Geometry. Slide 1 / 202. Slide 2 / 202. Slide 3 / 202. Analytic Geometry. Table of Contents

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1 Slide 1 / 202 Slide 2 / 202 Geometr Analtic Geometr Table of Contents Slide 3 / 202 Origin of Analtic Geometr 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 PARCC Sample Questions General Problems click on the topic to go to that section

2 Throughout this unit, the Standards for Mathematical Practice are used. MP1: Making sense of problems & persevere in solving them. MP2: Reason abstractl & quantitativel. MP3: Construct viable arguments and critique the reasoning of others. MP4: Model with mathematics. MP: Use appropriate tools strategicall. MP6: Attend to precision. MP7: Look for & make use of structure. MP8: Look for & express regularit in repeated reasoning. Slide 4 / 202 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 alread exist on a slide, then the specific MPs that the questions address are listed in the Pull-tab. Throughout this unit, the Standards for Mathematical Practice are used. MP1: Making sense of problems & persevere in solving them. MP2: Reason abstractl & quantitativel. MP3: Construct viable arguments and critique the reasoning of others. MP4: Model with mathematics. MP: Use appropriate tools strategicall. MP6: Attend to precision. MP7: Look for & make use of structure. MP8: Look for & express regularit in repeated reasoning. Math Practice Slide 4 () / 202 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 alread exist on a slide, then the specific MPs that the questions address are listed in the Pull-tab. Slide / 202 Origin of Analtic Geometr Return to Table of Contents

3 The Origin of Analtic Geometr Slide 6 / 202 Analtic Geometr is a powerful combination of geometr and algebra. Man jobs that are looking for emploees now, and will be in the future, rel on the process or results of analtic geometr. This includes jobs in medicine, veterinar science, biolog, chemistr, phsics, mathematics, engineering, financial analsis, economics, technolog, biotechnolog, etc. The Origin of Analtic Geometr Slide 7 / 202 Euclidean Geometr Was developed in Greece about 200 ears ago. Was lost to Europe for more than a thousand ears. Was maintained and refined during that time in the Islamic world. Its rediscover was a critical part of the European Renaissance. books-manuscripts/euclid-millietdechales-claude-francoismilliet details.aspx The Origin of Analtic Geometr Slide 8 / 202 Algebra Started b Diophantus in Alexandria about 1700 ears ago. Ongoing contributions from Bablon, Sria, Greece and Indians. Named from the Arabic word al-jabr which was used b al-khwarizmi in the title of his 7th centur book. Mathematics_in_medieval_Islam

4 The Origin of Analtic Geometr Slide 9 / 202 Analtic Geometr A powerful combination of algebra and geometr. Independentl developed, and published in 1637, b Rene Descartes and Pierre de Fermat in France. The Cartesian Plane is named for Descartes. The Origin of Analtic Geometr Slide / 202 How would ou describe to someone the location of these five points so the could draw them on another piece of paper without seeing our drawing? Discuss. The Origin of Analtic Geometr Slide () / 202 How would ou describe to someone the location of these five points so the The question on this slide addresses could draw them on MP2 & MP3. another piece of paper without seeing our drawing? Math Practice Discuss.

5 The Origin of Analtic Geometr Slide 11 / 202 Adding this Cartesian coordinate plane makes that task simple since the location of each point can be given b just two numbers: an x- and -coordinate, written as the ordered pair (x,) The Origin of Analtic Geometr Slide 12 / 202 With the Cartesian Plane providing a numerical description of locations on the plane, geometric figures can be analzed using algebra Slide 13 / 202 The Distance Formula Lab: Derivation of the Distance Formula Return to Table of Contents

6 Slide 13 () / 202 The lab link on this slide can be used as an activit where the students derive the distance formula in small groups. After the 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, MP, MP6, MP7 & MP8 Return to Table of Contents Slide 14 / 202 The Distance (x 2, 2) Formula (x 1, 1) Let's derive the formula to find the distance between an two points: call the points (x 1, 1) and (x 2, 2). - 0 x First, let's zoom in so we have more room to work. - - The Distance Formula Slide 1 / 202 (x 2, 2) Now, we'll construct two paths between the points. The first path will be directl between them. (x 1, 1) 0 x

7 The Distance Formula Slide 16 / 202 (x 2, 2) The second path will have a segment parallel the x-axis and a segment parallel to the -axis. That will be a big help since it's eas to read the length of each of those from the axis. (x 1, 1) 0 x The Distance Formula Slide 17 / 202 We now have a right triangle. (x 2, 2) For convenience, let's label the two legs "a" and "b" and the hpotenuse "c." c b Let's also label the point where the two legs meet b its coordinates: (x 2, 1). (x 1, 1) a (x 2, 1) Take a moment to see that those are the coordinates of that vertex of the triangle. 0 x Which formula relates the lengths of the sides of a right triangle? The Distance Formula Slide 17 () / 202 We now have a right triangle. (x 1, 1) (x 2, 2) Math Practice c a For convenience, let's label the two legs "a" and "b" and the hpotenuse "c." The question on this slide addresses Let's also label the point MP7. b where the two legs meet b its coordinates: (x 2, 1). (x 2, 1) Take a moment to see that those are the coordinates of that vertex of the triangle. 0 x Which formula relates the lengths of the sides of a right triangle?

8 The Distance Formula Slide 18 / 202 c (x 2, 2) b Did ou 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 subscript coordinates. Use the distance along each axis to find those. (x 1, 1) a (x 2, 1) The coordinates that we just added for the third vertex should help. 0 x The Distance Formula Slide 18 () / 202 Did ou get this? c 2 = a 2 + b 2 (x 2, 2) The next step is to write expressions for the lengths of sides a and b based on the x and 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 0 (x 1, 1) a (x 2, 1) x The coordinates that we just added for the third vertex should help. The Distance Formula Slide 19 / 202 Did ou get these? (x 2, 2) a = lx 2 - x 1l AND b = l 2-1l We could equall well write (x 1, 1) c a b (x 2, 1) a = lx 1 - x 2l AND b = l 1-2l We use absolute values since we are just concerned with lengths, which are alwas positive. That's wh the order doesn't matter and all of the above are OK. 0 x Substitute one pair of these into c 2 = a 2 + b 2

9 The Distance Formula Slide 19 () / 202 (x 1, 1) c (x 2, 2) Math Practice Did ou get these? a = lx 2 - x 1l AND b = l 2-1l The sentence, We "Substitute could equall one well pair..." write on this slide addresses a = lx 1 - x MP2. 2l AND b = l 1-2l b We use absolute values since we are just concerned with lengths, which are alwas positive. That's a [This wh object the is a pull order tab] doesn't matter and (x 2, 1) all of the above are OK. 0 x Substitute one pair of these into c 2 = a 2 + b 2 The Distance Formula Slide 20 / 202 We'll use the first pair: c (x 2, 2) b c 2 = (lx 2 - x 1l) 2 + (l 2-1l) 2 Since the quantities are squared, we don't need to indicate absolute value, the result will alwas be positive anwa. c 2 = (x 2 - x 1) 2 + ( 2-1) 2 0 (x 1, 1) a (x 2, 1) x Since "c" is the distance between the points, we can use "d" for distance and sa that d 2 = (x 2 - x 1) 2 + ( 2-1) 2 If we solve this equation for d, what will be the final formula? The Distance Formula Slide 20 () / 202 We'll use the first pair: c (x 2, 2) Math Practice c 2 = (lx 2 - x 1l) 2 + (l 2-1l) 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 alwas be positive anwa. c 2 = (x 2 - x 1) 2 + ( 2-1) 2 0 (x 1, 1) a (x 2, 1) x Since "c" is the distance between the points, we can use "d" for distance and sa that d 2 = (x 2 - x 1) 2 + ( 2-1) 2 If we solve this equation for d, what will be the final formula?

10 The Distance Formula Slide 21 / 202 (x 2, 2) d = ((x 2 - x 1) 2 + ( 2-1) 2 ) 1/2 OR c b d = (x 1, 1) a (x 2, 1) 0 x 1 What is the distance between the points: (4, 8) and (7, 3)? Round our answer to the nearest hundredth. Slide 22 / What is the distance between the points: (4, 8) and (7, 3)? Round our answer to the nearest hundredth. Slide 22 () / 202 d =.83

11 2 What is the distance between the points: (-4, 8) and (7, -3)? Round our answer to the nearest hundredth. Slide 23 / What is the distance between the points: (-4, 8) and (7, -3)? Round our answer to the nearest hundredth. Slide 23 () / 202 d = What is the distance between the points: (-2, -) and (-7, 3)? Round our answer to the nearest hundredth. Slide 24 / 202

12 3 What is the distance between the points: (-2, -) and (-7, 3)? Round our answer to the nearest hundredth. Slide 24 () / 202 d = What is the distance between the indicated points? Round our answer to the nearest hundredth. Slide 2 / What is the distance between the indicated points? Round our answer to the nearest hundredth. Slide 2 () / d =

13 What is the distance between the indicated points? Round our answer to the nearest hundredth. Slide 26 / What is the distance between the indicated points? Round our answer to the nearest hundredth. d =.20 Slide 26 () / What is the distance between the indicated points? Round our answer to the nearest hundredth. Slide 27 /

14 6 What is the distance between the indicated points? Round our answer to the nearest hundredth. d = Slide 27 () / Slide 28 / 202 The Midpoint Formula Lab - Midpoint Formula Return to Table of Contents The Midpoint Formula Slide 29 / 202 Another question which is easil solved using analtic geometr is to find the midpoint of a line. Once again, we make use of the fact that it's eas to determine distance parallel to an axis, so let's add those lines. 0 x

15 The Midpoint Formula Slide 30 / 202 The x-coordinate for the midpoint between (x 1, 1) and (x 2, 2) is halfwa between x 1 and x 2. Similarl, the -coordinate of that midpoint will be halfwa between 1 and 2. If ou're provided a graph of a line, and asked to mark the midpoint, ou can often do that without much calculating. 0 x The Midpoint Formula Slide 31 / 202 In this graph, x 1 = 1 and x 2 = 9. The are 8 units apart, so just go 4 units along the x-axis and go up until ou intersect the line. That will give ou the midpoint. 0 x In this case, that is at (,7), which can be read from the graph. We would get the same answer if we had done this along the - axis, as we do on the next slide. The Midpoint Formula Slide 31 () / 202 In this graph, x 1 = 1 and x 2 = 9. Math Practice 0 x The are 8 units apart, so just go 4 units along the x-axis and The sentence, go up "The until ou are intersect 8 units the line. apart..." on this slide addresses MP1. That will give ou the midpoint. In this case, that is at (,7), which can be read from the graph. We would get the same answer if we had done this along the - axis, as we do on the next slide.

16 The Midpoint Formula Slide 32 / 202 The -coordinates of the two points are 11 and 3, so the are also 8 units apart. 0 x So, just go up 4 from the lower -coordinate and then across to the line to also get (,7) to be the midpoint. So, the order doesn't matter and if ou have the graph and the numbers are eas to read, ou ma as well use it. If ou are not given a graph or the lines don't fall so that the values are eas to read, ou can do a quick calculation to get the same result. The Midpoint Formula Slide 32 () / 202 The -coordinates of the two points are 11 and 3, so the are also 8 units apart. Math Practice So, just go up 4 from the lower The first 2 paragraphs -coordinate on and this then slide across to address MP1. the line to also get (,7) to be the midpoint. So, the order doesn't matter and if ou have the graph and the numbers [This object is a are pull tab] eas to read, ou ma as well use it. 0 x If ou are not given a graph or the lines don't fall so that the values are eas to read, ou can do a quick calculation to get the same result. The Midpoint Formula Slide 33 / 202 (9,11) We can calculate the x-coordinate midwa between that of the two given points b finding their average. The same for the -coordinate. (1,3) 0 x Just add the two values and divide b two. The x-coordinate of the midpoint is (1+9)/2 = The -coordinate of the midpoint is (3+11)/2 = 7 That's the same answer: (,7)

17 The Midpoint Formula Slide 33 () / 202 (1,3) (9,11) Math Practice 0 x We can calculate the x-coordinate midwa between that of the two given points b finding their average. This The slide same addresses for the MP2. -coordinate. Just add the two values and divide b two. The x-coordinate of the midpoint is (1+9)/2 = The -coordinate of the midpoint is (3+11)/2 = 7 That's the same answer: (,7) The Midpoint Formula Slide 34 / 202 (x 2, 2) The more general solution is given below for an points: (x 1, 1) and (x 2, 2). The x-coordinate of the midpoint is given b x M = (x 1 + x 2) / 2 The -coordinate of the midpoint is given b M = ( 1 + 2) / 2 (x 1, 1) So, the coordinates of the midpoint are: 0 x 7 What is the midpoint between the indicated points? Slide 3 / 202 A (4, 9) B (-, -4) - 0 C (, 6) D (, 7) - -

18 7 What is the midpoint between the indicated points? Slide 3 () / 202 A (4, 9) D B (-, -4) - 0 C (, 6) D (, 7) What is the midpoint between the indicated points? Slide 36 / 202 A (0, 0) B (, ) - 0 C (, ) - D (, ) - 8 What is the midpoint between the indicated points? Slide 36 () / 202 A (0, 0) A B (, ) - 0 C (, ) D (, ) - -

19 9 What is the midpoint between the indicated points? Slide 37 / 202 A (3, 3) B (3, 4) - 0 C (4, 3) - D (, 3) - 9 What is the midpoint between the indicated points? A (3, 3) B (3, 4) - 0 A Slide 37 () / 202 C (4, 3) D (, 3) - - What is the midpoint between the points: (4, 8) and (7, 3)? Slide 38 / 202 A (8, 2) B (4, 7) C (.,.) D (6, )

20 What is the midpoint between the points: (4, 8) and (7, 3)? Slide 38 () / 202 A (8, 2) B (4, 7) C (.,.) D (6, ) C 11 What is the midpoint between the points: (-4, 8) and (4, -8)? Slide 39 / 202 A (8, 2) B (0, 0) C (12, 12) D (4, 4) 11 What is the midpoint between the points: (-4, 8) and (4, -8)? Slide 39 () / 202 A (8, 2) B (0, 0) C (12, 12) D (4, 4) B

21 12 What is the midpoint between the points: (-4, -8) and (-6, -4)? Slide 40 / 202 A (-, -6) B (-, -12) C (2, 4) D (, 6) 12 What is the midpoint between the points: (-4, -8) and (-6, -4)? Slide 40 () / 202 A (-, -6) B (-, -12) C (2, 4) D (, 6) A Finding the Coordinates of an E ndpoint of a Segment Slide 41 / 202 The midpoint of AB is M(1, 2). One endpoint is A(4, 6). Find the coordinates of the other endpoint B(x, ). You can use the midpoint formula to write equations using x and. Then, set up 2 equations and solve each one. B(x, ) M(1, 2) = A(4, 6) ( ) 4 + x,

22 M(1, 2) = 1 = Finding the Coordinates of an E ndpoint of a Segment 4 + x 2 2 = 4 + x -2 = x ( ) 4 + x, = = = M(1, 2) = A(4, 6) ( ) 4 + x, Slide 42 / 202 Therefore, the coordinates of B(-2, -2). B(x, ) Can ou find a shortcut to solve this problem? How would our shortcut make the problem easier? Finding the Coordinates of an E ndpoint of a Segment ( ) 4 + x, M(1, 2) = The first part of this example A(4, 6) 1 = 4 + x 2 = 6 addresses + MP = 4 + x 4 = The 6 + question at the bottom M(1, of 2) the = -2 = x -2 = slide addresses MP8. - 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). ) Math Practice ( ) 4 + x, Slide 42 () / 202 Can ou find a shortcut to solve this problem? How would our shortcut make the problem easier? Finding the Coordinates of an E ndpoint of a Segment Slide 43 / 202 Another wa of approaching this problem is to look for the pattern that occurs between the endpoint A (4, 6) and midpoint M (1, 2). Looking onl 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, ) left 3 M(1, 2) A(4, 6)

23 Finding the Coordinates of an E ndpoint of a Segment Slide 44 / 202 Similarl, if we line up the points verticall 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 ou use this method, alwas determine the operation required to get from the given endpoint to the midpoint. The reverse will not work. 13 Find the other endpoint of the segment with the endpoint (7, 2) and midpoint (3, 0) Slide 4 / 202 A (-1, -2) B (-2, -1) C (4, 2) D (2, 4) 13 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) M(x, ) = ( ( A M(3, 0) =, 3 =, 0 =, ) ) Slide 4 () / 202 (-1, -2)

24 14 Find the other endpoint of the segment with the endpoint (1, 4) and midpoint (, -2) Slide 46 / 202 A (11, -8) B (9, 0) C (9, -8) D (3, 1) 14 Find the other endpoint of the segment with the endpoint (1, 4) and midpoint (, -2) A (11, -8) B (9, 0) C (9, -8) D (3, 1) M(x, ) = M(, -2) = ( ( C,, ) ) Slide 46 () / 202 =, -2 = (9, -8) 1 Find the other endpoint of the segment with the endpoint (-4, -1) and midpoint (-2, 3). Slide 47 / 202 A (-6, -) B (-3, -2) C (0, 7) D (1, 9)

25 1 Find the other endpoint of the segment with the endpoint (-4, -1) and midpoint (-2, 3). Slide 47 () / 202 A (-6, -) B (-3, -2) C (0, 7) D (1, 9) C 16 Find the other endpoint of the segment with the endpoint (-2, ) and midpoint (0, 2). Slide 48 / 202 A (-1, -3.) B (-4, 8) C (1, 0.) D (2, -1) 16 Find the other endpoint of the segment with the endpoint (-2, ) and midpoint (0, 2). Slide 48 () / 202 A (-1, -3.) B (-4, 8) C (1, 0.) D (2, -1) D

26 Slide 49 / 202 Partitions of a Line Segment Return to Table of Contents Partitions of a Line Segment Slide 0 / 202 B Partitioning a line segment simpl 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 ou need to partition a segment so that its ratio is something different, for example 3:1, how can it be done? Partitions of a Line Segment Slide 0 () / 202 Math Practice Partitioning a line segment simpl 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, MP & MP7 The midpoint partitions a segment into 2 congruent segments, forming a ratio of 1:1. A But if ou need to partition a segment so that its ratio is something different, for example 3:1, how can it be done?

27 Partitions of a Line Segment Slide 1 / 202 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 A Partitions of a Line Segment B 0 x 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 sa 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 Slide 2 / 202 A Partitions of a Line Segment B 0 x 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, point the midpoint. that partitions AB - Click to form on midpoint the ratio to hide it Let's 3:1 sa that is (7, the 9) 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 Slide 2 () / 202

28 Partitions of a Line Segment Slide 3 / 202 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 A 0 x 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 Partitions of a Line Segment Slide 3 () / 202 A B 0 x Could we also determine the coordinates of our point if the ratio was reversed (1:3 instead of 3:1)? Therefore, the coordinates Yes, we can. Again, of one our midpoint that can partitions be eliminated. AB - Click to form on midpoint the ratio to hide it 1:3 is (3, ) 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 Partitions of a Line Segment A(1,3) B (9,11) 0 x We can also calculate both the x and coordinates between the two given points b using a formula that is similar to the midpoint formula. But instead of having a common ratio (1:1) and dividing b 2, what the midpoint formula has us do, we need to multipl the one set of coordinates b the first number in the ratio and the other set of coordinates b the second number in the ratio and divide b 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 Slide 4 / 202

29 Partitions of a Line Segment A(1,3) B (9,11) Math Practice 0 x We can also calculate both the x and coordinates between the two given points b 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. b 2, what the midpoint formula has us do, we need to multipl the one set of coordinates b the first number in the ratio and the other set of coordinates b the second number in the ratio and divide b 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 Slide 4 () / 202 Partitions of a Line Segment Slide / 202 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, ) 1(9) + 3(1) 1(11) + 3(3) A (1,3) 0 x Partitions of a Line Segment Slide () / 202 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, ) 1(9) + 3(1) 1(11) + 3(3) This example addresses 3 + MP A (1,3) 0 x

30 Partitions of a Line Segment Slide 6 / 202 (x 2, 2) The more general solution is given below for an points: (x 1, 1) and (x 2, 2). If a point P partitions a segment in a ratio of m:n, then the coordinates of P are P = (, ) mx2 + nx1 m + n m2 + n1 m + n (x 1, 1) Remember to also use the formula twice switching the m:n ratio to n:m. 0 x Partitions of a Line Segment Example with a graph: Line segment CD in the coordinate plane has endpoints with coordinates (-2, ) and (8, -). Graph CD and find two possible locations for a point P that divides CD into two parts with lengths in a ratio of 2:3. Slide 7 / x - Partitions of a Line Segment Math Practice Example with a graph: Line segment CD in the coordinate plane has endpoints with coordinates (-2, ) and (8, -). This example Graph (this CD slide and and find the two possible next 3) addresses locations MP1, for a MP4, point MP P that divides & MP7 CD into two parts with lengths in a ratio of 2:3. Slide 7 () / x -

31 C (-2, ) Partitions of a Line Segment Step #1: Graph the points. C(-2, ) and D(8, -) - Click on the points to show. Slide 8 / 202 (0, 7) (2, 4) (4, 1) 0 x (6, -2) - D (8, -) Step #2: Connect them w/ the segment CD. - Click on the segment to show. Step #3: Determine the total number of was required to divide the segment using the ratio 2: = Click Step #4: Plot the points on the graph that divide CD into the desired number of sections. - Click on the points to show. C (-2, ) (0, 7) Partitions of a Line Segment Step #: Since the ratio is 2:3, determine which points to eliminate (3 of them) leaving onl 1 point remaining so that, from left to right, the ratio is 2:3. Slide 9 / 202 (2, 4) (4, 1) 0 x (6, -2) - D (8, -) C (-2, ) (0, 7) Partitions of a Line Segment (2, 4) (4, 1) Step #: Since the ratio is 2:3, determine which points to eliminate (3 of them) leaving onl 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) Slide 9 () / x (6, -2) - D (8, -)

32 C (-2, ) (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 #, determine which points to eliminate (3 of them) leaving onl 1 point remaining so that, from left to right, the ratio is 3:2. Slide 60 / 202 (4, 1) 0 x (6, -2) - D (8, -) C (-2, ) (0, 7) Partitions of a Line Segment (2, 4) (4, 1) 0 x (6, -2) Step #6: Because the question asks for 2 points, we also need to consider the ratio 3:2. Similar to Step #, determine which points to eliminate (3 of them) leaving onl 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) Slide 60 () / D (8, -) Partitions of a Line Segment Slide 61 / 202 Example without a graph: Line segment EF in the coordinate plane has endpoints with coordinates (, -11) and (-4, ). Find two possible locations for a point P that divides EF into two parts with lengths in a ratio of :2.

33 Partitions of a Line Segment Slide 61 () / 202 Example without a graph: Line segment EF in the coordinate plane has endpoints with coordinates (, -11) and (-4, ). 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 :2. addresses MP2. Partitions of a Line Segment Slide 62 / 202 (, -11) and (-4, ) Step #1: Determine the total number of was required to divide the segment using the ratio : = 7 Click Step #2: Use the formula to determine the coordinates of our first point P. click P = P = mx2 + nx1 ( m2 + n1 m + n, m + n ) (, ) P = (-4) + 2() () + 2(-11) (-22) 7 7 P = (0, 4) Partitions of a Line Segment Slide 63 / 202 (, -11) and (-4, ) Step #3: Reverse the ratio 2: and reuse the formula to determine the coordinates of our second point P. click P = P = mx2 + nx1 ( m2 + n1 m + n, m + n ) (, ) P = 2(-4) + () 2() + (-11) (-) 7 7 P = (6, )

34 17 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. A (0., -3) B (3, -1) C (6, 1) D (-2, -) E (-, -7) - G (-7, -9) H(8, 3) 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 2:1. A (0., -3) B (3, -1) C (6, 1) D (-2, -) E (-, -7) - G (-7, -9) B & D H(8, 3) Slide 64 () / Line segment JK in the coordinate plane has endpoints with coordinates (, 12) & (-, -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) C (0, -0.) D (2, 2) E (6, 7) K(-, -13) J(, 12) Slide 6 / 202

35 18 Line segment JK in the coordinate plane has endpoints with coordinates (, 12) & (-, -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) C (0, -0.) D (2, 2) E (6, 7) K(-, -13) - A & E J(, 12) Slide 6 () / Line segment LM in L(-8, 13) 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, ) B (-4, 7) C (-2, 4) D (0, 1) E (2, -3) M(6, -8) Slide 66 / Line segment LM in L(-8, 13) 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, ) B (-4, 7) C (-2, 4) D (0, 1) E (2, -3) - C & D M(6, -8) Slide 66 () / 202

36 20 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. Slide 67 / 202 A (-6, ) B (-4, 7) C (-2, 4) D (0, 1) E (4, -6) 20 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. Slide 67 () / 202 A (-6, ) B (-4, 7) C (-2, 4) D (0, 1) E (4, -6) A & E 21 Line segment NO in the coordinate plane has endpoints with coordinates (, 12) & (-, -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. Slide 68 / 202 A (-6, -8) B (-2, -3) C (0, -0.) D (2, 2) E (6, 7)

37 21 Line segment NO in the coordinate plane has endpoints with coordinates (, 12) & (-, -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. Slide 68 () / 202 A (-6, -8) B (-2, -3) C (0, -0.) D (2, 2) E (6, 7) B & D 22 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 :3. Slide 69 / 202 A (3, -4) B (0, -1) C (-3, 2) D (-6, ) E (-9, 8) 22 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 :3. Slide 69 () / 202 A (3, -4) B (0, -1) C (-3, 2) D (-6, ) E (-9, 8) A & C

38 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. Slide 70 / 202 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) 23 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) Slide 71 / 202 B (-1, 8) C (0, 7) D (4, 3) E (, 2) K (8, -1) F (6, 1) PARCC Released Question (EOY) - Part 2 - Response Format 23 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) Slide 71 () / 202 B (-1, 8) C (0, 7) D (4, 3) E (, 2) K (8, -1) F (6, 1) B & E 2 PARCC Released Question (EOY) - Part 2 - Response Format

39 24 Point Q lie on ST, where point S is located at (-2, -6) and point T is located at (, 8). If SQ:QT = :2, where is point Q on ST? Slide 72 / 202 A (-1, -4) B (0, -2) C (1, 0) D (2, 2) E (3, 4) F (4, 6) PARCC Released Question (PBA) 24 Point Q lie on ST, where point S is located at (-2, -6) and point T is located at (, 8). If SQ:QT = :2, where is point Q on ST? Slide 72 () / 202 A (-1, -4) B (0, -2) C (1, 0) D (2, 2) E (3, 4) F (4, 6) E (3, 4) PARCC Released Question (PBA) Slide 73 / 202 Slopes of Parallel & Perpendicular Lines Return to Table of Contents

40 Slope Slide 74 / 202 The slope of a line indicates the angle it makes with the x-axis. The smbol for slope is "m". 0 x Slope Slide 7 / x A horizontal line has a slope of zero. A vertical line has an undefined slope. A line which rises as ou move from left to right has a positive slope. A line which falls as ou move from left to right has a negative slope. 2 The slope of the indicated line is: Slide 76 / 202 A negative B positive C zero D undefined x

41 2 The slope of the indicated line is: Slide 76 () / 202 A negative B positive C zero D undefined B x 26 The slope of the indicated line is: Slide 77 / 202 A negative B positive C zero D undefined x 26 The slope of the indicated line is: Slide 77 () / 202 A negative B positive C zero D undefined B x

42 27 The slope of the indicated line is: Slide 78 / 202 A negative B positive C zero D undefined x 27 The slope of the indicated line is: Slide 78 () / 202 A negative B positive C zero D undefined A x 28 The slope of the indicated line is: Slide 79 / 202 A negative B positive C zero D undefined x

43 28 The slope of the indicated line is: Slide 79 () / 202 A negative B positive C zero D undefined D x 29 The slope of the indicated line is: Slide 80 / 202 A negative B positive C zero D undefined x 29 The slope of the indicated line is: Slide 80 () / 202 A negative B positive C zero D undefined C x

44 30 The slope of the indicated line is: Slide 81 / 202 A negative B positive C zero D undefined x 30 The slope of the indicated line is: Slide 81 () / 202 A negative B positive C zero D undefined C x Slope Slide 82 / 202 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 anwhere along the line, so an two points on the line can be used to calculate the slope. 0 x

45 Slope Slide 83 / 202 (x 1, 2) "rise" (x 1, 1) x "run" (x 2, 2) The "rise" is the change in the value of the -coordinate while the "run" is the change in the value of the x-coordinate. The smbol for change is the Greek letter delta, " ", which just means "change in". So the slope is equal to the change in divided b the change in x, or divided b x...delta over delta x. 0 x m = Δ 2-1 Δx = x 2-x 1 Slope Slide 84 / 202 x (2, 11) "run" (8,11) m = Δ 2-1 Δx = x 2-x 1 In this case: The rise is from 4 to 11 "rise" = 11-4 = 7 (2,4) And the run is from 2 to 8, x = 8-2 = 6 0 x So the slope is m = = 7 x 6 Slope Slide 8 / 202 x x (x 2, 2) An points on the line can be used to calculate its slope, since the slope of a line is the same everwhere. 0 (x 1, 1) x The values of and x ma 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 = Δ 2-1 Δx = x 2-x 1

46 Using Slope to Draw a Line Slide 86 / 202 The slope also allows us to quickl 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 m = Δ 2-1 Δx = x 2-x 1 Using Slope to Draw a Line Slide 87 / 202 I do this b recognizing that the slope of 2 means that if I go up 2 units on the -axis I have to go 1 unit to the right on the x-axis. x 0 x m = Δ 2-1 Δx = x 2-x 1 Using Slope to Draw a Line Slide 88 / 202 x I do this b recognizing that the slope of 2 means that if I go up 2 units on the -axis I have to go 1 unit to the right on the x-axis. Or if I go up, I have to go over units, etc. 0 x m = Δ 2-1 Δx = x 2-x 1

47 Using Slope to Draw a Line Slide 89 / 202 x Then I draw the line through an two of those points. This method is the easiest to use if ou just have to draw a line given a point and slope. The same approach works for writing the equation of a line. 0 x m = Δ 2-1 Δx = x 2-x 1 Slopes of Parallel Lines Slide 90 / 202 Parallel lines have the same slope. With what we've learned about parallel lines this is eas 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. Slopes of Parallel Lines Slide 90 () / 202 Parallel lines have the same slope. Math Practice With what we've learned about parallel lines this is eas 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.

48 31 What is the name of this pair of angles, formed b a transversal intersecting two lines. Slide 91 / 202 A Alternate Interior Angles B Corresponding Angles C Alternate Exterior Angles D Same Side Interior Angles What is the name of this pair of angles, formed b a transversal intersecting two lines. Slide 91 () / 202 A Alternate Interior Angles B Corresponding Angles C Alternate Exterior Angles D Same Side Interior Angles B We know those angles must then be: Slide 92 / 202 A Supplementar B Complementar C Equal D Adjacent 1 2

49 32 We know those angles must then be: Slide 92 () / 202 A Supplementar B Complementar C Equal D Adjacent C That means that the slopes of parallel lines must be: Slide 93 / 202 A Reciprocals B Inverses C Equal D Nothing special That means that the slopes of parallel lines must be: Slide 93 () / 202 A Reciprocals B Inverses C Equal D Nothing special C 1 2

50 34 If one line has a slope of 4, what must be the slope of an line parallel to it? Slide 94 / If one line has a slope of 4, what must be the slope of an line parallel to it? Slide 94 () / If one line passes through the points (0, 0) and (2, 2) what must be the slope of an line parallel to that first line? Slide 9 / 202

51 3 If one line passes through the points (0, 0) and (2, 2) what must be the slope of an line parallel to that first line? Slide 9 () / If one line passes through the points (-, 9) and (, 8) what must be the slope of an line parallel to that first line? Slide 96 / If one line passes through the points (-, 9) and (, 8) what must be the slope of an line parallel to that first line? Slide 96 () / 202 _ 1

52 37 If one line passes through the points (2, 2) and (, ) and a parallel line passes through the point (1, ) which of these points could lie on that second line? Slide 97 / 202 A (2, 2) B (4, 4) C (, 6) D (-1, 3) 37 If one line passes through the points (2, 2) and (, ) and a parallel line passes through the point (1, ) which of these points could lie on that second line? Slide 97 () / 202 A (2, 2) B (4, 4) C (, 6) D (-1, 3) D 38 If one line passes through the points (-3, 4) and (0, ) and a parallel line passes through the point (-1, -4) which of these points could lie on that second line? Slide 98 / 202 A (0, -2) B (2, -) C (3, 1) D (-1, 3)

53 38 If one line passes through the points (-3, 4) and (0, ) and a parallel line passes through the point (-1, -4) which of these points could lie on that second line? Slide 98 () / 202 A (0, -2) B (2, -) C (3, 1) D (-1, 3) A 39 If one line passes through the points (3, ) and (, -1) and a parallel line passes through the point (-1, -1) which of these points could lie on that second line? Slide 99 / 202 A (0, 1) B (-2, 2) C (4, 8) D (-4, -4) 39 If one line passes through the points (3, ) and (, -1) and a parallel line passes through the point (-1, -1) which of these points could lie on that second line? Slide 99 () / 202 A (0, 1) B (-2, 2) C (4, 8) D (-4, -4) B

54 Slopes of Perpendicular Lines The slopes of perpendicular lines are negative (or opposite) reciprocals. Slide 0 / 202 There are three was of expressing this smbolicall: m 1m 2 = -1 m 1 = m 2 = -1 m 2-1 m 1 These all have the same meaning. Slopes of Perpendicular Lines The slopes of perpendicular lines are negative (or opposite) reciprocals. Slide 1 / 202 This will be useful in cases where ou need to prove lines perpendicular, including proving that triangles are right triangles. First, let's prove this is true. Slopes of Perpendicular Lines The slopes of perpendicular lines are negative (or opposite) reciprocals. Slide 1 () / 202 Math Practice This will be useful in cases where ou 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.

55 Slopes of Perpendicular Lines First, let's move our lines to the origin so our work gets easier and we can focus on the important parts. Slide 2 / 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. Slopes of Perpendicular Lines Slide 3 / (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 -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 -value must increase b the amount of the slope, m (1,m 2) The same for the second line, whose slope is m 2. Slopes of Perpendicular Lines Slide 4 / Perpendicular lines obe the Pthagorean 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 b (1,m 2) 1 c Side "c" is hpotenuse and is the distance between the points along the vertical line x=1. It has length m 1-m 2 (in this case, that effectivel adds their magnitudes since m 2 is negative). We can use the distance formula to find the length of each leg.

56 Slopes of Perpendicular Lines Slide / a (1,m 1) The lengths of legs "a" and "b" can be found using the distance formula. d 2 = (x 2-x 1) 2 + ( 2-1) 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 b squaring our result for c from the prior slide. -1 (1,m 2) c 2 = (m 1-m 2) 2 = m 1 2-2m 1m 2 + m 2 2 Slopes of Perpendicular Lines Slide 6 / (1,m 1) We're going to substitute the expressions from the prior slide into the Pthagorean Theorem. a We'll color code them so we can keep track. 0 1 c c 2 = m 1 2-2m 1m 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 1m 2 + m 2 2 = 1 + m m 2 2 (1,m 2) Slopes of Perpendicular Lines Slide 7 / 202 m 1 2-2m 1m 2 + m 2 2 = 1 + m m a (1,m 1) Now, let's identif like terms m 1 2-2m 1m 2 + m 2 2 = 1 + m m 2 2 Notice that we have m 1 2 and m 2 2 on both sides so the can be canceled, and that = 2. So, 0 1 c -2m 1m 2 = 2 m 1m 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

57 40 If one line has a slope of 4, what must be the slope of an line perpendicular to it? Slide 8 / If one line has a slope of 4, what must be the slope of an line perpendicular to it? Slide 8 () / 202 _ If one line has a slope of -1/2, what must be the slope of an line perpendicular to it? Slide 9 / 202

58 41 If one line has a slope of -1/2, what must be the slope of an line perpendicular to it? Slide 9 () / If one line passes through the points (0, 0) and (4, 2) what must be the slope of an line perpendicular to that first line? Slide 1 / If one line passes through the points (0, 0) and (4, 2) what must be the slope of an line perpendicular to that first line? Slide 1 () / 202-2

59 43 If one line passes through the points (-, 9) and (, 8) what must be the slope of an line perpendicular to that first line? Slide 111 / If one line passes through the points (-, 9) and (, 8) what must be the slope of an line perpendicular to that first line? Slide 111 () / If one line passes through the points (1, 2) and (, 6) and a perpendicular line passes through the point (1, ) which of these points could lie on that second line? Slide 112 / 202 A (2, 2) B (4, 4) C (2, 4) D (-1, 3)

60 44 If one line passes through the points (1, 2) and (, 6) and a perpendicular line passes through the point (1, ) which of these points could lie on that second line? Slide 112 () / 202 A (2, 2) B (4, 4) C (2, 4) D (-1, 3) C 4 If one line passes through the points (-3, 4) and (0, ) and a perpendicular line passes through the point (-1, -4) which of these points could lie on that second line? Slide 113 / 202 A (0, -2) B (2, -) C (3, 1) D (-3, -) 4 If one line passes through the points (-3, 4) and (0, ) and a perpendicular line passes through the point (-1, -4) which of these points could lie on that second line? Slide 113 () / 202 A (0, -2) B (2, -) C (3, 1) D (-3, -) D

61 46 If one line passes through the points (3, ) and (, -1) and a perpendicular line passes through the point (-1, -1) which of these points could lie on that second line? Slide 114 / 202 A (, 0) B (2, 0) C (4, 8) D (-4, -4) 46 If one line passes through the points (3, ) and (, -1) and a perpendicular line passes through the point (-1, -1) which of these points could lie on that second line? Slide 114 () / 202 A (, 0) B (2, 0) C (4, 8) D (-4, -4) B Slide 11 / 202 Equations of Parallel & Perpendicular Lines Return to Table of Contents

62 Writing the Equation of a Line x m = 2-1 x 2-x 1 Let's start with the above definition of slope, multipl both sides b (x 2-x 1), and rearrange to get: Slide 116 / x ( 2-1) = m(x 2-x 1) Now, if I enter one point (x 1, 1) this defines the infinite locus of other points on the line. - 1 = m(x-x 1) Point-slope form Then, I can add 1 to both sides to isolate the variable. = m(x-x 1)+ 1 Writing the Equation of a Line Slide 117 / 202 x = m(x-x 1)+ 1 The term in red is just our steps to the right along the x-axis, our "run". 0 x Multipling b slope tells ou how man steps up ou must take along the -axis, our "rise". Those steps are added to the term in green, our original position on the -axis to find our final -coordinate. That tells ou the -coordinate on the line for the given x- coordinate. Slope Intercept Equation of a Line Slide 118 / 202 x = m(x-x 1)+ 1 A common simplification of this is to use the -intercept for (x 1, 1). The -intercept is the point where the line crosses the - axis. Its smbol is "b" so the coordinates of that point are (0,b), since x = 0 on the -axis. 0 x Just substitute (0,b) in the above equation for (x 1, 1).

63 Slope Intercept Equation of a Line Slide 119 / 202 x = m(x-x 1)+ 1 = m(x-0) + b = mx + b This is ver useful since both the slope and the -intercepts often have important meaning when we are solving real problems. 0 x In the graph to the left, b= 0, so the equation is = 2x. Slope Intercept Equation of a Line Slide 119 () / 202 Teacher Note = m(x-x 1)+ 1 If xthe students need to review how to find the equation = of m(x-0) a line, + b use the link below. = mx + b algebra-i/graphing-linear- equations/graphing-linear- equations-presentation/ the slope and the -intercepts This is ver 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 0 x In the graph to the left, b= 0, so the equation is = 2x. Writing Equations of Parallel Lines Slide 120 / Find an equation of the line passing through the point (-4, ) and parallel to the line whose equation is -3x + 2 = -1. Step 1: Identif the information given in the problem. Contains the point (-4, ) & is parallel to the line -3x + 2 = -1 Step 2: Identif what information ou 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 + 2 = -1 Click Click 2 = 3x - 1 = 3/2 x - 1/2 Therefore m = 3/2 click

64 Writing Equations of Parallel Lines Slide 120 () / Find an equation of the line passing through the point (-4, ) and parallel to the line whose equation is -3x + 2 = -1. Step 1: Identif the information given in the problem. Math Practice Contains the point (-4, This ) & example is parallel (this to slide the line and -3x the + next) 2 = -1 addresses MP1& MP2. Step 2: Identif what information ou 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 + 2 = -1 Click Click 2 = 3x - 1 = 3/2 x - 1/2 Therefore m = 3/2 click Writing Equations of Parallel Lines Slide 121 / 202 Step 3: Create the equation. - 1= m(x - x 1) - = 3/2 (x +4) Click - = 3/2 x + 6 Click = 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 was to write the same linear equations. Writing Equations of Perpendicular Lines 2. Write the equation of the line passing through the point (-2, ) and perpendicular to the line = 1/2 x + 3 Step 1: Identif the slope according to the given equation. Given equation: m = 1/2 click Perpendicular Line: m = -2 click Slide 122 / 202 Step 2: Use the given point and the point-slope formula to write the equation of the perpendicular line. - = -2 (x + 2) Click - = -2x - 4 Click = -2x + 1 Click Point-Slope Form Slope-Intercept Form

65 Writing Equations of Perpendicular Lines 2. Write the equation of the line passing through the point (-2, ) and perpendicular to the line = 1/2 x + 3 Step 1: Identif the slope according to the given equation. Given equation: m = 1/2 click This example addresses MP1& MP2. Perpendicular Line: m = -2 Step 2: Use the given point and the point-slope formula to write the equation of the perpendicular line. - = -2 (x + 2) Click - = -2x - 4 Click = -2x + 1 Click Math Practice click Point-Slope Form Slope-Intercept Form Slide 122 () / 202 Writing Equations of Perpendicular Lines Slide 123 / Write the equation of the line passing through the point (4, 7) and perpendicular to the line x - = 0 Writing Equations of Perpendicular Lines Slide 123 () / Write Original the equation Equation of the in line slope-intercept passing through the point form: (4, 7) and = perpendicular 1/ x - to the line x - = 0 m = 1/, so Perpendicular m = - & Math Practice - 7 = -(x - 4) point-slope form - 7 = -x + 20 = -x + 27 slope-intercept form This example addresses MP1& MP2.

66 47 What is the equation of the line passing through (6, -2) and parallel to the line whose equation is = 2x - 3? Slide 124 / 202 A = 2x + 2 B = -2x + C = 1/2 x - D = 2x What is the equation of the line passing through (6, -2) and parallel to the line whose equation is = 2x - 3? Slide 124 () / 202 A = 2x + 2 B = -2x + C = 1/2 x - D = 2x - 14 D 48 Which is the equation of a line parallel to the line represented b: = -x - 22? A x - = 22 B - x = 22 C + x = -17 D 2 + x = -22 Slide 12 / 202

67 48 Which is the equation of a line parallel to the line represented b: = -x - 22? A x - = 22 B - x = 22 C + x = -17 D 2 + x = -22 C Slide 12 () / Two lines are represented b the equation: Slide 126 / = 12x - 14 and = kx + 14 For which value of k will the lines be parallel? A 12 B -14 C 3 D Two lines are represented b the equation: Slide 126 () / = 12x - 14 and = kx + 14 For which value of k will the lines be parallel? A 12 B -14 C 3 D -4 D

68 0 Which equation represents a line parallel to the line whose equation is: 3 + 4x = 21 A x = 12 B 3-4x = 22 C 3 = 4x + 21 D 4 + 3x = 21 Slide 127 / Which equation represents a line parallel to the line whose equation is: 3 + 4x = 21 A x = 12 B 3-4x = 22 C 3 = 4x + 21 D 4 + 3x = 21 A Slide 127 () / What is the equation of a line that passes through (9, 3) and is perpendicular to the line whose equation is 4x - = 20? A - 3 = -/4(x - 9) B - 3 = 4/(x - 9) Slide 128 / 202 C - 3 = 4(x - 9) D - 3 = -(x - 9)

69 1 What is the equation of a line that passes through (9, 3) and is perpendicular to the line whose equation is 4x - = 20? A - 3 = -/4(x - 9) B - 3 = 4/(x - 9) Slide 128 () / 202 C - 3 = 4(x - 9) D - 3 = -(x - 9) A 2 What is an equation of the line that passes through point (6, -2) and is parallel to the line whose equation is Slide 129 / 202 = x +? A = x + B = x + 2 C = - x + 2 D = -2x + 2 E = x 2 What is an equation of the line that passes through point (6, -2) and is parallel to the line whose equation is Slide 129 () / 202 = x +? A = x + B = x + 2 C = - x + 2 B D = -2x + 2 E = x

70 3 What is an equation of the line that passes through the point (, -2) and is parallel to the line: 9x - 3 = 12 Slide 130 / 202 A = 3x - 17 B = x C = - x + 17 D = -3x What is an equation of the line that passes through the point (, -2) and is parallel to the line: 9x - 3 = 12 Slide 130 () / 202 A = 3x - 17 B = x C = - x + 17 D = -3x + 1 A 4 What is an equation of the line that contains the point (-4, 1) and is perpendicular to the line whose equation is = -2x - 3? Slide 131 / 202 A = 2x + 1 B = 1/2x + 3 C = -2x - 1 D = - 1/2x + 3

71 4 What is an equation of the line that contains the point (-4, 1) and is perpendicular to the line whose equation is = -2x - 3? Slide 131 () / 202 A = 2x + 1 B = 1/2x + 3 C = -2x - 1 D = - 1/2x + 3 B Two lines are represented b the given equations. What would be the best statement to describe these two lines? Slide 132 / 202 2x + =1 (x + 1) = 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. Two lines are represented b the given equations. What would be the best statement to describe these two lines? Slide 132 () / 202 A B C 2x + =1 (x + 1) = The lines are parallel. The lines are the same line. The lines are perpendicular. D The lines intersect at an angle other than 90. D

72 Slide 133 / 202 Triangle Coordinate Proofs Return to Table of Contents Triangle Coordinate Proof Slide 134 / 202 Coordinate Proofs place figures on the Cartesian Plane to make use of the coordinates of ke 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 ou do some proofs. Triangle Coordinate Proof Slide 134 () / 202 Coordinate Proofs This place lesson figures addresses on the MP1, Cartesian MP2, MP3 Plane to & MP6. make use of the coordinates of ke 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 ou given? something. (MP1) What do ou need to (MP1) Can ou represent an side lengths with The use of the coordinates smbols and is numbers? an extra first (MP2) step in conducting the proof. What Math language will help ou prove our 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] ou do some proofs. Math Practice

73 Triangle Coordinate Proof Slide 13 / 202 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 strateg to accomplish the proof. Example Slide 136 / 202 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 ours? A(0,4) C(-3,0) Q(0,0) B(3,0) If not, take a moment to see if this is correct. Looking at this, our strateg becomes clear. 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. Example Slide 137 / 202 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 the 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 = (2) 1/2 = AC = ((0-(-3)) 2 + (4-0) 2 ) 1/2 = (2) 1/2 =

74 Example Slide 138 / 202 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 4 You should be able to see that these are two right triangles with identical length sides, so the must be congruent. But, let's work out the proof as good practice. C 3 Q 3 B Given: Coordinates of vertices of ABC Coordinates of vertices AQB Prove: QA bisects CAB A 4 Slide 139 / 202 Statements 1. CQ = 3 and BQ = 3 2. AC = and AB = 3. QC QB 4. AQ QA. AC AB 6. ΔAQC ΔAQB 7. CAQ BAQ 8. QA bisects CAB C 3 Q 3 Reasons 1. Given in graph 2. Distance Formula 3. segments have equal measure 4. Reflexive Propert of. segments have equal measure 6. SSS triangle congruence 7. CPCTC 8. Definition of angle bisector B Triangle Coordinate Proof Slide 140 / 202 Given: The points A(4, -1), B(, 6), and C(1, 3) Prove: ABC is an isosceles right triangle Make a sketch and think of a strateg for this proof.

75 Triangle Coordinate Proof B Given: The points A(4, -1), B(, 6), and C(1, 3) Prove: ABC is an isosceles right triangle Slide 141 / 202 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 x Once we do two sides we ma as well do three and then use Pthagorean Theorem to prove it both isosceles and right. Triangle Coordinate Proof Slide 142 / 202 B Let's use the distance formula to find the lengths: d = ((x 2-x 1) 2 + ( 2-1) 2 ) 1/2 AC = ((4-1) 2 +(-1-3) 2 ) 1/2 = (9+16) 1/2 = (2) 1/2 = C BC = ((-1) 2 + (6-3) 2 ) 1/2 = (16+9) 1/2 = (2) 1/2 = AB = ((4-) 2 + (-1-6) 2 ) 1/2 = (1+49) 1/2 + (0) 1/2 = #2 0 x A Triangle Coordinate Proof B Given: The points A(4, -1), B(, 6), and C(1, 3) Prove: ABC is an isosceles right triangle The lengths are AC = ; BC = ; AB = #2 Slide 143 / 202 C Since AC = BC this is an isosceles triangle. Since (#2) 2 = = 0 this is a right triangle 0 A x So, we have proven this to be a right isosceles triangle

76 Triangle Coordinate Proof Slide 144 / 202 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 strateg. 0 A Triangle Coordinate Proof Now we just have to prove that two sides of this triangle are B perpendicular. x C Given: Points A(1, 1), B(4, 4) & C(6, 2) Prove: ABC is a right triangle At a glance, it'd seem like AB and BC are the likel 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 b finding if their slopes are negative (or opposite) reciprocals. Slide 14 / 202 Triangle Coordinate Proof Slide 146 / 202 Given: Points A(1, 1), B(4, 4) & C(6, 2) Prove: ABC is a right triangle B m = Δ 2-1 Δx = x 2-x 1 m AB = x = = = 1 2-x C m BC = x = = = -1 2-x A x Since -1 is the negative (or opposite) reciprocal of 1, AB is perpendicular to BC and ABC is a right triangle

77 Slide 147 / 202 Equation of a Circle & Completing the Square Return to Table of Contents The Circle as a Locus of Points Slide 148 / 202 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." r (4, 8) 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). Slide 149 / 202 (0, 0) Let's use the distance formula to find the lengths: d = ((x 2-x 1) 2 + ( 2-1) 2 ) 1/2 r = ((4-0) 2 + (8-0) 2 ) 1/2 click = (16+64) 1/2 = (80) 1/2 = 4# click

78 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 ou given? is (MP1) located at about (4,8). What do ou need to find? (MP1) Can (0, 0) ou represent an side lengths with smbols and numbers? (MP2) Let's use the distance formula to find the lengths: Slide 149 () / 202 d = ((x 2-x 1) 2 + ( 2-1) 2 ) 1/2 r = ((4-0) 2 + (8-0) 2 ) 1/2 click = (16+64) 1/2 = (80) 1/2 = 4# click (0,0) r (x,) The Circle as a Locus of Points We can also solve in general for an point (x,) 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 ever point on the circumference must satisf this equation. Slide / 202 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 + ( 2-1) 2 click (0,0) r (x,) The Circle as a Locus of Points d 2 = (x 2-x 1) 2 + ( 2-1) 2 Substituting r 2 for d 2 (x, ) for (x 2, 2) (0, 0) for (x 1, 1) What will be resulting equation after the substitution? Slide 11 / 202 r 2 = ((x-0) 2 + (-0) 2 ) = x click The sides are usuall swapped to ield: click x = r 2 The equation of a circle whose center is at the origin.

79 6 What is the radius of the circle whose equation is x = 2? Slide 12 / What is the radius of the circle whose equation is x = 2? Slide 12 () / 202 The radius equal 2 = 7 If the coordinate of a point on the circle x = 2 is, what is the x coordinate? Slide 13 / 202

80 7 If the coordinate of a point on the circle x = 2 is, what is the x coordinate? Slide 13 () / 202 The x-coordinate would be 0. 8 How man points on the circle x = 2 have an x-coordinate of 3? Slide 14 / How man points on the circle x = 2 have an x-coordinate of 3? Substitute 3 in for x and solve for : (3 2 ) + 2 = 2 2 = 2-9 = 16 = ± 4 Slide 14 () / 202 Therefore there are TWO points, (3, 4) and (3, -4)

81 9 How man points on the circle x = 2 have an -coordinate of 6? Slide 1 / How man points on the circle x = 2 have an -coordinate of 6? Slide 1 () / 202 None because it is outside of the radius of. The Circle as a Locus of Points Slide 16 / 202 r In general, an circle centered on the origin will have an equation x = r 2 (0,0) If a point is on the circle, it must satisf this equation. How about circles whose center is not on the origin (0, 0).

82 The Circle as a Locus of Points Slide 17 / 202 r (4,8) If a circle is not centered on the origin, the equation has to be shifted b the amount it is awa from the origin. (0,0) For example, let's shift the center of this circle to (2, 3). (6,11) The Circle as a Locus of Points Slide 18 / 202 (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 coordinate. r (6,11) 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. Slide 19 / 202 (2,3) The new equation will be (x - 2) 2 + ( - 3) 2 = r 2 We can check to see if we still get the same radius.

83 r (6,11) The Circle as a Locus of Points r 2 = (6-2) 2 + (11-3) 2 r 2 = (4) 2 + (8) 2 r 2 = = 80 Slide 160 / 202 (2,3) 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. The Circle as a Locus of Points Slide 161 / 202 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 + ( - k) 2 = r 2 (h,k) Keep in mind that ou are subtracting the x or coordinate of the center of the circle. So, if the center is at (3, ) and the radius is 4, the equation becomes (x - 3) 2 + ( - ) 2 = 16 Example Slide 162 / 202 Write the equation of a circle with center (-2, 3) & radius 3.8.

84 Example Slide 162 () / 202 & Math Practice Write the equation of a circle with center (-2, 3) & radius 3.8. (x - h) 2 + ( - k) 2 = r 2 (x - (-2)) 2 + ( - (3)) 2 = (x + 2) 2 + ( - 3) 2 = This example addresses MP1 & MP2. Additional Q's to address MP standards What information are ou given? (MP1) What do ou need to find? (MP1) Create an equation to represent the circle in this example. (MP2) 60 What is the radius of the circle whose equation is (x - ) 2 + ( - 3) 2 = 36? Slide 163 / What is the radius of the circle whose equation is (x - ) 2 + ( - 3) 2 = 36? Slide 163 () / 202 The radius equals 36 = 6

85 61 What is the radius of the circle whose equation is (x + 3) 2 + ( - 4) 2 = 67? Slide 164 / What is the radius of the circle whose equation is (x + 3) 2 + ( - 4) 2 = 67? Slide 164 () / 202 The radius equals 67 = What is the x-coordinate of the center of the circle whose equation is (x - ) 2 + ( - 3) 2 = 47? Slide 16 / 202

86 62 What is the x-coordinate of the center of the circle whose equation is (x - ) 2 + ( - 3) 2 = 47? Slide 16 () / 202 It is equal to. 63 What is the center and radius of the circle whose equation is (x + 3) 2 + ( - 4) 2 = 30? Slide 166 / What is the center and radius of the circle whose equation is (x + 3) 2 + ( - 4) 2 = 30? Slide 166 () / 202 The center is (-3, 4) and the radius is approximatel.48.

87 64 What is the center and the radius of the circle whose equation is (x - ) 2 + ( - 3) 2 = 7? Slide 167 / What is the center and the radius of the circle whose equation is (x - ) 2 + ( - 3) 2 = 7? Slide 167 () / 202 First, we need to find the center of the circle: (,3) Then we find the radius of the circle b approximating 7 which is What is the center and the radius of the circle whose equation is (x + 3) 2 + ( - 4) 2 = 6.36? Slide 168 / 202

88 6 What is the center and the radius of the circle whose equation is (x + 3) 2 + ( - 4) 2 = 6.36? Slide 168 () / 202 First, we need to find the center of the circle: (-3,4) Then we find the radius of the circle b approximating 6.36 which is Example Slide 169 / 202 The point (-, 6) is on a circle with center (-1, 3). Write the standard equation of the circle. Slide 169 () / 202

89 Example Slide 170 / 202 The equation of a circle is (x - 4) 2 + ( + 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. Example Slide 170 () / 202 The equation of a circle is (x - 4) 2 + ( + 2) 2 = 36. Math Practice Graph the circle. Additional Q's to address MP standards What information are Center: ou given? (4,(MP1) -2) Click to reveal What do ou need to find? (MP1) How do ou graph the Radius circle? = (MP4 36 & = 6 MP) Click to reveal (4, -2) Click on the center of the circle to reveal it in the graph. 66 Which is the standard equation of the circle below? Slide 171 / 202 A B C D x = 400 (x - ) 2 + ( - ) 2 = 400 (x + ) 2 + ( - ) 2 = 400 (x - ) 2 + ( + ) 2 =

90 66 Which is the standard equation of the circle below? A B x = 400 (x - ) 2 + ( - ) 2 = 400 A Slide 171 () / 202 C D (x + ) 2 + ( - ) 2 = 400 (x - ) 2 + ( + ) 2 = Which is the standard equation of the circle? Slide 172 / 202 A (x - 4) 2 + ( - 3) 2 = 81 B (x - 4) 2 + ( - 3) 2 = 9 C (x + 4) 2 + ( + 3) 2 = 81 D (x + 4) 2 + ( + 3) 2 = Which is the standard equation of the circle? Slide 172 () / 202 A (x - 4) 2 + ( - 3) 2 = 81 D B (x - 4) 2 + ( - 3) 2 = 9 C (x + 4) 2 + ( + 3) 2 = 81 D (x + 4) 2 + ( + 3) 2 = 9 [This object is a pull 2tab] 4 6 8

91 68 What is the center of (x - 4) 2 + ( - 2) 2 = 64? Slide 173 / 202 A (0, 0) B (4, 2) C (-4, -2) D (4, -2) 68 What is the center of (x - 4) 2 + ( - 2) 2 = 64? Slide 173 () / 202 A (0, 0) B (4, 2) C (-4, -2) D (4, -2) B 69 What is the radius of (x - 4) 2 + ( - 2) 2 = 89? Slide 174 / 202

92 69 What is the radius of (x - 4) 2 + ( - 2) 2 = 89? Slide 174 () / 202 r = What is the diameter of a circle whose equation is Slide 17 / 202 (x - 2) 2 + ( + 1) 2 = 16? A 2 B 4 C 8 D What is the diameter of a circle whose equation is Slide 17 () / 202 (x - 2) 2 + ( + 1) 2 = 16? A 2 B 4 C 8 D 16 C

93 71 Which point does not lie on the circle described b the equation (x + 2) 2 + ( - 4) 2 = 2? Slide 176 / 202 A (-2, -1) B (1, 8) C (3, 4) D (0, ) 71 Which point does not lie on the circle described b the equation (x + 2) 2 + ( - 4) 2 = 2? Slide 176 () / 202 A (-2, -1) B (1, 8) C (3, 4) D (0, ) D Completing the Square Slide 177 / 202 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 b this equation. x 2-4x = 0

94 Completing the Square Slide 177 () / 202 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 ou and given? the (MP1) radius. What do ou 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? b this equation. (MP2) What do ou know about perfect x 2-4x square = 0 trinomials that can appl to this situation? (MP7) Math Practice Completing the Square Slide 178 / 202 x 2-4x = 0 To find the radius and the coordinates of the center, we need to transform this into the form (x - h) 2 + ( - k) 2 = r 2 The first step is to separate groups of terms which have x, which have, and are constants. Just moving those around makes this equation: [x 2-4x] + 2 = 12 Take a moment to confirm that this is true. Completing the Square Slide 179 / 202 [x 2-4x] + 2 = 12 x 2-4x = 0 We alread see that the -coordinate of the center is 0 (k = 0), since 2 is b 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.

95 Completing the Square Slide 180 / 202 If ou recall, when ou square a binomial, ou 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 b -2 To make the expressions equivalent, subtract h 2 from the binomial Completing the Square [x 2-4x] + 2 = 12 Slide 181 / 202 Dividing the coefficient -4 b -2 ields 2, so h = 2 Then -h 2 = -4 [x 2-4x] + 2 = 12 [(x - 2) 2-4] + 2 = 12 (x - 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. Completing the Square [x 2-4x] + 2 = 12 Slide 181 () / 202 Dividing the coefficient -4 b -2 ields 2, so h = 2 The extra step for Then -h 2 = -4 completing the square is shown in bold below: [x 2-4x] + 2 = 12 [x 2-4x] + 2 = 12 [(x - 2) 2-4] + 2 = 12 [(x (x - 2) = x + 4) - 4] + 2 = 12 Teacher Notes The center is at [(x(2, - 2) 0) 2 and - 4] the + radius 2 = 12is 4. The same steps are used to find k, when needed, as in the next example.

96 Example of Completing the Square Slide 182 / 202 Determine the radius and center of this circle. x x = 0 [x 2-2x] + [ 2 + 6] = -6 [x 2-2x] [ 2 + 6] h = -2/(-2) = 1 k = +6/(-2) = -3 x 2-2x = (x - h) [ 2 + 6] = ( -(-3)) 2 - (3) 2 x 2-2x = (x - 1) 2-1 [ 2 + 6] = ( + 3) 2-9 (x - 1) ( + 3) 2-9 = -6 (x - 1) 2 + ( + 3) 2 = 4 The center is (1,-3) and the radius is 2 Example of Completing the Square Slide 182 () / 202 Determine the radius and center of this circle. This example addresses MP2 & MP7. x x = 0 Additional Q's to address MP standards What information [x 2 are - 2x] ou + given? [ 2 + 6] (MP1) = -6 What do ou need to find? (MP1) [x 2-2x] What do the numbers represent in the [ 2 + 6] problem? (MP2) h = -2/(-2) What = 1 do ou know about perfect square k = +6/(-2) = -3 trinomials that can appl to this x 2-2x = (x situation? - h) (MP7) [ 2 + 6] = ( -(-3)) 2 - (3) 2 Math Practice x 2-2x = (x - 1) 2-1 [ 2 + 6] = ( + 3) 2-9 (x - 1) ( + 3) 2-9 = -6 (x - 1) 2 + ( + 3) 2 = 4 The center is (1,-3) and the radius is 2 72 What is the radius of the circle described b this equation? x x = 0 Slide 183 / 202

97 72 What is the radius of the circle described b this equation? x x = 0 Slide 183 () / What is the x-coordinate of the center of the circle described b this equation? x x = 0 Slide 184 / What is the x-coordinate of the center of the circle described b this equation? x x = 0 Slide 184 () / 202 1

98 74 What is the x-coordinate of the center of the circle described b this equation? x x = 0 Slide 18 / What is the x-coordinate of the center of the circle described b this equation? x x = 0 Slide 18 () / What is the radius of the circle described b this equation? x x = 0 Slide 186 / 202

99 7 What is the radius of the circle described b this equation? x x = 0 Slide 186 () / 202 The radius equal 2 = 76 What is the radius of the circle described b this equation? x x = 0 Slide 187 / What is the radius of the circle described b this equation? x x = 0 Slide 187 () / 202 The radius equal 11 = 3.32

100 77 What is the -coordinate of the center of the circle described b this equation? x x = 0 Slide 188 / What is the -coordinate of the center of the circle described b this equation? x x = 0 Slide 188 () / Part A The equation x x + 2 = b describes a circle. Determine the -coordinate of the center of the circle. Slide 189 / 202 PARCC Released Question (EOY)

101 78 Part A The equation x x + 2 = b describes a circle. Determine the x 2 + -coordinate 2-4x + 2 = bof the center of the circle. [x 2-4x] + [ 2 + 2] = b [(x-2) 2-4] + [(+1) 2-1] = b (x-2) (+1) 2-1 = b (x-2) 2 + (+1) 2 = b + Center (2,-1) Slide 189 () / 202 The -coordinate is -1 PARCC Released Question (EOY) 79 Part B The equation x x + 2 = b describes a circle. The radius of the circle is 7 units. What is the value of b in the equation? Slide 190 / 202 PARCC Released Question (EOY) 79 Part B The equation x x + 2 = b describes a circle. The radius of (x-2) the 2 circle + (+1) is 7 2 = units. b + What is the value of b in the equation? Radius is (b+) 1/2 r = 7 = (b+) 1/2 49 = b + Slide 190 () / 202 b = 44 PARCC Released Question (EOY)

102 80 The equation x 2-8x + 2 = 9 defines a circle in the xcoordinate plane. To find the radius of the circle, the equation can be rewritten as ( ) =. Slide 191 / 202 (Select two answers.) A x + 4 B x - 4 C x + 16 D x - 16 E 2 F 13 G 9 H PARCC Released Question (EOY) 80 The equation x 2-8x + 2 = 9 defines a circle in the xcoordinate plane. To find the radius of the circle, the equation can be rewritten as ( ) =. x 2-8x = 9 A x + 4 E 2 [x 2-8x] + [-0] 2 = 9 (Select two answers.) [(x-4) 2-16] + B 2 = x 9-4 F 13 (x-4) = 2 C x + 16 G 9 Center (4,0) Radius = (2) D x /2 = H b. x - 4 e. 2 Slide 191 () / 202 PARCC Released Question (EOY) Slide 192 / 202 PARCC Sample Questions Return to Table of Contents

103 Question /7 Topic: Partitions of a Line Segment Slide 193 / 202 PARCC Released Question (EOY) Question /7 Topic: Partitions of a Line Segment Slide 193 () / 202 P1 P2 PARCC Released Question (EOY) Question 6/7 Topic: Equation of a Circle Slide 194 / 202 PARCC Released Question (EOY)

104 Question 6/7 Topic: Equation of a Circle Slide 194 () / 202 PARCC Released Question (EOY) Question 7/2 Topic: Equation of a Circle Slide 19 / 202 a. x + 4 b. x - 4 c. x + 16 d. x - 16 e. 2 f. 13 g. 9 h. PARCC Released Question (EOY) Question 7/2 Topic: Equation of a Circle Slide 19 () / 202 a. x + 4 b. x - 4 c. x + 16 d. x - 16 e. 2 f. 13 g. 9 h. b. x - 4 e. 2 PARCC Released Question (EOY)