. Ongoing contributions from Babylon, Syria, Greece and Indians. Named from the Arabic word al-jabr which was used by al-khwarizmi in the title of his 7th century book. Mathematics_in_medieval_Islam

10 Slide 9 / 202 The Origin of Analytic Geometry Analytic Geometry A powerful combination of algebra and geometry. Independently developed, and published in 1637, by Rene Descartes and Pierre de Fermat in France. The Cartesian Plane is named for Descartes.

11 Slide 10 / 202 The Origin of Analytic Geometry How would you describe to someone the location of these five points so they could draw them on another piece of paper without seeing your drawing? Discuss.

12 Slide 10 (Answer) / 202 The Origin of Analytic Geometry How would you describe to someone the location of these five points so they The question on this slide addresses could draw them on MP2 & MP3. another piece of paper without seeing your drawing? Math Practice Discuss. [This object is a pull tab]

13 Slide 11 / 202 The Origin of Analytic Geometry 10 Adding this Cartesian coordinate plane makes that task simple since the location of each point can be given by just two numbers: an x- and y-coordinate, written as the ordered pair (x,y)

14 Slide 12 / 202 The Origin of Analytic Geometry 10 With the Cartesian Plane providing a numerical description of locations on the plane, geometric figures can be analyzed using algebra

15 Slide 13 / 202 The Distance Formula Lab: Derivation of the Distance Formula Return to Table of Contents

16 Slide 13 (Answer) / 202 The lab link on this slide can be used as an activity where the students derive the distance formula in small groups. After they complete the lab, use the next Lab: 8 Derivation slides to review of the the steps used Distance in the derivation. Formula The Distance Formula Teacher Note This lab addresses MP1, MP2, MP4, MP5, MP6, MP7 & MP8 [This object is a pull tab] Return to Table of Contents

17 Slide 14 / 202 y 10 (x 2, y 2 ) The Distance Formula 5 (x 1, y 1 ) Let's derive the formula to find the distance between any two points: call the points (x 1, y 1 ) and (x 2, y 2 ) x First, let's zoom in so we have more room to work

18 Slide 15 / 202 The Distance Formula 10 y (x 2, y 2 ) Now, we'll construct two paths between the points. The first path will be directly between them. 5 (x 1, y 1 ) 0 5 x

19 Slide 16 / 202 The Distance Formula 10 5 y (x 2, y 2 ) The second path will have a segment parallel the x-axis and a segment parallel to the y-axis. That will be a big help since it's easy to read the length of each of those from the axis. (x 1, y 1 ) 0 5 x

20 Slide 17 / 202 The Distance Formula y We now have a right triangle. 10 (x 2, y 2 ) For convenience, let's label the two legs "a" and "b" and the hypotenuse "c." 5 c b Let's also label the point where the two legs meet by its coordinates: (x 2, y 1 ). (x 1, y 1 ) a (x 2, y 1 ) Take a moment to see that those are the coordinates of that vertex of the triangle. 0 5 x Which formula relates the lengths of the sides of a right triangle?

21 Slide 17 (Answer) / 202 The Distance Formula y We now have a right triangle (x 1, y 1 ) (x 2, y 2 ) Math Practice c a For convenience, let's label the two legs "a" and "b" and the hypotenuse "c." The question on this slide addresses Let's also label the point MP7. b where the two legs meet by its coordinates: (x 2, y 1 ). (x 2, y 1 ) Take a moment to see that those are the coordinates of that vertex of the triangle. [This object is a pull tab] 0 5 x Which formula relates the lengths of the sides of a right triangle?

22 10 5 y (x 1, y 1 ) c (x 2, y 2 ) Slide 18 / 202 The Distance Formula a b (x 2, y 1 ) Did you get this? c 2 = a 2 + b 2 The next step is to write expressions for the lengths of sides a and b based on the x and y subscript coordinates. Use the distance along each axis to find those. The coordinates that we just added for the third vertex should help. 0 5 x

23 y (x 1, y 1 ) 5 a (x 2, y 1 ) x Slide 18 (Answer) / 202 The Distance Formula Did you get this? c 2 = a 2 + b 2 (x 2, y 2 ) The next step is to write expressions for the lengths of sides a and b based on the x and y c The sentence, subscript "Use the coordinates. distance..." b on this slide addresses MP2. Use the distance along each axis to find those. Math Practice The coordinates that we just added for the third vertex should help. [This object is a pull tab]

24 Slide 19 / 202 The Distance Formula y Did you get these? 10 (x 2, y 2 ) a = lx 2 - x 1 l AND b = ly 2 - y 1 l We could equally well write 5 (x 1, y 1 ) c a b (x 2, y 1 ) a = lx 1 - x 2 l AND b = ly 1 - y 2 l We use absolute values since we are just concerned with lengths, which are always positive. That's why the order doesn't matter and all of the above are OK. 0 5 x Substitute one pair of these into c 2 = a 2 + b 2

25 Slide 19 (Answer) / 202 The Distance Formula y Did you get these? 10 5 (x 1, y 1 ) c (x 2, y 2 ) Math Practice a = lx 2 - x 1 l AND b = ly 2 - y 1 l The sentence, We "Substitute could equally one well pair..." write on this slide addresses a = lx 1 - x 2 l MP2. AND b = ly 1 - y 2 l b We use absolute values since we are just concerned with lengths, which are always positive. That's a [This why object the is a pull order tab] doesn't matter and (x 2, y 1 ) all of the above are OK. 0 5 x Substitute one pair of these into c 2 = a 2 + b 2

26 Slide 20 / 202 The Distance Formula y We'll use the first pair: 10 c (x 2, y 2 ) b c 2 = (lx 2 - x 1 l) 2 + (ly 2 - y 1 l) 2 Since the quantities are squared, we don't need to indicate absolute value, the result will always be positive anyway. 5 c 2 = (x 2 - x 1 ) 2 + (y 2 - y 1 ) 2 (x 1, y 1 ) a (x 2, y 1 ) Since "c" is the distance between the points, we can use "d" for distance and say that 0 5 x d 2 = (x 2 - x 1 ) 2 + (y 2 - y 1 ) 2 If we solve this equation for d, what will be the final formula?

27 Slide 20 (Answer) / 202 The Distance Formula y We'll use the first pair: 10 5 c (x 2, y 2 ) Math Practice c 2 = (lx 2 - x 1 l) 2 + (ly 2 - y 1 l) 2 Since the quantities are squared, The question we on this don't slide need addresses to indicate MP2. absolute value, the result will b always be positive anyway. c 2 = (x 2 - x 1 ) 2 + (y 2 - y 1 ) 2 0 (x 1, y 1 ) 5 a (x 2, y 1 ) x Since "c" is the distance between the points, we can use "d" for distance and say that [This object is a pull tab] d 2 = (x 2 - x 1 ) 2 + (y 2 - y 1 ) 2 If we solve this equation for d, what will be the final formula?

28 Slide 21 / 202 The Distance Formula 10 y (x 2, y 2 ) d = ((x 2 - x 1 ) 2 + (y 2 -y 1 ) 2 ) 1/2 OR 5 c b d = (x 1, y 1 ) a (x 2, y 1 ) 0 5 x

29 Slide 22 / What is the distance between the points: (4, 8) and (7, 3)? Round your answer to the nearest hundredth.

30 Slide 22 (Answer) / What is the distance between the points: (4, 8) and (7, 3)? Round your answer to the nearest hundredth. Answer d = 5.83 [This object is a pull tab]

31 Slide 23 / What is the distance between the points: (-4, 8) and (7, -3)? Round your answer to the nearest hundredth.

32 Slide 23 (Answer) / What is the distance between the points: (-4, 8) and (7, -3)? Round your answer to the nearest hundredth. Answer d = [This object is a pull tab]

33 Slide 24 / What is the distance between the points: (-2, -5) and (-7, 3)? Round your answer to the nearest hundredth.

34 Slide 24 (Answer) / What is the distance between the points: (-2, -5) and (-7, 3)? Round your answer to the nearest hundredth. Answer d = 9.43 [This object is a pull tab]

35 Slide 25 / What is the distance between the indicated points? Round your answer to the nearest hundredth

36 Slide 25 (Answer) / What is the distance between the indicated points? Round your answer to the nearest hundredth Answer 5 d = [This object is a pull tab] -10

37 Slide 26 / What is the distance between the indicated points? Round your answer to the nearest hundredth

38 10 5 Answer Slide 26 (Answer) / What is the distance between the indicated points? Round your answer to the nearest hundredth. d = [This object is a pull tab] -10

39 Slide 27 / What is the distance between the indicated points? Round your answer to the nearest hundredth

40 10 5 Answer Slide 27 (Answer) / What is the distance between the indicated points? Round your answer to the nearest hundredth. d = [This object is a pull tab] -10

41 Slide 28 / 202 The Midpoint Formula Lab - Midpoint Formula Return to Table of Contents

42 Slide 29 / 202 The Midpoint Formula 10 5 y Another question which is easily solved using analytic geometry is to find the midpoint of a line. Once again, we make use of the fact that it's easy to determine distance parallel to an axis, so let's add those lines. 0 x 5

43 Slide 30 / 202 The Midpoint Formula 10 5 y The x-coordinate for the midpoint between (x 1, y 1 ) and (x 2, y 2 ) is halfway between x 1 and x 2. Similarly, the y-coordinate of that midpoint will be halfway between y 1 and y 2. If you're provided a graph of a line, and asked to mark the midpoint, you can often do that without much calculating. 0 x 5

44 Slide 31 / 202 The Midpoint Formula y In this graph, x 1 = 1 and x 2 = They are 8 units apart, so just go 4 units along the x-axis and go up until you intersect the line. That will give you the midpoint. 5 In this case, that is at (5,7), which can be read from the graph. 0 x 5 We would get the same answer if we had done this along the y- axis, as we do on the next slide.

45 Slide 31 (Answer) / 202 The Midpoint Formula 10 5 y In this graph, x 1 = 1 and x 2 = 9. Math Practice 0 x 5 They are 8 units apart, so just go 4 units along the x-axis and The sentence, go up "They until you are intersect 8 units the line. apart..." on this slide addresses MP1. That will give you the midpoint. In this case, that is at (5,7), which can be read from the graph. [This object is a pull tab] We would get the same answer if we had done this along the y- axis, as we do on the next slide.

46 Slide 32 / 202 The Midpoint Formula 10 5 y 0 x 5 The y-coordinates of the two points are 11 and 3, so they are also 8 units apart. So, just go up 4 from the lower y-coordinate and then across to the line to also get (5,7) to be the midpoint. So, the order doesn't matter and if you have the graph and the numbers are easy to read, you may as well use it. If you are not given a graph or the lines don't fall so that the values are easy to read, you can do a quick calculation to get the same result.

47 Slide 32 (Answer) / 202 The Midpoint Formula y The y-coordinates of the two points are 11 and 3, so they are also 8 units apart Math Practice So, just go up 4 from the lower The first 2 paragraphs y-coordinate on and this then slide across to address MP1. the line to also get (5,7) to be the midpoint. So, the order doesn't matter and if you have the graph and the numbers [This object is a are pull tab] easy to read, you may as well use it. 0 x 5 If you are not given a graph or the lines don't fall so that the values are easy to read, you can do a quick calculation to get the same result.

48 Slide 33 / 202 The Midpoint Formula 10 y (9,11) We can calculate the x-coordinate midway between that of the two given points by finding their average. The same for the y-coordinate. 5 (1,3) 0 x 5 Just add the two values and divide by two. The x-coordinate of the midpoint is (1+9)/2 = 5 The y-coordinate of the midpoint is (3+11)/2 = 7 That's the same answer: (5,7)

49 10 5 y (1,3) Slide 33 (Answer) / 202 The Midpoint Formula (9,11) Math Practice 0 x 5 We can calculate the x-coordinate midway between that of the two given points by finding their average. This The slide same addresses for the MP2. y-coordinate. Just add the two values and divide by two. The x-coordinate of the midpoint [This object is a pull tab] is (1+9)/2 = 5 The y-coordinate of the midpoint is (3+11)/2 = 7 That's the same answer: (5,7)

50 Slide 34 / 202 The Midpoint Formula 10 5 y (x 2, y 2 ) The more general solution is given below for any points: (x 1, y 1 ) and (x 2, y 2 ). The x-coordinate of the midpoint is given by x M = (x 1 + x 2 ) / 2 The y-coordinate of the midpoint is given by y M = (y 1 + y 2 ) / 2 (x 1, y 1 ) So, the coordinates of the midpoint are: 0 5 x

51 Slide 35 / What is the midpoint between the indicated points? 10 5 A (4, 9) B (-5, -4) C (5, 6) D (5, 7) -5-10

52 Slide 35 (Answer) / What is the midpoint between the indicated points? 10 A (4, 9) Answer 5 D B (-5, -4) C (5, 6) D (5, 7) [This object is a pull tab] -5-10

53 Slide 36 / What is the midpoint between the indicated points? 10 5 A (0, 0) B (5, 10) C (5, 5) -5 D (10, 10) -10

54 Slide 36 (Answer) / What is the midpoint between the indicated points? 10 A (0, 0) Answer 5 A B (5, 10) C (5, 5) D (10, 10) [This object is a pull tab] -5-10

55 Slide 37 / What is the midpoint between the indicated points? 10 5 A (3, 3) B (3, 4) C (4, 3) -5 D (5, 3) -10

56 Slide 37 (Answer) / What is the midpoint between the indicated points? 10 A (3, 3) Answer 5 A B (3, 4) C (4, 3) [This object is a pull tab] -5 D (5, 3) -10

57 Slide 38 / What is the midpoint between the points: (4, 8) and (7, 3)? A (8, 2) B (4, 7) C (5.5, 5.5) D (6, 5)

58 Slide 38 (Answer) / What is the midpoint between the points: (4, 8) and (7, 3)? A (8, 2) B (4, 7) C (5.5, 5.5) D (6, 5) Answer C [This object is a pull tab]

59 Slide 39 / What is the midpoint between the points: (-4, 8) and (4, -8)? A (8, 2) B (0, 0) C (12, 12) D (4, 4)

60 Slide 39 (Answer) / What is the midpoint between the points: (-4, 8) and (4, -8)? A (8, 2) B (0, 0) C (12, 12) D (4, 4) Answer B [This object is a pull tab]

61 Slide 40 / What is the midpoint between the points: (-4, -8) and (-6, -4)? A (-5, -6) B (-10, -12) C (2, 4) D (5, 6)

62 Slide 40 (Answer) / What is the midpoint between the points: (-4, -8) and (-6, -4)? A (-5, -6) B (-10, -12) C (2, 4) D (5, 6) Answer A [This object is a pull tab]

63 Slide 41 / 202 Finding the Coordinates of an E ndpoint of a Segment The midpoint of AB is M(1, 2). One endpoint is A(4, 6). Find the coordinates of the other endpoint B(x, y). You can use the midpoint formula to write equations using x and y. Then, set up 2 equations and solve each one. B(x, y) M(1, 2) = A(4, 6) ( 4 + x, 6 + y) 2 2

64 Slide 42 / 202 Finding the Coordinates of an E ndpoint of a Segment M(1, 2) = 1 = 4 + x 2 2 = 4 + x -2 = x ( 4 + x, 6 + y) = 6 + y 2 4 = 6 + y -2 = y M(1, 2) = A(4, 6) ( 4 + x, 6 + y) 2 2 Therefore, the coordinates of B(-2, -2). B(x, y) Can you find a shortcut to solve this problem? How would your shortcut make the problem easier?

65 ( ) 4 + x, 6 + y 2 2 Slide 42 (Answer) / 202 Finding the Coordinates of an E ndpoint of a Segment M(1, 2) = The first part of this example A(4, 6) 1 = 4 + x 2 = 6 addresses + y MP = 4 + x 4 = The 6 + yquestion at the bottom M(1, of 2) the = -2 = x -2 = slide y addresses MP8. - Answer is found on the next 2 slides Therefore, the coordinates (1st using the graph & 2nd using the of B(-2, -2). pattern w/ the B(x, points/numbers). y) Math Practice [This object is a pull tab] ( 4 + x, 6 + y) 2 2 Can you find a shortcut to solve this problem? How would your shortcut make the problem easier?

66 Slide 43 / 202 Finding the Coordinates of an E ndpoint of a Segment Another way of approaching this problem is to look for the pattern that occurs between the endpoint A (4, 6) and midpoint M (1, 2). Looking only at our points, we can determine that we traveled left 3 units and down 4 units to get from A to M. If we travel the same units in the same direction starting at M, we will get to B(-2, -2). down 4 down 4 left 3 B(x, y) left 3 M(1, 2) A(4, 6)

67 Slide 44 / 202 Finding the Coordinates of an E ndpoint of a Segment Similarly, if we line up the points vertically and determine the pattern of the numbers, without a graph, we can calculate the coordinates for our missing endpoint. A (4, 6) -3 M (1, 2) -3 B (-2, -2) -4-4 If you use this method, always determine the operation required to get from the given endpoint to the midpoint. The reverse will not work.

68 Slide 45 / Find the other endpoint of the segment with the endpoint (7, 2) and midpoint (3, 0) A (-1, -2) B (-2, -1) C (4, 2) D (2, 4)

69 Slide 45 (Answer) / Find the other endpoint of the segment with the endpoint (7, 2) and midpoint (3, 0) A (-1, -2) B (-2, -1) C (4, 2) D (2, 4) Answer M(x, y) = ( ( A M(3, 0) =, 3 =, 0 = (-1, -2) [This object is a pull tab], ) )

70 Slide 46 / Find the other endpoint of the segment with the endpoint (1, 4) and midpoint (5, -2) A (11, -8) B (9, 0) C (9, -8) D (3, 1)

71 Slide 46 (Answer) / Find the other endpoint of the segment with the endpoint (1, 4) and midpoint (5, -2) A (11, -8) B (9, 0) C (9, -8) D (3, 1) Answer M(x, y) = M(5, -2) = ( ( C 5 =, -2 =,, ) ) (9, -8) [This object is a pull tab]

72 Slide 47 / Find the other endpoint of the segment with the endpoint (-4, -1) and midpoint (-2, 3). A (-6, -5) B (-3, -2) C (0, 7) D (1, 9)

73 Slide 47 (Answer) / Find the other endpoint of the segment with the endpoint (-4, -1) and midpoint (-2, 3). A (-6, -5) B (-3, -2) C (0, 7) D (1, 9) Answer C [This object is a pull tab]

74 Slide 48 / Find the other endpoint of the segment with the endpoint (-2, 5) and midpoint (0, 2). A (-1, -3.5) B (-4, 8) C (1, 0.5) D (2, -1)

75 Slide 48 (Answer) / Find the other endpoint of the segment with the endpoint (-2, 5) and midpoint (0, 2). A (-1, -3.5) B (-4, 8) C (1, 0.5) D (2, -1) Answer D [This object is a pull tab]

76 Slide 49 / 202 Partitions of a Line Segment Return to Table of Contents

77 Slide 50 / 202 Partitions of a Line Segment B Partitioning a line segment simply means to divide the segment into 2 or more parts, based on a given ratio. The midpoint partitions a segment into 2 congruent segments, forming a ratio of 1:1. A But if you need to partition a segment so that its ratio is something different, for example 3:1, how can it be done?

78 Slide 50 (Answer) / 202 Partitions of a Line Segment Math Practice Partitioning a line segment simply means to divide the B segment into 2 or more parts, based on a given ratio. This slide & the next 3 slides address MP1, MP4, MP5 & MP7 The midpoint partitions a segment into 2 congruent segments, forming a ratio of 1:1. A But if you need to partition a [This object is a pull tab] segment so that its ratio is something different, for example 3:1, how can it be done?

79 Slide 51 / 202 Partitions of a Line Segment B In order to divide the segment in the ratio of 3:1, think of dividing the segment into 3 + 1, or 4 congruent pieces. Plot the points that would divide AB into 4 congruent pieces. - Click on one of the points in the grid to show them all. A

80 10 5 y A Slide 52 / 202 Partitions of a Line Segment 0 x 5 B If we add a coordinate plane to our segment, could we also determine the coordinates of our points? Yes, we can. We can also eliminate one of our points that does not divide our segment into the ratio 3:1. In our case, the midpoint. - Click on midpoint to hide it Let's say that the ratio of the two segments that we're looking for from left to right is 3:1. Which other point should be eliminated? - Click on that point to hide it

81 10 5 y A 0 x 5 Slide 52 (Answer) / 202 Partitions of a Line Segment Answer B If we add a coordinate plane to our segment, could we also determine the coordinates of our points? Yes, we can. We can also eliminate Therefore, one of the our points that does not divide our segment coordinates into the ratio 3:1. of In one our case, the midpoint. - Click on midpoint to hide it point that partitions AB to form the ratio 3:1 is (7, 9) Let's say that the ratio of the two segments that we're looking for from left to right is 3:1. Which [This object is a pull tab] other point should be eliminated? - Click on that point to hide it

82 Slide 53 / 202 Partitions of a Line Segment 10 y B Could we also determine the coordinates of our point if the ratio was reversed (1:3 instead of 3:1)? Yes, we can. Again, our midpoint can be eliminated. - Click on midpoint to hide it 5 A 0 x 5 Now, the ratio of the two segments that we're looking for from left to right is 1:3. Which other point should be eliminated? - Click on that point to hide it

83 Slide 53 (Answer) / 202 Partitions of a Line Segment 10 5 y A Answer 0 x 5 B Could we also determine the coordinates of our point if the ratio was reversed (1:3 instead of 3:1)? Therefore, the coordinates of one point that partitions AB to form the ratio 1:3 is (3, 5) Yes, we can. Again, our midpoint can be eliminated. - Click on midpoint to hide it Now, the ratio of the two segments that we're looking for from left to right is 1:3. Which [This object is a pull tab] other point should be eliminated? - Click on that point to hide it

84 Partitions of a Line Segment 10 5 y A(1,3) B (9,11) 0 x 5 Slide 54 / 202 We can also calculate both the x and y coordinates between the two given points by using a formula that is similar to the midpoint formula. But instead of having a common ratio (1:1) and dividing by 2, what the midpoint formula has us do, we need to multiply the one set of coordinates by the first number in the ratio and the other set of coordinates by the second number in the ratio and divide by the number of segments that are required for our ratio 3:1, or = 4. ( 3(9) + 1(1), 3(11) + 1(3) ) ( , )= (7, 9) 4 4

85 Partitions of a Line Segment 10 5 y A(1,3) B (9,11) Math Practice 0 x 5 Slide 54 (Answer) / 202 We can also calculate both the x and y coordinates between the two given points by using a formula that is similar to the midpoint formula. But instead of having a common This ratio example (1:1) and addresses dividing MP2. by 2, what the midpoint formula has us do, we need to multiply the one set of coordinates by the first number in the ratio and the other set of coordinates by the second number in the ratio and divide by [This object is a pull tab] the number of segments that are required for our ratio 3:1, or = 4. ( 3(9) + 1(1), 3(11) + 1(3) ) ( , )= (7, 9) 4 4

86 Slide 55 / 202 Partitions of a Line Segment 10 5 y B (9,11) If we use the same formula, but switch to the other ratio 1:3, we will get the other point that partitions AB. (, ) (, ) = (3, 5) 1(9) + 3(1) 1(11) + 3(3) A (1,3) 0 x 5

87 Slide 55 (Answer) / 202 Partitions of a Line Segment 10 5 y B (9,11) Math Practice If we use the same formula, but switch to the other ratio 1:3, we will get the other point that partitions AB. (, ) (, ) = (3, 5) 1(9) + 3(1) 1(11) + 3(3) This example addresses 3 + MP A (1,3) [This object is a pull tab] 0 x 5

88 Slide 56 / 202 Partitions of a Line Segment 10 5 y (x 2, y 2 ) The more general solution is given below for any points: (x 1, y 1 ) and (x 2, y 2 ). If a point P partitions a segment in a ratio of m:n, then the coordinates of P are (, ) P = mx 2 + nx 1 my 2 + ny 1 m + n m + n (x 1, y 1 ) Remember to also use the formula twice switching the m:n ratio to n:m. 0 5 x

89 Slide 57 / 202 y 10 5 Partitions of a Line Segment Example with a graph: Line segment CD in the coordinate plane has endpoints with coordinates (-2, 10) and (8, -5). Graph CD and find two possible locations for a point P that divides CD into two parts with lengths in a ratio of 2: x -5

90 Slide 57 (Answer) / 202 y Partitions of a Line Segment Example with a graph: 10 5 Math Practice Line segment CD in the coordinate plane has endpoints with coordinates (-2, 10) and (8, -5). This example Graph (this CD slide and and find the two possible next 3) addresses locations MP1, for a MP4, point MP5 P that divides & MP7 CD into two parts with lengths in a ratio of 2: x [This object is a pull tab] -5

91 y C (-2, 10) 10 Slide 58 / 202 Partitions of a Line Segment Step #1: Graph the points. C(-2, 10) and D(8, -5) - Click on the points to show. (0, 7) 5 (2, 4) (4, 1) 0 5 x (6, -2) -5 D (8, -5) Step #2: Connect them w/ the segment CD. - Click on the segment to show. Step #3: Determine the total number of ways required to divide the segment using the ratio 2: = 5 Click Step #4: Plot the points on the graph that divide CD into the desired number of sections. - Click on the points to show.

92 Slide 59 / 202 y C (-2, 10) 10 (0, 7) Partitions of a Line Segment Step #5: Since the ratio is 2:3, determine which points to eliminate (3 of them) leaving only 1 point remaining so that, from left to right, the ratio is 2:3. 5 (2, 4) (4, 1) 0 5 x (6, -2) -5 D (8, -5)

93 Slide 59 (Answer) / 202 y C (-2, 10) 10 5 (0, 7) Partitions of a Line Segment (2, 4) (4, 1) Answer Step #5: Since the ratio is 2:3, determine which points to eliminate (3 of them) leaving only 1 point remaining so that, from left to right, the ratio is 2:3. Therefore, the coordinates of one point that partitions CD to form the ratio 2:3 is (2, 4) 0 5 x [This object is a pull tab] (6, -2) -5 D (8, -5)

94 Slide 60 / 202 y C (-2, 10) 10 5 (0, 7) Partitions of a Line Segment (2, 4) Step #6: Because the question asks for 2 points, we also need to consider the ratio 3:2. Similar to Step #5, determine which points to eliminate (3 of them) leaving only 1 point remaining so that, from left to right, the ratio is 3:2. (4, 1) 0 5 x (6, -2) -5 D (8, -5)

95 y C (-2, 10) 10 5 (0, 7) Slide 60 (Answer) / 202 Partitions of a Line Segment (2, 4) (4, 1) Answer 0 5 x (6, -2) Step #6: Because the question asks for 2 points, we also need to consider the ratio 3:2. Similar to Step #5, determine which points to eliminate (3 of them) leaving only 1 point remaining so that, from left to right, Therefore, the ratio is 3:2. the coordinates of one point that partitions CD to form the ratio 3:2 is (4, 1) [This object is a pull tab] -5 D (8, -5)

96 Slide 61 / 202 Partitions of a Line Segment Example without a graph: Line segment EF in the coordinate plane has endpoints with coordinates (10, -11) and (-4, 10). Find two possible locations for a point P that divides EF into two parts with lengths in a ratio of 5:2.

97 Slide 61 (Answer) / 202 Partitions of a Line Segment Example without a graph: Line segment EF in the coordinate plane has endpoints with coordinates (10, -11) and (-4, 10). Math Practice Find two possible locations for This a point example P that (slides divides #61-63) EF into two parts with lengths in a ratio of 5:2. addresses MP2. [This object is a pull tab]

98 Slide 62 / 202 Partitions of a Line Segment (10, -11) and (-4, 10) Step #1: Determine the total number of ways required to divide the segment using the ratio 5: = 7 Click Step #2: Use the formula to determine the coordinates of our first point P. click (, ) P = mx 2 + nx 1 my 2 + ny 1 m + n m + n P = P = (, ) 5(-4) + 2(10) 5(10) + 2(-11) (-22) 7 7 P = (0, 4)

99 Slide 63 / 202 Partitions of a Line Segment (10, -11) and (-4, 10) Step #3: Reverse the ratio 2:5 and reuse the formula to determine the coordinates of our second point P. (, ) (, ) P = mx 2 + nx 1 my 2 + ny 1 m + n m + n click P = P = 2(-4) + 5(10) 2(10) + 5(-11) (-55) 7 7 P = (6, 5)

100 Slide 64 / Line segment GH in the coordinate plane has endpoints with coordinates (-7, -9) & (8, 3), shown in the graph below. Find 2 possible locations for a point P that divides GH into two parts with lengths in a ratio of 2: H(8, 3) A (0.5, -3) B (3, -1) C (6, 1) D (-2, -5) G (-7, -9) E (-5, -7)

101 Slide 64 (Answer) / Line segment GH in the coordinate plane has endpoints with coordinates (-7, -9) & (8, 3), shown in the graph below. Find 2 possible locations for a point P that divides GH into two parts with lengths in a ratio of 2:1. Answer B & D 0 5 H(8, 3) A (0.5, -3) B (3, -1) C (6, 1) D (-2, -5) G (-7, -9) -5 [This object is a pull tab] -10 E (-5, -7)

102 Slide 65 / Line segment JK in the coordinate plane has endpoints with coordinates (10, 12) & (-10, -13), shown in the graph below. Find 2 possible locations for a point P that divides JK into two parts with lengths in a ratio of 4:1. A (-6, -8) J(10, 12) B (-2, -3) C (0, -0.5) D (2, 2) E (6, 7) K(-10, -13) -10

103 Slide 65 (Answer) / Line segment JK in the coordinate plane has endpoints with coordinates (10, 12) & (-10, -13), shown in the graph below. Find 2 possible locations for a point P that divides JK into two parts with lengths in a ratio of 4:1. A (-6, -8) B (-2, -3) Answer A & E -5 0 [This object is a pull tab] 5 J(10, 12) C (0, -0.5) D (2, 2) E (6, 7) K(-10, -13) -10

104 Slide 66 / Line segment LM in the coordinate plane has endpoints with coordinates (-8, 13) & (6, -8), shown in the graph below. Find 2 possible locations for a point P that divides LM into two parts with lengths in a ratio of 3:4. A (-6, 10) B (-4, 7) C (-2, 4) D (0, 1) L(-8, 13) M(6, -8) E (2, -3)

105 Slide 66 (Answer) / Line segment LM in the coordinate plane has endpoints with coordinates (-8, 13) & (6, -8), shown in the graph below. Find 2 possible locations for a point P that divides LM into two parts with lengths in a ratio of 3:4. A (-6, 10) B (-4, 7) C (-2, 4) D (0, 1) L(-8, 13) Answer C & D -5 0 [This object is a pull tab] M(6, -8) E (2, -3)

106 Slide 67 / Line segment LM in the coordinate plane has endpoints with coordinates (-8, 13) & (6, -8), shown in the graph below. Find 2 possible locations for a point P that divides LM into two parts with lengths in a ratio of 6:1. A (-6, 10) B (-4, 7) C (-2, 4) D (0, 1) E (4, -6)

107 Slide 67 (Answer) / Line segment LM in the coordinate plane has endpoints with coordinates (-8, 13) & (6, -8), shown in the graph below. Find 2 possible locations for a point P that divides LM into two parts with lengths in a ratio of 6:1. A (-6, 10) B (-4, 7) C (-2, 4) D (0, 1) E (4, -6) Answer A & E [This object is a pull tab]

108 Slide 68 / Line segment NO in the coordinate plane has endpoints with coordinates (10, 12) & (-10, -13), shown in the graph below. Find 2 possible locations for a point P that divides NO into two parts with lengths in a ratio of 3:2. A (-6, -8) B (-2, -3) C (0, -0.5) D (2, 2) E (6, 7)

109 Slide 68 (Answer) / Line segment NO in the coordinate plane has endpoints with coordinates (10, 12) & (-10, -13), shown in the graph below. Find 2 possible locations for a point P that divides NO into two parts with lengths in a ratio of 3:2. A (-6, -8) B (-2, -3) C (0, -0.5) D (2, 2) E (6, 7) Answer B & D [This object is a pull tab]

110 Slide 69 / Line segment QR in the coordinate plane has endpoints with coordinates (-12, 11) & (12, -13), shown in the graph below. Find 2 possible locations for a point P that divides QR into two parts with lengths in a ratio of 5:3. A (3, -4) B (0, -1) C (-3, 2) D (-6, 5) E (-9, 8)

111 Slide 69 (Answer) / Line segment QR in the coordinate plane has endpoints with coordinates (-12, 11) & (12, -13), shown in the graph below. Find 2 possible locations for a point P that divides QR into two parts with lengths in a ratio of 5:3. A (3, -4) B (0, -1) C (-3, 2) D (-6, 5) E (-9, 8) Answer A & C [This object is a pull tab]

112 Slide 70 / 202 Line segment JK in the coordinate plane has endpoints with coordinates (-4, 11) & (8, -1). Graph JK and find two possible locations for point M so that M divides JK into two parts with J(-4, 11) lengths in a ratio of 1:3. To graph a line segment, click the locations for the two points. Then click in between the two points to make the segment. K(8, -1)

113 Slide 71 / Find two possible locations for point M so that M divides JK into two parts with lengths in a ratio of 1:3. A (-2, 9) J (-4, 11) B (-1, 8) C (0, 7) D (4, 3) E (5, 2) K (8, -1) F (6, 1) PARCC Released Question (EOY) - Part 2 - Response Format

114 Slide 71 (Answer) / Find two possible locations for point M so that M divides JK into two parts with lengths in a ratio of 1:3. A (-2, 9) J (-4, 11) B (-1, 8) C (0, 7) D (4, 3) Answer E (5, 2) K (8, -1) F (6, 1) B & E 2 [This object is a pull tab] PARCC Released Question (EOY) - Part 2 - Response Format

115 Slide 72 / Point Q lie on ST, where point S is located at (-2, -6) and point T is located at (5, 8). If SQ:QT = 5:2, where is point Q on ST? A (-1, -4) B (0, -2) C (1, 0) D (2, 2) E (3, 4) F (4, 6) PARCC Released Question (PBA)

116 Slide 72 (Answer) / Point Q lie on ST, where point S is located at (-2, -6) and point T is located at (5, 8). If SQ:QT = 5:2, where is point Q on ST? A (-1, -4) B (0, -2) C (1, 0) D (2, 2) E (3, 4) F (4, 6) Answer E [This object is a pull tab] (3, 4) PARCC Released Question (PBA)

117 Slide 73 / 202 Slopes of Parallel & Perpendicular Lines Return to Table of Contents

118 Slide 74 / 202 Slope 10 y The slope of a line indicates the angle it makes with the x-axis. The symbol for slope is "m". 5 0 x 5

119 Slide 75 / 202 Slope 10 5 y 0 x 5 A horizontal line has a slope of zero. A vertical line has an undefined slope. A line which rises as you move from left to right has a positive slope. A line which falls as you move from left to right has a negative slope.

120 Slide 76 / The slope of the indicated line is: A negative y B positive C zero D undefined x

121 Slide 76 (Answer) / The slope of the indicated line is: A negative B positive C zero D undefined Answer y B [This object is a pull tab] x

122 Slide 77 / The slope of the indicated line is: A negative y B positive C zero D undefined x

123 Slide 77 (Answer) / The slope of the indicated line is: A negative B positive C zero D undefined Answer y B [This object is a pull tab] x

124 Slide 78 / The slope of the indicated line is: A negative y B positive C zero D undefined x

125 Slide 78 (Answer) / The slope of the indicated line is: A negative y B positive C zero D undefined Answer A [This object is a pull tab] x

126 Slide 79 / The slope of the indicated line is: A negative y B positive C zero D undefined x

127 Slide 79 (Answer) / The slope of the indicated line is: A negative B positive C zero D undefined Answer y D [This object is a pull tab] x

128 Slide 80 / The slope of the indicated line is: A negative y B positive C zero D undefined x

129 Slide 80 (Answer) / The slope of the indicated line is: A negative y B positive C zero D undefined Answer C [This object is a pull tab] x

130 Slide 81 / The slope of the indicated line is: A negative y B positive C zero D undefined x

131 Slide 81 (Answer) / The slope of the indicated line is: A negative B positive C zero D undefined Answer y C [This object is a pull tab] x

132 Slide 82 / 202 Slope 10 5 y The slope of a line is not given in degrees. Rather, it is given as the ratio of "rise" over "run". The slope of a line is the same anywhere along the line, so any two points on the line can be used to calculate the slope. 0 x 5

133 Slide 83 / 202 Slope 10 5 y (x 1, y 2 ) y "rise" (x 1, y 1 ) x "run" (x 2, y 2 ) The "rise" is the change in the value of the y-coordinate while the "run" is the change in the value of the x-coordinate. The symbol for change is the Greek letter delta, " ", which just means "change in". So the slope is equal to the change in y divided by the change in x, or y divided by x...delta y over delta x. 0 5 x m = Δy y 2-y 1 Δx = x 2 -x 1

134 Slide 84 / 202 Slope 10 y x (2, 11) "run" y (8,11) In this case: m = Δy y 2-y 1 Δx = x 2 -x 1 The rise is from 4 to 11 "rise" y = 11-4 = 7 5 (2,4) And the run is from 2 to 8, x = 8-2 = x So the slope is m = y = 7 x 6

135 Slide 85 / 202 Slope y 10 y x x (x 2, y 2 ) Any points on the line can be used to calculate its slope, since the slope of a line is the same everywhere. y 5 0 (x 1, y 1 ) 5 x The values of y and x may be different for other points, but their ratio will be the same. You can check that with the red and green triangles shown here. m = Δy y 2-y 1 Δx = x 2 -x 1

136 Slide 86 / 202 Using Slope to Draw a Line 10 5 y The slope also allows us to quickly graph a line, given one point on the line. For instance, if I know one point on a line is (1, 1) and that the slope of the line is 2, I can find a second point, and then draw the line. 0 x 5 m = Δy y 2-y 1 Δx = x 2 -x 1

137 Slide 87 / 202 Using Slope to Draw a Line 10 y I do this by recognizing that the slope of 2 means that if I go up 2 units on the y-axis I have to go 1 unit to the right on the x-axis. 5 x y 0 x 5 m = Δy y 2-y 1 Δx = x 2 -x 1

138 Slide 88 / 202 Using Slope to Draw a Line 10 y x I do this by recognizing that the slope of 2 means that if I go up 2 units on the y-axis I have to go 1 unit to the right on the x-axis. 5 y Or if I go up 10, I have to go over 5 units, etc. 0 x 5 m = Δy y 2-y 1 Δx = x 2 -x 1

139 Slide 89 / 202 Using Slope to Draw a Line 10 5 y y x Then I draw the line through any two of those points. This method is the easiest to use if you just have to draw a line given a point and slope. The same approach works for writing the equation of a line. 0 x 5 m = Δy y 2-y 1 Δx = x 2 -x 1

140 Slide 90 / 202 Slopes of Parallel Lines Parallel lines have the same slope. With what we've learned about parallel lines this is easy to understand. The slope of a line is related to the angle it makes with the x-axis. Now, think of the x-axis as a transversal.

141 Slide 90 (Answer) / 202 Slopes of Parallel Lines Parallel lines have the same slope. Math Practice With what we've learned about parallel lines this is easy to This example (and understand. the next 3 questions) addresses MP3 & MP7. The slope of a line is related to the angle it makes with the x-axis. [This object Now, is a pull think tab] of the x-axis as transversal.

142 Slide 91 / What is the name of this pair of angles, formed by a transversal intersecting two lines. A Alternate Interior Angles B Corresponding Angles C Alternate Exterior Angles D Same Side Interior Angles 1 2

143 Slide 91 (Answer) / What is the name of this pair of angles, formed by a transversal intersecting two lines. A Alternate Interior Angles B Corresponding Angles C Alternate Exterior Angles D Same Side Interior Angles Answer B [This object is a pull tab] 1 2

144 Slide 92 / We know those angles must then be: A Supplementary B Complementary C Equal D Adjacent 1 2

145 Slide 92 (Answer) / We know those angles must then be: A Supplementary B Complementary C Equal D Adjacent Answer C [This object is a pull tab] 1 2

146 Slide 93 / That means that the slopes of parallel lines must be: A Reciprocals B Inverses C Equal D Nothing special 1 2

147 Slide 93 (Answer) / That means that the slopes of parallel lines must be: A Reciprocals B Inverses C Equal D Nothing special Answer C [This object is a pull tab] 1 2

148 Slide 94 / If one line has a slope of 4, what must be the slope of any line parallel to it?

149 Slide 94 (Answer) / If one line has a slope of 4, what must be the slope of any line parallel to it? Answer 4 [This object is a pull tab]

150 Slide 95 / If one line passes through the points (0, 0) and (2, 2) what must be the slope of any line parallel to that first line?

151 Slide 95 (Answer) / If one line passes through the points (0, 0) and (2, 2) what must be the slope of any line parallel to that first line? Answer 2 [This object is a pull tab]

152 Slide 96 / If one line passes through the points (-5, 9) and (5, 8) what must be the slope of any line parallel to that first line?

153 Slide 96 (Answer) / If one line passes through the points (-5, 9) and (5, 8) what must be the slope of any line parallel to that first line? Answer _ 1 10 [This object is a pull tab]

154 Slide 97 / If one line passes through the points (2, 2) and (5, 5) and a parallel line passes through the point (1, 5) which of these points could lie on that second line? A (2, 2) B (4, 4) C (5, 6) D (-1, 3)

155 Slide 97 (Answer) / If one line passes through the points (2, 2) and (5, 5) and a parallel line passes through the point (1, 5) which of these points could lie on that second line? A (2, 2) B (4, 4) C (5, 6) D (-1, 3) Answer D [This object is a pull tab]

156 Slide 98 / If one line passes through the points (-3, 4) and (0, 10) and a parallel line passes through the point (-1, -4) which of these points could lie on that second line? A (0, -2) B (2, -5) C (3, 1) D (-1, 3)

157 Slide 98 (Answer) / If one line passes through the points (-3, 4) and (0, 10) and a parallel line passes through the point (-1, -4) which of these points could lie on that second line? A (0, -2) B (2, -5) C (3, 1) D (-1, 3) Answer A [This object is a pull tab]

158 Slide 99 / If one line passes through the points (3, 5) and (5, -1) and a parallel line passes through the point (-1, -1) which of these points could lie on that second line? A (0, 1) B (-2, 2) C (4, 8) D (-4, -4)

159 Slide 99 (Answer) / If one line passes through the points (3, 5) and (5, -1) and a parallel line passes through the point (-1, -1) which of these points could lie on that second line? A (0, 1) B (-2, 2) C (4, 8) D (-4, -4) Answer B [This object is a pull tab]

160 Slide 100 / 202 Slopes of Perpendicular Lines The slopes of perpendicular lines are negative (or opposite) reciprocals. There are three ways of expressing this symbolically: m 1 m 2 = -1 m 1 = m 2 = -1 m 2-1 m 1 These all have the same meaning.

161 Slide 101 / 202 Slopes of Perpendicular Lines The slopes of perpendicular lines are negative (or opposite) reciprocals. This will be useful in cases where you need to prove lines perpendicular, including proving that triangles are right triangles. First, let's prove this is true.

162 Slide 101 (Answer) / 202 Slopes of Perpendicular Lines The slopes of perpendicular lines are negative (or opposite) reciprocals. Math Practice This will be useful in cases where you need to prove lines perpendicular, including proving that triangles are right triangles. This example addresses First, let's MP2, prove MP3 this is true. & MP7. [This object is a pull tab]

163 Slide 102 / Slopes of Perpendicular Lines 5 First, let's move our lines to the origin so our work gets easier and we can focus on the important parts We can move them since we can just think of drawing new parallel lines through the origin that have the same slopes as these lines. We earlier showed that parallel lines have the same slope, so the slopes of these new lines will be the same as that of the original lines.

164 Slide 103 / 202 Slopes of Perpendicular Lines 1 (1,m 1 ) Now, let's zoom in and focus on the lines between x = 0 and x = When the lines are at x = 1, their y-coordinates are m 1 for the first line and m 2 for the second line. That's because if the slope of the first line is m 1, then when we move +1 along the x-axis, the y-value must increase by the amount of the slope, m (1,m 2 ) The same for the second line, whose slope is m 2.

165 Slide 104 / 202 Slopes of Perpendicular Lines 1 Perpendicular lines obey the Pythagorean Theorem: a (1,m 1 ) c 2 = a 2 + b 2 Let's find expressions for those three terms so we can substitute them in. 0 b 1 c Side "c" is hypotenuse and is the distance between the points along the vertical line x=1. It has length m 1 -m 2 (in this case, that effectively adds their magnitudes since m 2 is negative). -1 (1,m 2 ) We can use the distance formula to find the length of each leg.

166 Slide 105 / 202 Slopes of Perpendicular Lines 1 (1,m 1 ) The lengths of legs "a" and "b" can be found using the distance formula. a d 2 = (x 2 -x 1 ) 2 + (y 2 -y 1 ) c a 2 = (1-0) 2 + (m 1-0) 2 = 1 + m 1 2 b 2 = (1-0) 2 + (m 2-0) 2 = 1 + m 2 2 b And, we get c 2 by squaring our result for c from the prior slide. -1 (1,m 2 ) c 2 = (m 1 -m 2 ) 2 = m 1 2-2m 1 m 2 + m 2 2

167 Slide 106 / 202 Slopes of Perpendicular Lines 1 (1,m 1 ) We're going to substitute the expressions from the prior slide into the Pythagorean Theorem. a We'll color code them so we can keep track. 0 1 c c 2 = m 1 2-2m 1 m 1 + m 2 2 a 2 = 1 + m 1 2 b 2 = 1 + m 2 2 b c 2 = a 2 + b 2-1 m 1 2-2m 1 m 2 + m 2 2 = 1 + m m 2 2 (1,m 2 )

168 Slide 107 / 202 Slopes of Perpendicular Lines m 1 2-2m 1 m 2 + m 2 2 = 1 + m m a (1,m 1 ) Now, let's identify like terms m 1 2-2m 1 m 2 + m 2 2 = 1 + m m 2 2 Notice that we have m 1 2 and m 2 2 on both sides so they can be canceled, and that = 2. So, 0 1 c -2m 1 m 2 = 2 m 1 m 2 = -1 b which is what we set out to prove Notice that this also be written as: -1 (1,m 2 ) m 1 = m 2 = -1 m 2-1 m 1

169 Slide 108 / If one line has a slope of 4, what must be the slope of any line perpendicular to it?

170 Slide 108 (Answer) / If one line has a slope of 4, what must be the slope of any line perpendicular to it? Answer _ 1 4 [This object is a pull tab]

171 Slide 109 / If one line has a slope of -1/2, what must be the slope of any line perpendicular to it?

172 Slide 109 (Answer) / If one line has a slope of -1/2, what must be the slope of any line perpendicular to it? Answer 2 [This object is a pull tab]

173 Slide 110 / If one line passes through the points (0, 0) and (4, 2) what must be the slope of any line perpendicular to that first line?

174 Slide 110 (Answer) / If one line passes through the points (0, 0) and (4, 2) what must be the slope of any line perpendicular to that first line? Answer -2 [This object is a pull tab]

175 Slide 111 / If one line passes through the points (-5, 9) and (5, 8) what must be the slope of any line perpendicular to that first line?

176 Slide 111 (Answer) / If one line passes through the points (-5, 9) and (5, 8) what must be the slope of any line perpendicular to that first line? Answer 10 [This object is a pull tab]

177 Slide 112 / If one line passes through the points (1, 2) and (5, 6) and a perpendicular line passes through the point (1, 5) which of these points could lie on that second line? A (2, 2) B (4, 4) C (2, 4) D (-1, 3)

178 Slide 112 (Answer) / If one line passes through the points (1, 2) and (5, 6) and a perpendicular line passes through the point (1, 5) which of these points could lie on that second line? A (2, 2) B (4, 4) C (2, 4) D (-1, 3) Answer C [This object is a pull tab]

179 Slide 113 / If one line passes through the points (-3, 4) and (0, 10) and a perpendicular line passes through the point (-1, -4) which of these points could lie on that second line? A (0, -2) B (2, -5) C (3, 1) D (-3, -5)

180 Slide 113 (Answer) / If one line passes through the points (-3, 4) and (0, 10) and a perpendicular line passes through the point (-1, -4) which of these points could lie on that second line? A (0, -2) B (2, -5) C (3, 1) D (-3, -5) Answer D [This object is a pull tab]

181 Slide 114 / If one line passes through the points (3, 5) and (5, -1) and a perpendicular line passes through the point (-1, -1) which of these points could lie on that second line? A (5, 0) B (2, 0) C (4, 8) D (-4, -4)

182 Slide 114 (Answer) / If one line passes through the points (3, 5) and (5, -1) and a perpendicular line passes through the point (-1, -1) which of these points could lie on that second line? A (5, 0) B (2, 0) C (4, 8) D (-4, -4) Answer B [This object is a pull tab]

183 Slide 115 / 202 Equations of Parallel & Perpendicular Lines Return to Table of Contents

184 10 y Slide 116 / 202 Writing the Equation of a Line x m = y 2-y 1 x 2 -x 1 Let's start with the above definition of slope, multiply both sides by (x 2 -x 1 ), and rearrange to get: 5 y 0 x 5 (y 2 -y 1 ) = m(x 2 -x 1 ) Now, if I enter one point (x 1,y 1 ) this defines the infinite locus of other points on the line. y-y 1 = m(x-x 1 ) Point-slope form Then, I can add y 1 to both sides to isolate the variable y. y = m(x-x 1 )+y 1

185 Slide 117 / 202 Writing the Equation of a Line 10 y x y = m(x-x 1 )+y 1 The term in red is just your steps to the right along the x-axis, your "run". 5 y 0 x 5 Multiplying by slope tells you how many steps up you must take along the y-axis, your "rise". Those steps are added to the term in green, your original position on the y-axis to find your final y-coordinate. That tells you the y-coordinate on the line for the given x- coordinate.

186 Slide 118 / 202 Slope Intercept Equation of a Line 10 y x y = m(x-x 1 )+y 1 A common simplification of this is to use the y-intercept for (x 1,y 1 ). 5 y 0 x 5 The y-intercept is the point where the line crosses the y- axis. Its symbol is "b" so the coordinates of that point are (0,b), since x = 0 on the y-axis. Just substitute (0,b) in the above equation for (x 1,y 1 ).

187 Slide 119 / 202 Slope Intercept Equation of a Line 10 y x y = m(x-x 1 )+y 1 y = m(x-0) + b y = mx + b 5 y 0 x 5 This is very useful since both the slope and the y-intercepts often have important meaning when we are solving real problems. In the graph to the left, b= 0, so the equation is y = 2x.

188 Slide 119 (Answer) / 202 Slope Intercept Equation of a Line 10 5 y Teacher Note y y = m(x-x 1 )+y 1 If xthe students need to review how to find the equation y = of m(x-0) a line, + b use the link below. y = mx + b algebra-i/graphing-linearequations/graphing-linearequations-presentation/ the slope and the y-intercepts This is very useful since both often have important meaning Click the weblink in the when bottom we left are solving real corner to access the URL problems. above [This object is a pull tab] 0 x 5 In the graph to the left, b= 0, so the equation is y = 2x.

189 Slide 120 / 202 Writing Equations of Parallel Lines 1. Find an equation of the line passing through the point (-4, 5) and parallel to the line whose equation is -3x + 2y = -1. Step 1: Identify the information given in the problem. Contains the point (-4, 5) & is parallel to the line -3x + 2y = -1 Step 2: Identify what information you still need to create the equation and choose the method to obtain it. The slope: Use the equation of the parallel line to determine the slope -3x + 2y = -1 Click Click 2y = 3x - 1 y = 3/2 x - 1/2 Therefore m = 3/2 click

190 Slide 120 (Answer) / 202 Writing Equations of Parallel Lines 1. Find an equation of the line passing through the point (-4, 5) and parallel to the line whose equation is -3x + 2y = -1. Step 1: Identify the information given in the problem. Math Practice Contains the point (-4, This 5) & example is parallel (this to slide the line and -3x the + next) 2y = -1 addresses MP1& MP2. Step 2: Identify what information you still need to create the equation and choose the method to obtain it. The slope: Use the equation of the parallel line to determine the slope -3x + 2y = -1 Click Click [This object is a pull tab] 2y = 3x - 1 y = 3/2 x - 1/2 Therefore m = 3/2 click

191 Step 3: Create the equation. Slide 121 / 202 Writing Equations of Parallel Lines y - y 1 = m(x - x 1 ) y - 5 = 3/2 (x +4) Click y - 5 = 3/2 x + 6 Click y = 3/2 x + 11 Click Point-Slope Form Slope-Intercept Form The correct solution to the original problem is either form of the equation. Point-Slope Form and Slope-Intercept Form are two ways to write the same linear equations.

192 Slide 122 / 202 Writing Equations of Perpendicular Lines 2. Write the equation of the line passing through the point (-2, 5) and perpendicular to the line y = 1/2 x + 3 Step 1: Identify the slope according to the given equation. Given equation: m = 1/2 click Perpendicular Line: m = -2 click Step 2: Use the given point and the point-slope formula to write the equation of the perpendicular line. y - 5 = -2 (x + 2) Click y - 5 = -2x - 4 Click y = -2x + 1 Click Point-Slope Form Slope-Intercept Form

193 Slide 122 (Answer) / 202 Writing Equations of Perpendicular Lines 2. Write the equation of the line passing through the point (-2, 5) and perpendicular to the line y = 1/2 x + 3 Step 1: Identify the slope according to the given equation. Given equation: m = 1/2 Math Practice click This example addresses MP1& MP2. Perpendicular Line: m = -2 click Step 2: Use the given point and the point-slope formula to write the equation of the perpendicular line. y - 5 = -2 (x + 2) Click y - 5 = -2x - 4 Click y = -2x + 1 Click Point-Slope Form [This object is a pull tab] Slope-Intercept Form

194 Slide 123 / 202 Writing Equations of Perpendicular Lines 3. Write the equation of the line passing through the point (4, 7) and perpendicular to the line x - 5y = 50

195 Slide 123 (Answer) / 202 Writing Equations of Perpendicular Lines 3. Write Original the equation Equation of the in line slope-intercept passing through the point form: (4, 7) and y = perpendicular 1/5 x - 10 to the line x - 5y = 50 m = 1/5, so Perpendicular m = -5 Answer & Math Practice y - 7 = -5(x - 4) point-slope form y - 7 = -5x + 20 y = -5x + 27 slope-intercept form This example addresses MP1& MP2. [This object is a pull tab]

196 Slide 124 / What is the equation of the line passing through (6, -2) and parallel to the line whose equation is y = 2x - 3? A y = 2x + 2 B y = -2x + 10 C y = 1/2 x - 5 D y = 2x - 14

197 Slide 124 (Answer) / What is the equation of the line passing through (6, -2) and parallel to the line whose equation is y = 2x - 3? A y = 2x + 2 B y = -2x + 10 C y = 1/2 x - 5 D y = 2x - 14 Answer D [This object is a pull tab]

198 Slide 125 / Which is the equation of a line parallel to the line represented by: y = -x - 22? A x - y = 22 B y - x = 22 C y + x = -17 D 2y + x = -22

199 Slide 125 (Answer) / Which is the equation of a line parallel to the line represented by: y = -x - 22? A x - y = 22 B y - x = 22 C y + x = -17 D 2y + x = -22 Answer C [This object is a pull tab]

200 Slide 126 / Two lines are represented by the equation: -3y = 12x - 14 and y = kx + 14 For which value of k will the lines be parallel? A 12 B -14 C 3 D -4

201 Slide 126 (Answer) / Two lines are represented by the equation: -3y = 12x - 14 and y = kx + 14 For which value of k will the lines be parallel? A 12 B -14 C 3 D -4 Answer D [This object is a pull tab]

202 Slide 127 / Which equation represents a line parallel to the line whose equation is: 3y + 4x = 21 A 12y + 16x = 12 B 3y - 4x = 22 C 3y = 4x + 21 D 4y + 3x = 21

203 Slide 127 (Answer) / Which equation represents a line parallel to the line whose equation is: 3y + 4x = 21 A 12y + 16x = 12 B 3y - 4x = 22 C 3y = 4x + 21 D 4y + 3x = 21 Answer A [This object is a pull tab]

204 Slide 128 / What is the equation of a line that passes through (9, 3) and is perpendicular to the line whose equation is 4x - 5y = 20? A y - 3 = -5/4(x - 9) B y - 3 = 4/5(x - 9) C y - 3 = 4(x - 9) D y - 3 = -5(x - 9)

205 Slide 128 (Answer) / What is the equation of a line that passes through (9, 3) and is perpendicular to the line whose equation is 4x - 5y = 20? A y - 3 = -5/4(x - 9) B y - 3 = 4/5(x - 9) C y - 3 = 4(x - 9) Answer D y - 3 = -5(x - 9) A [This object is a pull tab]

206 Slide 129 / What is an equation of the line that passes through point (6, -2) and is parallel to the line whose equation is y = x + 5? A y = x + 5 B y = x + 2 C y = - x + 2 D y = -2x + 2 E y = x

207 Slide 129 (Answer) / What is an equation of the line that passes through point (6, -2) and is parallel to the line whose equation is y = x + 5? A y = x + 5 B y = x + 2 C y = - x + 2 D y = -2x + 2 Answer B [This object is a pull tab] E y = x

208 Slide 130 / What is an equation of the line that passes through the point (5, -2) and is parallel to the line: 9x - 3y = 12 A y = 3x - 17 B y = x C y = - x + 17 D y = -3x + 15

209 Slide 130 (Answer) / What is an equation of the line that passes through the point (5, -2) and is parallel to the line: 9x - 3y = 12 A y = 3x - 17 B y = x C y = - x + 17 D y = -3x + 15 Answer A [This object is a pull tab]

210 Slide 131 / What is an equation of the line that contains the point (-4, 1) and is perpendicular to the line whose equation is y = -2x - 3? A y = 2x + 1 B y = 1/2x + 3 C y = -2x - 1 D y = - 1/2x + 3

211 Slide 131 (Answer) / What is an equation of the line that contains the point (-4, 1) and is perpendicular to the line whose equation is y = -2x - 3? A y = 2x + 1 B y = 1/2x + 3 C y = -2x - 1 D y = - 1/2x + 3 Answer B [This object is a pull tab]

212 Slide 132 / Two lines are represented by the given equations. What would be the best statement to describe these two lines? 2x + 5y =15 5(x + 1) = -2y + 20 A The lines are parallel. B The lines are the same line. C The lines are perpendicular. D The lines intersect at an angle other than 90.

213 Slide 132 (Answer) / Two lines are represented by the given equations. What would be the best statement to describe these two lines? A B C 2x + 5y =15 5(x + 1) = -2y + 20 Answer The lines are parallel. The lines are the same line. The lines are perpendicular. D The lines intersect at an angle other than 90. [This object is a pull tab] D

214 Slide 133 / 202 Triangle Coordinate Proofs Return to Table of Contents

215 Slide 134 / 202 Triangle Coordinate Proof Coordinate Proofs place figures on the Cartesian Plane to make use of the coordinates of key features of the figure, combined with formulae (e.g. distance formula, midpoint formula and slope formula) to help prove something. The use of the coordinates is an extra first step in conducting the proof. We'll provide a few examples and then have you do some proofs.

216 Slide 134 (Answer) / 202 Triangle Coordinate Proof Coordinate Proofs This place lesson figures addresses on the MP1, Cartesian MP2, MP3 Plane to & MP6. make use of the coordinates of key features of the figure, combined with formulae Additional (e.g. Q's distance to address formula, MP standards midpoint formula and slope What formula) information to help are prove you given? something. (MP1) What do you need to (MP1) Can you represent any side lengths with The use of the coordinates symbols and is numbers? an extra first (MP2) step in conducting the proof. What Math language will help you prove your answer? (MP3) What is the reason for this step? (MP6) We'll provide a few examples [This and object then is a pull have tab] you do some proofs. Math Practice

217 Slide 135 / 202 Triangle Coordinate Proof Given: The coordinates A(0, 4); B(3, 0); C(-3, 0) and Q(0, 0) are the vertices of ABC and AQB Prove: QA bisects CAB Sketch the triangles on some graph paper and then discuss a strategy to accomplish the proof.

218 Slide 136 / 202 Example Given: The coordinates A(0, 4); B(3, 0); C(-3, 0) and Q(0, 0) are the vertices of ABC and AQB Prove: QA bisects CAB Does this sketch look like yours? A(0,4) If not, take a moment to see if this is correct. Looking at this, our strategy becomes clear. C(-3,0) Q(0,0) B(3,0) If we can prove that AQC AQB, then we could prove CAQ BAQ which would mean that segment QA bisects CAB: our goal. If that makes sense, let's get to work.

219 Slide 137 / 202 Example Given: The coordinates A(0, 4); B(3, 0); C(-3, 0) and Q(0, 0) are the vertices of ABC and AQB Prove: QA bisects CAB A(0,4) We don't need to use the distance formula to find these lengths since they can be read off the graph: CQ = BQ = 3 & AQ = 4 We can use the distance formula to find these lengths: C(-3,0) Q(0,0) B(3,0) AB = ((3-0) 2 + (0-4) 2 ) 1/2 = (25) 1/2 = 5 AC = ((0-(-3)) 2 + (4-0) 2 ) 1/2 = (25) 1/2 = 5

220 Slide 138 / 202 Example Given: The coordinates A(0, 4); B(3, 0); C(-3, 0) and Q(0, 0) are the vertices of ABC and AQB Prove: QA bisects CAB 5 A 5 You should be able to see that these are two right triangles with identical length sides, so they must be congruent. 4 But, let's work out the proof as good practice. C 3 Q 3 B

221 Slide 139 / 202 Given: Coordinates of vertices of ABC A Coordinates of vertices AQB Prove: QA bisects CAB Statements Reasons C 3 Q 3 B 1. CQ = 3 and BQ = 3 2. AC = 5 and AB = 5 3. QC QB 4. AQ QA 5. AC AB 6. ΔAQC ΔAQB 7. CAQ BAQ 8. QA bisects CAB 1. Given in graph 2. Distance Formula 3. segments have equal measure 4. Reflexive Property of 5. segments have equal measure 6. SSS triangle congruence 7. CPCTC 8. Definition of angle bisector

222 Slide 140 / 202 Triangle Coordinate Proof Given: The points A(4, -1), B(5, 6), and C(1, 3) Prove: ABC is an isosceles right triangle Make a sketch and think of a strategy for this proof.

223 Slide 141 / y Triangle Coordinate Proof B Given: The points A(4, -1), B(5, 6), and C(1, 3) Prove: ABC is an isosceles right triangle C If we just had to prove this a right triangle, we could just show that the slope of BC and AC are negative (or opposite) reciprocals. But, we also have to show this is an isosceles triangle, so we'd still have to determine the lengths of the sides. 0 A 5 x Once we do two sides we may as well do three and then use Pythagorean Theorem to prove it both isosceles and right.

224 Slide 142 / 202 Triangle Coordinate Proof 5 y B Let's use the distance formula to find the lengths: d = ((x 2 -x 1 ) 2 + (y 2 -y 1 ) 2 ) 1/2 AC = ((4-1) 2 +(-1-3) 2 ) 1/2 = (9+16) 1/2 = (25) 1/2 = 5 C BC = ((5-1) 2 + (6-3) 2 ) 1/2 = (16+9) 1/2 = (25) 1/2 = 5 AB = ((4-5) 2 + (-1-6) 2 ) 1/2 = (1+49) 1/2 + (50) 1/2 = 5#2 0 5 x A

225 Slide 143 / y Triangle Coordinate Proof B Given: The points A(4, -1), B(5, 6), and C(1, 3) Prove: ABC is an isosceles right triangle The lengths are AC = 5; BC = 5; AB = 5#2 C Since AC = BC this is an isosceles triangle. Since (5#2) 2 = = 50 this is a right triangle 0 A 5 x So, we have proven this to be a right isosceles triangle

226 Slide 144 / 202 Triangle Coordinate Proof Given: The points A(1, 1), B(4, 4), and C(6, 2) Prove: ABC is a right triangle Make a sketch and think of a strategy.

227 Slide 145 / 202 y Triangle Coordinate Proof Given: Points A(1, 1), B(4, 4) & C(6, 2) Prove: ABC is a right triangle 5 Now we just have to prove that two sides of this triangle are B perpendicular. 0 A 5 x C At a glance, it'd seem like AB and BC are the likely ones, so let's check them first. If that failed, we'd have to check the other possibilities, but we let's check the obvious ones first. We do this by finding if their slopes are negative (or opposite) reciprocals.

228 Slide 146 / 202 y Triangle Coordinate Proof Given: Points A(1, 1), B(4, 4) & C(6, 2) Prove: ABC is a right triangle 5 B m = Δy y 2-y 1 Δx = x 2 -x 1 m AB = y 2-y x 2 -x = = = C m BC = y 2-y x 2 -x = = = A 5 x Since -1 is the negative (or opposite) reciprocal of 1, AB is perpendicular to BC and ABC is a right triangle

229 Slide 147 / 202 Equation of a Circle & Completing the Square Return to Table of Contents

230 Slide 148 / 202 The Circle as a Locus of Points r A Circle is a set of points that are all the same distance from the center of the circle. The distance from the center to the circumference is the radius, "r."

231 r (4, 8) Slide 149 / 202 The Circle as a Locus of Points Placing the circle on a coordinate plane we can see that if the center of the circle is at (0,0) this particular point is located at about (4,8). (0, 0) Let's use the distance formula to find the lengths: d = ((x 2 -x 1 ) 2 + (y 2 -y 1 ) 2 ) 1/2 r = ((4-0) 2 + (8-0) 2 ) 1/2 click = (16+64) 1/2 = (80) 1/2 = 4#5 click

232 Slide 149 (Answer) / 202 Math Practice The Circle as a Locus of Points (4, 8) Placing the circle on a This derivation example (this and the coordinate plane we can see next 3 slides) addresses MP1 & MP2. r that if the center of the circle Additional Q's to address MP standards is at (0,0) this particular point What information are you given? is (MP1) located at about (4,8). What do you need to find? (MP1) Can (0, 0) you represent any side lengths with symbols and numbers? (MP2) Let's use the distance formula to find the lengths: [This object is a pull tab] d = ((x 2 -x 1 ) 2 + (y 2 -y 1 ) 2 ) 1/2 r = ((4-0) 2 + (8-0) 2 ) 1/2 click = (16+64) 1/2 = (80) 1/2 = 4#5 click

233 (0,0) r (x,y) Slide 150 / 202 The Circle as a Locus of Points We can also solve in general for any point (x,y) which lies on the circumference of the circle whose center is located at the origin (0,0). This will give us the equation of a circle with its center at the origin, since every point on the circumference must satisfy this equation. If we start with the distance formula d = and square both sides, what will be the resulting equation? d 2 = (x 2 -x 1 ) 2 + (y 2 -y 1 ) 2 click

234 (0,0) r (x,y) Slide 151 / 202 The Circle as a Locus of Points d 2 = (x 2 -x 1 ) 2 + (y 2 -y 1 ) 2 Substituting r 2 for d 2 (x, y) for (x 2, y 2 ) (0, 0) for (x 1, y 1 ) What will be resulting equation after the substitution? r 2 = ((x-0) 2 + (y-0) 2 ) = x 2 +y 2 click The sides are usually swapped to yield: click x 2 + y 2 = r 2 The equation of a circle whose center is at the origin.

235 Slide 152 / What is the radius of the circle whose equation is x 2 + y 2 = 25?

236 Slide 152 (Answer) / What is the radius of the circle whose equation is x 2 + y 2 = 25? Answer The radius equal 25 = 5 [This object is a pull tab]

237 Slide 153 / If the y coordinate of a point on the circle x 2 + y 2 = 25 is 5, what is the x coordinate?

238 Slide 153 (Answer) / If the y coordinate of a point on the circle x 2 + y 2 = 25 is 5, what is the x coordinate? Answer The x-coordinate would be 0. [This object is a pull tab]

239 Slide 154 / How many points on the circle x 2 + y 2 = 25 have an x-coordinate of 3?

240 Slide 154 (Answer) / How many points on the circle x 2 + y 2 = 25 have an x-coordinate of 3? Answer Substitute 3 in for x and solve for y: (3 2 ) + y 2 = 25 y 2 = 25-9 y = 16 y = ± 4 Therefore there are TWO points, (3, 4) and (3, -4) [This object is a pull tab]

241 Slide 155 / How many points on the circle x 2 + y 2 = 25 have an y-coordinate of 6?

242 Slide 155 (Answer) / How many points on the circle x 2 + y 2 = 25 have an y-coordinate of 6? Answer None because it is outside of the radius of 5. [This object is a pull tab]

243 Slide 156 / 202 The Circle as a Locus of Points r In general, any circle centered on the origin will have an equation x 2 + y 2 = r 2 (0,0) If a point is on the circle, it must satisfy this equation. How about circles whose center is not on the origin (0, 0).

244 r (4,8) Slide 157 / 202 The Circle as a Locus of Points If a circle is not centered on the origin, the equation has to be shifted by the amount it is away from the origin. (0,0) For example, let's shift the center of this circle to (2, 3).

245 Slide 158 / 202 (6,11) The Circle as a Locus of Points (2,3) r Shifting the center of this circle from (0, 0) to (2, 3): You can see that the point on the circle that was at (4, 8) is now at (6, 11) Moving the center of the circle right 2 and up 3 will add that amount to each x and y coordinate.

246 r (6,11) Slide 159 / 202 The Circle as a Locus of Points But the distance from the center to each point on the circle has not changed. So, our equation for this circle has to reflect that. (2,3) The new equation will be (x - 2) 2 + (y - 3) 2 = r 2 We can check to see if we still get the same radius.

247 (2,3) r (6,11) Slide 160 / 202 The Circle as a Locus of Points r 2 = (6-2) 2 + (11-3) 2 r 2 = (4) 2 + (8) 2 r 2 = = 80 These are the same values we had before when the circle was centered on (0, 0) and that point was located at (4, 8). So, this translation of the center did not change the circle or its radius.

248 Slide 161 / 202 The Circle as a Locus of Points r In general, if the center of a circle is located at (h, k) and its radius is r, the equation for the circle is (x - h) 2 + (y - k) 2 = r 2 (h,k) Keep in mind that you are subtracting the x or y coordinate of the center of the circle. So, if the center is at (3, 5) and the radius is 4, the equation becomes (x - 3) 2 + (y - 5) 2 = 16

249 Slide 162 / 202 Example Write the equation of a circle with center (-2, 3) & radius 3.8.

250 Slide 162 (Answer) / 202 Example Answer & Math Practice Write the equation of a circle with center (-2, 3) & radius 3.8. (x - h) 2 + (y - k) 2 = r 2 (x - (-2)) 2 + (y - (3)) 2 = (x + 2) 2 + (y - 3) 2 = This example addresses MP1 & MP2. Additional Q's to address MP standards What information are you given? (MP1) What do you need to find? (MP1) Create an equation to represent the circle in this example. (MP2) [This object is a pull tab]

251 Slide 163 / What is the radius of the circle whose equation is (x - 5) 2 + (y - 3) 2 = 36?

252 Slide 163 (Answer) / What is the radius of the circle whose equation is (x - 5) 2 + (y - 3) 2 = 36? Answer The radius equals 36 = 6 [This object is a pull tab]

253 Slide 164 / What is the radius of the circle whose equation is (x + 3) 2 + (y - 4) 2 = 67?

254 Slide 164 (Answer) / What is the radius of the circle whose equation is (x + 3) 2 + (y - 4) 2 = 67? Answer The radius equals 67 = 8.19 [This object is a pull tab]

255 Slide 165 / What is the x-coordinate of the center of the circle whose equation is (x - 5) 2 + (y - 3) 2 = 47?

256 Slide 165 (Answer) / What is the x-coordinate of the center of the circle whose equation is (x - 5) 2 + (y - 3) 2 = 47? Answer It is equal to 5. [This object is a pull tab]

257 Slide 166 / What is the center and radius of the circle whose equation is (x + 3) 2 + (y - 4) 2 = 30?

258 Slide 166 (Answer) / What is the center and radius of the circle whose equation is (x + 3) 2 + (y - 4) 2 = 30? Answer The center is (-3, 4) and the radius is approximately [This object is a pull tab]

259 Slide 167 / What is the center and the radius of the circle whose equation is (x - 5) 2 + (y - 3) 2 = 57?

260 Slide 167 (Answer) / What is the center and the radius of the circle whose equation is (x - 5) 2 + (y - 3) 2 = 57? Answer First, we need to find the center of the circle: (5,3) Then we find the radius of the circle by approximating 57 which is [This object is a pull tab]

261 Slide 168 / What is the center and the radius of the circle whose equation is (x + 3) 2 + (y - 4) 2 = 65.36?

262 Slide 168 (Answer) / What is the center and the radius of the circle whose equation is (x + 3) 2 + (y - 4) 2 = 65.36? Answer First, we need to find the center of the circle: (-3,4) Then we find the radius of the circle by approximating which is [This object is a pull tab]

263 Slide 169 / 202 Example The point (-5, 6) is on a circle with center (-1, 3). Write the standard equation of the circle.

264 Slide 169 (Answer) / 202

265 Slide 170 / 202 Example The equation of a circle is (x - 4) 2 + (y + 2) 2 = 36. Graph the circle. Center: (4, -2) Click to reveal Radius = 36 = 6 Click to reveal (4, -2) Click on the center of the circle to reveal it in the graph.

266 Slide 170 (Answer) / 202 Example The equation of a circle is (x - 4) 2 + (y + 2) 2 = 36. Math Practice Graph the circle. Additional Q's to address MP standards What information are Center: you given? (4,(MP1) -2) Click to reveal What do you need to find? (MP1) How do you graph the Radius circle? = (MP4 36 & = 6 MP5) Click to reveal (4, -2) [This object is a pull tab] Click on the center of the circle to reveal it in the graph.

267 Slide 171 / Which is the standard equation of the circle below? A B C D x 2 + y 2 = 400 (x - 10) 2 + (y - 10) 2 = 400 (x + 10) 2 + (y - 10) 2 = 400 (x - 10) 2 + (y + 10) 2 =

268 Slide 171 (Answer) / Which is the standard equation of the circle below? A B x 2 + y 2 = 400 Answer (x - 10) 2 + (y - 10) 2 = 400 A C D (x + 10) 2 + (y - 10) 2 = 400 (x - 10) 2 + (y + 10) 2 = 400 [This object is a pull tab]

269 Slide 172 / Which is the standard equation of the circle? A B C D (x - 4) 2 + (y - 3) 2 = 81 (x - 4) 2 + (y - 3) 2 = 9 (x + 4) 2 + (y + 3) 2 = 81 (x + 4) 2 + (y + 3) 2 =

270 Slide 172 (Answer) / Which is the standard equation of the circle? A B C D (x - 4) 2 + (y - 3) 2 = 81 Answer (x - 4) 2 + (y - 3) 2 = 9 (x + 4) 2 + (y + 3) 2 = 81 (x + 4) 2 + (y + 3) 2 = 9 D [This object is a pull 2tab] 4 6 8

271 Slide 173 / What is the center of (x - 4) 2 + (y - 2) 2 = 64? A (0, 0) B (4, 2) C (-4, -2) D (4, -2)

272 Slide 173 (Answer) / What is the center of (x - 4) 2 + (y - 2) 2 = 64? A (0, 0) B (4, 2) C (-4, -2) D (4, -2) Answer B [This object is a pull tab]

273 Slide 174 / What is the radius of (x - 4) 2 + (y - 2) 2 = 89?

274 Slide 174 (Answer) / What is the radius of (x - 4) 2 + (y - 2) 2 = 89? Answer r = 9.43 [This object is a pull tab]

275 Slide 175 / What is the diameter of a circle whose equation is (x - 2) 2 + (y + 1) 2 = 16? A 2 B 4 C 8 D 16

276 Slide 175 (Answer) / What is the diameter of a circle whose equation is (x - 2) 2 + (y + 1) 2 = 16? A 2 B 4 C 8 D 16 Answer C [This object is a pull tab]

277 Slide 176 / Which point does not lie on the circle described by the equation (x + 2) 2 + (y - 4) 2 = 25? A (-2, -1) B (1, 8) C (3, 4) D (0, 5)

278 Slide 176 (Answer) / Which point does not lie on the circle described by the equation (x + 2) 2 + (y - 4) 2 = 25? A (-2, -1) B (1, 8) C (3, 4) D (0, 5) Answer D [This object is a pull tab]

279 Slide 177 / 202 Completing the Square You're sometimes going to be given the equation of a circle which is not in standard form. You need to be able to transform the equation to standard form in order to find the location of the center and the radius. For instance, it's not clear what the radius and center are of the circle described by this equation. x 2-4x + y 2-12 = 0

280 Slide 177 (Answer) / 202 Completing the Square You're sometimes This going example to be (this given slide the and equation the next of a 4) circle which is not in standard addresses form. MP2 & MP7. You need to be Additional able to transform Q's to address the equation MP standards to standard form in order to find the What location information of the are center you and given? the (MP1) radius. What do you need to find? (MP1) For instance, it's What not clear do the what numbers the radius represent and center in the are of the circle described problem? by this equation. (MP2) What do you know about perfect x 2-4x + y 2-12 square = 0 trinomials that can apply to this situation? (MP7) Math Practice [This object is a pull tab]

281 Slide 178 / 202 Completing the Square x 2-4x + y 2-12 = 0 To find the radius and the coordinates of the center, we need to transform this into the form (x - h) 2 + (y - k) 2 = r 2 The first step is to separate groups of terms which have x, which have y, and are constants. Just moving those around makes this equation: [x 2-4x] + y 2 = 12 Take a moment to confirm that this is true.

282 [x 2-4x] + y 2 = 12 Slide 179 / 202 Completing the Square x 2-4x + y 2-12 = 0 We already see that the y-coordinate of the center is 0 (k = 0), since y 2 is by itself. But what to do with the expression (x 2-4x)? We have to convert that into the form (x - h) 2 to find the x- coordinate of the center...and then the radius.

283 Slide 180 / 202 Completing the Square If you recall, when you square a binomial, you get a trinomial. (x-h) 2 = x 2-2hx + h 2 Our problem starts with an expression in the form of x 2-2hx, so let's solve for that so we can see what can replace it: x 2-2hx = (x-h) 2 - h 2 So The coefficient (-2h) of x is -2h. The constant of the trinomial (-h 2 ) is -(h) 2. So, to get h, divide the coefficient of x by -2 To make the expressions equivalent, subtract h 2 from the binomial

284 Slide 181 / 202 Completing the Square [x 2-4x] + y 2 = 12 Dividing the coefficient -4 by -2 yields 2, so h = 2 Then -h 2 = -4 [x 2-4x] + y 2 = 12 [(x - 2) 2-4] + y 2 = 12 (x - 2) 2 + y 2 = 16 The center is at (2, 0) and the radius is 4. The same steps are used to find k, when needed, as in the next example.

285 Slide 181 (Answer) / 202 Completing the Square [x 2-4x] + y 2 = 12 Dividing the coefficient -4 by -2 yields 2, so h = 2 Then -h 2 = -4 Teacher Notes [x 2-4x] + y 2 = 12 [(x - 2) 2-4] + y 2 = 12 (x - 2) 2 + y 2 = 16 The extra step for completing the square is shown in bold below: [x 2-4x] + y 2 = 12 [(x 2-4x + 4) - 4] + y 2 = 12 [(x - 2) 2-4] + y 2 = 12 The center is at (2, 0) and the radius is 4. [This object is a pull tab] The same steps are used to find k, when needed, as in the next example.

286 Slide 182 / 202 Example of Completing the Square Determine the radius and center of this circle. x 2 + y 2-2x + 6y + 6 = 0 [x 2-2x] + [y 2 + 6y] = -6 [x 2-2x] [y 2 + 6y] h = -2/(-2) = 1 k = +6/(-2) = -3 x 2-2x = (x - h) [y 2 + 6y] = (y -(-3)) 2 - (3) 2 x 2-2x = (x - 1) 2-1 [y 2 + 6y] = (y + 3) 2-9 (x - 1) (y + 3) 2-9 = -6 (x - 1) 2 + (y + 3) 2 = 4 The center is (1,-3) and the radius is 2

287 Slide 182 (Answer) / 202 Example of Completing the Square Determine the radius and center of this circle. This example addresses MP2 & MP7. x 2 + y 2-2x + 6y + 6 = 0 Additional Q's to address MP standards What information [x 2 are - 2x] you + given? [y 2 + 6y] (MP1) = -6 What do you need to find? (MP1) [x 2-2x] What do the numbers represent in the [y 2 + 6y] problem? (MP2) h = -2/(-2) What = 1 do you know about perfect square k = +6/(-2) = -3 trinomials that can apply to this x 2-2x = (x situation? - h) (MP7) [y 2 + 6y] = (y -(-3)) 2 - (3) 2 Math Practice x 2-2x = (x - 1) 2-1 [y 2 + 6y] = (y + 3) 2-9 [This object is a pull tab] (x - 1) (y + 3) 2-9 = -6 (x - 1) 2 + (y + 3) 2 = 4 The center is (1,-3) and the radius is 2

288 Slide 183 / What is the radius of the circle described by this equation? x 2 + y 2-2x + 6y + 6 = 0

289 Slide 183 (Answer) / What is the radius of the circle described by this equation? x 2 + y 2-2x + 6y + 6 = 0 Answer 2 [This object is a pull tab]

290 Slide 184 / What is the x-coordinate of the center of the circle described by this equation? x 2 + y 2-2x + 6y + 6 = 0

291 Slide 184 (Answer) / What is the x-coordinate of the center of the circle described by this equation? x 2 + y 2-2x + 6y + 6 = 0 Answer 1 [This object is a pull tab]

292 Slide 185 / What is the x-coordinate of the center of the circle described by this equation? x 2 + y 2-8x + 4y - 5 = 0

293 Slide 185 (Answer) / What is the x-coordinate of the center of the circle described by this equation? x 2 + y 2-8x + 4y - 5 = 0 Answer 4 [This object is a pull tab]

294 Slide 186 / What is the radius of the circle described by this equation? x 2 + y 2-8x + 4y - 5 = 0

295 Slide 186 (Answer) / What is the radius of the circle described by this equation? x 2 + y 2-8x + 4y - 5 = 0 Answer The radius equal 25 = 5 [This object is a pull tab]

296 Slide 187 / What is the radius of the circle described by this equation? x 2 + y x - 22y = 0

297 Slide 187 (Answer) / What is the radius of the circle described by this equation? x 2 + y x - 22y = 0 Answer The radius equal 11 = 3.32 [This object is a pull tab]

298 Slide 188 / What is the y-coordinate of the center of the circle described by this equation? x 2 + y x - 22y = 0

299 Slide 188 (Answer) / What is the y-coordinate of the center of the circle described by this equation? x 2 + y x - 22y = 0 Answer 11 [This object is a pull tab]

300 Slide 189 / Part A The equation x 2 + y 2-4x + 2y = b describes a circle. Determine the y-coordinate of the center of the circle. PARCC Released Question (EOY)

301 Slide 189 (Answer) / Part A The equation x 2 + y 2-4x + 2y = b describes a circle. Determine the y-coordinate of the center of the circle. Answer x 2 + y 2-4x + 2y = b [x 2-4x] + [y 2 + 2y] = b [(x-2) 2-4] + [(y+1) 2-1] = b (x-2) (y+1) 2-1 = b (x-2) 2 + (y+1) 2 = b + 5 Center (2,-1) The y-coordinate is -1 [This object is a pull tab] PARCC Released Question (EOY)

302 Slide 190 / Part B The equation x 2 + y 2-4x + 2y = b describes a circle. The radius of the circle is 7 units. What is the value of b in the equation? PARCC Released Question (EOY)

303 Slide 190 (Answer) / Part B The equation x 2 + y 2-4x + 2y = b describes a circle. The radius of (x-2) the 2 circle + (y+1) is 7 2 = units. b + 5What is the value of b in the equation? Radius is (b+5) 1/2 Answer r = 7 = (b+5) 1/2 49 = b + 5 b = 44 [This object is a pull tab] PARCC Released Question (EOY)

304 Slide 191 / The equation x 2-8x + y 2 = 9 defines a circle in the xycoordinate plane. To find the radius of the circle, the equation can be rewritten as ( ) 2 + y 2 =. (Select two answers.) A x + 4 B x - 4 C x + 16 D x - 16 E 25 F 13 G 9 H 5 PARCC Released Question (EOY)

305 Slide 191 (Answer) / The equation x 2-8x + y 2 = 9 defines a circle in the xycoordinate plane. To find the radius of the circle, the equation can be rewritten as ( ) 2 + y 2 =. (Select two answers.) Answer x 2-8x + + y 2 = 9 A x + 4 [x 2-8x] + [y-0] 2 = 9 B x - 4 [(x-4) 2-16] + y 2 = 9 (x-4) 2 + y 2 = 25 C x + 16 E 25 F 13 G 9 Center (4,0) Radius = (25) 1/2 = 5 D x - 16 b. x - 4 e. 25 [This object is a pull tab] H 5 PARCC Released Question (EOY)

306 Slide 192 / 202 PARCC Sample Questions Return to Table of Contents

307 Question 5/7 Slide 193 / 202 Topic: Partitions of a Line Segment PARCC Released Question (EOY)

308 Question 5/7 Slide 193 (Answer) / 202 Topic: Partitions of a Line Segment P 1 P 2 Answer [This object is a pull tab] PARCC Released Question (EOY)

309 Slide 194 / 202 Question 6/7 Topic: Equation of a Circle PARCC Released Question (EOY)

310 Slide 194 (Answer) / 202 Question 6/7 Topic: Equation of a Circle Answer [This object is a pull tab] PARCC Released Question (EOY)