Geometry. Origin of Analytic Geometry. Slide 1 / 202 Slide 2 / 202. Slide 4 / 202. Slide 3 / 202. Slide 5 / 202. Slide 6 / 202.

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1 Slide 1 / Slide / Geometr Analtic Geometr Slide 3 / Table of Contents 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 Slide 4 / Throughout this unit, the Standards for Mathematical Practice are used. MP1: Making sense of problems & persevere in solving them. MP: 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 & epress regularit 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 alread eist on a slide, then the specific MPs that the questions address are listed in the Pull-tab. Slide / Slide 6 / The Origin of Analtic Geometr Origin of Analtic Geometr 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. Return to Table of Contents

2 Slide 7 / The Origin of Analtic Geometr Slide 8 / The Origin of Analtic Geometr Euclidean Geometr Algebra Was developed in Greece about ears ago. Started b Diophantus in Aleandria about 17 ears ago. Was lost to Europe for more than a thousand ears. Ongoing contributions from Bablon, Sria, Greece and Indians. 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.asp 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 Slide 9 / The Origin of Analtic Geometr Slide / The Origin of Analtic Geometr 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. 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. Slide 11 / Slide 1 / The Origin of Analtic Geometr The Origin of Analtic Geometr Adding this Cartesian coordinate plane makes that task simple since the location of each point can be given b just two numbers: an - and -coordinate, written as the ordered pair (,). - - With the Cartesian Plane providing a numerical description of locations on the plane, geometric figures can be analzed using algebra

3 Slide 13 / Slide 14 / (, ) The Distance Formula The Distance Formula Lab: Derivation of the Distance Formula - ( 1, 1) Let's derive the formula to find the distance between an two points: call the points ( 1, 1) and (, ). First, let's zoom in so we have more room to work. - Return to Table of Contents - Slide 1 / The Distance Formula Slide 16 / The Distance Formula (, ) Now, we'll construct two paths between the points. The first path will be directl between them. (, ) The second path will have a segment parallel the -ais and a segment parallel to the -ais. That will be a big help since it's eas to read the length of each of those from the ais. ( 1, 1) ( 1, 1) Slide 17 / Slide 18 / The Distance Formula The Distance Formula ( 1, 1) c (, ) a b (, 1) We now have a right triangle. For convenience, let's label the two legs "a" and "b" and the hpotenuse "c." Let's also label the point where the two legs meet b its coordinates: (, 1). Take a moment to see that those are the coordinates of that verte of the triangle. ( 1, 1) c (, ) a b (, 1) Did ou get this? c = a + b The net step is to write epressions for the lengths of sides a and b based on the and subscript coordinates. Use the distance along each ais to find those. The coordinates that we just added for the third verte should help. Which formula relates the lengths of the sides of a right triangle?

4 Slide 19 / Slide / The Distance Formula The Distance Formula Did ou get these? We'll use the first pair: ( 1, 1) (, ) c b a (, 1) a = l - 1l AND b = l - 1l We could equall well write a = l 1 - l AND b = l 1 - l 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. Substitute one pair of these into c = a + b ( 1, 1) (, ) c b a (, 1) c = (l - 1l) + (l - 1l) Since the quantities are squared, we don't need to indicate absolute value, the result will alwas be positive anwa. c = ( - 1) + ( - 1) Since "c" is the distance between the points, we can use "d" for distance and sa that d = ( - 1) + ( - 1) If we solve this equation for d, what will be the final formula? Slide 1 / The Distance Formula d = (( - 1) + ( - 1) ) 1/ (, ) OR Slide / 1 What is the distance between the points: (4, 8) and (7, 3)? Round our answer to the nearest hundredth. c b d = ( 1, 1) a (, 1) Slide 3 / What is the distance between the points: (-4, 8) and (7, -3)? Round our answer to the nearest hundredth. Slide 4 / 3 What is the distance between the points: (-, -) and (-7, 3)? Round our answer to the nearest hundredth.

5 Slide / Slide 6 / 4 What is the distance between the indicated points? Round our answer to the nearest hundredth. What is the distance between the indicated points? Round our answer to the nearest hundredth Slide 7 / Slide 8 / 6 What is the distance between the indicated points? Round our answer to the nearest hundredth. The Midpoint Formula - - Lab - Midpoint Formula - Return to Table of Contents Slide 9 / Slide 3 / The Midpoint Formula The Midpoint Formula 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 ais, so let's add those lines. The -coordinate for the midpoint between ( 1, 1) and (, ) is halfwa between 1 and. Similarl, the -coordinate of that midpoint will be halfwa between 1 and. If ou're provided a graph of a line, and asked to mark the midpoint, ou can often do that without much calculating.

6 Slide 31 / The Midpoint Formula Slide 3 / The Midpoint Formula In this graph, 1 = 1 and = 9. The are 8 units apart, so just go 4 units along the -ais and go up until ou intersect the line. 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 - ais, as we do on the net slide. The -coordinates of the two points are 11 and 3, so the are also 8 units apart. 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. Slide 33 / The Midpoint Formula Slide 34 / The Midpoint Formula (1,3) (9,11) We can calculate the -coordinate midwa between that of the two given points b finding their average. The same for the -coordinate. Just add the two values and divide b two. The -coordinate of the midpoint is (1+9)/ = The -coordinate of the midpoint is (3+11)/ = 7 That's the same answer: (,7) ( 1, 1) (, ) The more general solution is given below for an points: ( 1, 1) and (, ). The -coordinate of the midpoint is given b M = ( 1 + ) / The -coordinate of the midpoint is given b M = ( 1 + ) / So, the coordinates of the midpoint are: Slide 3 / Slide 36 / 7 What is the midpoint between the indicated points? 8 What is the midpoint between the indicated points? A (4, 9) A (, ) B (-, -4) - B (, ) - C (, 6) D (, 7) - C (, ) D (, ) - - -

7 9 What is the midpoint between the indicated points? Slide 37 / Slide 38 / What is the midpoint between the points: (4, 8) and (7, 3)? A (8, ) B (4, 7) A (3, 3) C (.,.) B (3, 4) - D (6, ) C (4, 3) - D (, 3) - Slide 39 / 11 What is the midpoint between the points: (-4, 8) and (4, -8)? A (8, ) B (, ) C (1, 1) D (4, 4) Slide 4 / 1 What is the midpoint between the points: (-4, -8) and (-6, -4)? A (-, -6) B (-, -1) C (, 4) D (, 6) Slide 41 / Finding the Coordinates of an E ndpoint of a Segment The midpoint of AB is M(1, ). One endpoint is A(4, 6). Find the coordinates of the other endpoint B(, ). You can use the midpoint formula to write equations using and. Then, set up equations and solve each one. B(, ) M(1, ) = A(4, 6) ( ) 4 +, 6 + M(1, ) = 1 = 4 + = = ( ) 4 +, 6 + = = = Slide 4 / Finding the Coordinates of an E ndpoint of a Segment Therefore, the coordinates of B(-, -). B(, ) M(1, ) = A(4, 6) ( ) 4 +, 6 + Can ou find a shortcut to solve this problem? How would our shortcut make the problem easier?

8 Slide 43 / Finding the Coordinates of an E ndpoint of a Segment Slide 44 / Finding the Coordinates of an E ndpoint of a Segment Another wa of approaching this problem is to look for the pattern that occurs between the endpoint A (4, 6) and midpoint M (1, ). 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(-, -). down 4 down 4 left 3 B(, ) left 3 M(1, ) A(4, 6) 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. -3 A (4, 6) M (1, ) -3 B (-, -) -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. Slide 4 / 13 Find the other endpoint of the segment with the endpoint (7, ) and midpoint (3, ) A (-1, -) B (-, -1) C (4, ) D (, 4) Slide 46 / 14 Find the other endpoint of the segment with the endpoint (1, 4) and midpoint (, -) A (11, -8) B (9, ) C (9, -8) D (3, 1) Slide 47 / 1 Find the other endpoint of the segment with the endpoint (-4, -1) and midpoint (-, 3). A (-6, -) B (-3, -) C (, 7) D (1, 9) Slide 48 / 16 Find the other endpoint of the segment with the endpoint (-, ) and midpoint (, ). A (-1, -3.) B (-4, 8) C (1,.) D (, -1)

9 Slide 49 / Slide / Partitions of a Line Segment Partitions of a Line Segment Return to Table of Contents A B Partitioning a line segment simpl means to divide the segment into or more parts, based on a given ratio. The midpoint partitions a segment into congruent segments, forming a ratio of 1:1. But if ou need to partition a segment so that its ratio is something different, for eample 3:1, how can it be done? A Slide 1 / 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 Slide / Partitions of a Line Segment 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 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 A Slide 3 / Partitions of a Line Segment 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 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) Slide 4 / We can also calculate both the 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, 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

10 Slide / Partitions of a Line Segment Slide 6 / Partitions of a Line Segment 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) (, ) The more general solution is given below for an points: ( 1, 1) and (, ). If a point P partitions a segment in a ratio of m:n, then the coordinates of P are P = (, ) m + n1 m + n m + n1 m + n A (1,3) ( 1, 1) Remember to also use the formula twice switching the m:n ratio to n:m. Slide 7 / Partitions of a Line Segment - C (-, ) (, 7) Eample with a graph: Line segment CD in the coordinate plane has endpoints with coordinates (-, ) 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 :3. Slide 9 / Partitions of a Line Segment (, 4) Step #: Since the ratio is :3, determine which points to eliminate (3 of them) leaving onl 1 point remaining so that, from left to right, the ratio is :3. C (-, ) Slide 8 / Partitions of a Line Segment (6, -) - C (-, ) (, 7) (, 7) (, 4) (4, 1) D (8, -) Step #1: Graph the points. C(-, ) and D(8, -) - Click on the points to show. Step #: 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 : = Click Step #4: Plot the points on the graph that divide CD into the desired number of sections. - Click on the points to show. Slide 6 / Partitions of a Line Segment (, 4) Step #6: Because the question asks for points, we also need to consider the ratio 3:. 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:. (4, 1) (6, -) (4, 1) (6, -) - D (8, -) - D (8, -)

11 Slide 61 / Partitions of a Line Segment Eample 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 :. Slide 6 / Partitions of a Line Segment Step #1: Determine the total number of was required to divide the segment using the ratio :. + = 7 Click Step #: Use the formula to determine the coordinates of our first point P. click P = P = (, -11) and (-4, ) m + n1 ( m + n1 m + n, m + n ) (, ) P = (-4) + () () + (-11) (-) 7 7 P = (, 4) Slide 63 / Slide 64 / Partitions of a Line Segment Step #3: Reverse the ratio : and reuse the formula to determine the coordinates of our second point P. click P = P = (, -11) and (-4, ) m + n1 ( m + n1 m + n, m + n ) (, ) P = (-4) + () () + (-11) (-) 7 7 P = (6, ) 17 Line segment GH in the coordinate plane has endpoints with coordinates (-7, -9) & (8, 3), shown in the graph below. Find possible locations for a point P that divides GH into two parts with lengths in a ratio of :1. A (., -3) B (3, -1) C (6, 1) D (-, -) E (-, -7) - G (-7, -9) - - H(8, 3) Slide 6 / Slide 66 / 18 Line segment JK in the coordinate plane has endpoints with coordinates (, 1) & (-, -13), shown in the graph below. Find possible locations for a point P that divides JK into two parts with lengths in a ratio of 4:1. A (-6, -8) B (-, -3) C (, -.) D (, ) E (6, 7) K(-, -13) J(, 1) 19 Line segment LM in L(-8, 13) the coordinate plane has endpoints with coordinates (-8, 13) & (6, -8), shown in the graph below. Find 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 (-, 4) D (, 1) E (, -3) M(6, -8)

12 Slide 67 / Line segment LM in the coordinate plane has endpoints with coordinates (-8, 13) & (6, -8), shown in the graph below. Find possible locations for a point P that divides LM into two parts with lengths in a ratio of 6:1. Slide 68 / 1 Line segment NO in the coordinate plane has endpoints with coordinates (, 1) & (-, -13), shown in the graph below. Find possible locations for a point P that divides NO into two parts with lengths in a ratio of 3:. A (-6, ) B (-4, 7) C (-, 4) D (, 1) E (4, -6) A (-6, -8) B (-, -3) C (, -.) D (, ) E (6, 7) Slide 69 / Line segment QR in the coordinate plane has endpoints with coordinates (-1, 11) & (1, -13), shown in the graph below. Find possible locations for a point P that divides QR into two parts with lengths in a ratio of :3. Slide 7 / 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. A (3, -4) B (, -1) C (-3, ) D (-6, ) E (-9, 8) 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) Slide 71 / 3 Find two possible locations for point M so that M divides JK into two parts with lengths in a ratio of 1:3. A (-, 9) B (-1, 8) C (, 7) D (4, 3) J (-4, 11) Slide 7 / 4 Point Q lie on ST, where point S is located at (-, -6) and point T is located at (, 8). If SQ:QT = :, where is point Q on ST? A (-1, -4) B (, -) C (1, ) E (, ) F (6, 1) K (8, -1) D (, ) E (3, 4) F (4, 6) PARCC Released Question (EOY) - Part - Response Format PARCC Released Question (PBA)

13 Slide 73 / Slopes of Parallel & Perpendicular Lines Slide 74 / Slope The slope of a line indicates the angle it makes with the -ais. The smbol for slope is "m". Return to Table of Contents Slide 7 / Slope 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. Slide 76 / The slope of the indicated line is: A negative B positive C zero D undefined A line which falls as ou move from left to right has a negative slope. Slide 77 / 6 The slope of the indicated line is: Slide 78 / 7 The slope of the indicated line is: A negative A negative B positive B positive C zero C zero D undefined D undefined

14 Slide 79 / 8 The slope of the indicated line is: Slide 8 / 9 The slope of the indicated line is: A negative A negative B positive B positive C zero C zero D undefined D undefined Slide 81 / 3 The slope of the indicated line is: A negative B positive C zero D undefined Slide 8 / Slope 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. Slide 83 / Slide 84 / Slope Slope ( 1, ) "rise" ( 1, 1) "run" (, ) The "rise" is the change in the value of the -coordinate while the "run" is the change in the value of the -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, or divided b...delta over delta. (, 11) "run" "rise" (,4) (8,11) m = Δ -1 Δ = - 1 In this case: The rise is from 4 to 11 = 11-4 = 7 And the run is from to 8, = 8 - = 6 m = Δ -1 Δ = - 1 So the slope is m = = 7 6

15 Slide 8 / Slope Slide 86 / Using Slope to Draw a Line ( 1, 1) (, ) An points on the line can be used to calculate its slope, since the slope of a line is the same everwhere. The values of and 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 -1 Δ = - 1 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, I can find a second point, and then draw the line. m = Δ -1 Δ = - 1 Slide 87 / Using Slope to Draw a Line Slide 88 / Using Slope to Draw a Line I do this b recognizing that the slope of means that if I go up units on the -ais I have to go 1 unit to the right on the -ais. I do this b recognizing that the slope of means that if I go up units on the -ais I have to go 1 unit to the right on the -ais. Or if I go up, I have to go over units, etc. m = Δ -1 Δ = - 1 m = Δ -1 Δ = - 1 Slide 89 / Using Slope to Draw a Line Slide 9 / Slopes of Parallel Lines 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. 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 -ais. m = Δ -1 Δ = - 1 Now, think of the -ais as a transversal.

16 Slide 91 / 31 What is the name of this pair of angles, formed b a transversal intersecting two lines. A Alternate Interior Angles B Corresponding Angles C Alternate Eterior Angles D Same Side Interior Angles Slide 9 / 3 We know those angles must then be: A Supplementar C Equal B Complementar D Adjacent 1 1 Slide 93 / 33 That means that the slopes of parallel lines must be: A Reciprocals C Equal B Inverses D Nothing special Slide 94 / 34 If one line has a slope of 4, what must be the slope of an line parallel to it? 1 Slide 9 / 3 If one line passes through the points (, ) and (, ) what must be the slope of an line parallel to that first line? Slide 96 / 36 If one line passes through the points (-, 9) and (, 8) what must be the slope of an line parallel to that first line?

17 Slide 97 / 37 If one line passes through the points (, ) and (, ) and a parallel line passes through the point (1, ) which of these points could lie on that second line? A (, ) B (4, 4) C (, 6) D (-1, 3) Slide 98 / 38 If one line passes through the points (-3, 4) and (, ) and a parallel line passes through the point (-1, -4) which of these points could lie on that second line? A (, -) B (, -) C (3, 1) D (-1, 3) Slide 99 / 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? A (, 1) B (-, ) C (4, 8) D (-4, -4) Slide / Slopes of Perpendicular Lines The slopes of perpendicular lines are negative (or opposite) reciprocals. There are three was of epressing this smbolicall: m 1m = -1 m 1 = m = -1 m -1 m 1 These all have the same meaning. Slide 1 / Slide / Slopes of Perpendicular Lines The slopes of perpendicular lines are negative (or opposite) reciprocals. 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 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.

18 Slide 3 / Slopes of Perpendicular Lines Slide 4 / Slopes of Perpendicular Lines 1 1 (1,m 1) Now, let's zoom in and focus on the lines between = and = 1. When the lines are at = 1, their -coordinates are m 1 for the first line and m for the second line. That's because if the slope of the first line is m 1, then when we move +1 along the -ais, the -value must increase b the amount of the slope, m 1. 1 b a 1 (1,m 1) c Perpendicular lines obe the Pthagorean Theorem: c = a + b Let's find epressions for those three terms so we can substitute them in. Side "c" is hpotenuse and is the distance between the points along the vertical line =1. It has length m 1-m (in this case, that effectivel adds their magnitudes since m is negative). -1 (1,m ) The same for the second line, whose slope is m. -1 (1,m ) We can use the distance formula to find the length of each leg. Slide / Slopes of Perpendicular Lines Slide 6 / Slopes of Perpendicular Lines 1 b a 1 (1,m 1) c The lengths of legs "a" and "b" can be found using the distance formula. d = ( - 1) + ( - 1) a = (1-) + (m 1-) = 1 + m 1 b = (1-) + (m -) = 1 + m And, we get c b squaring our result for c from the prior slide. 1 b a 1 (1,m 1) c We're going to substitute the epressions from the prior slide into the Pthagorean Theorem. We'll color code them so we can keep track. c = m 1 - m 1m 1 + m a = 1 + m 1 b = 1 + m c = a + b -1 (1,m ) c = (m 1-m ) = m 1 - m 1m + m -1 (1,m ) m 1 - m 1m + m = 1 + m m 1 Slide 7 / Slopes of Perpendicular Lines a 1 (1,m 1) c m 1 - m 1m + m = 1 + m m Now, let's identif like terms m 1 - m 1m + m = 1 + m m Notice that we have m 1 and m on both sides so the can be canceled, and that =. So, -m 1m = m 1m = -1 Slide 8 / 4 If one line has a slope of 4, what must be the slope of an line perpendicular to it? b which is what we set out to prove Notice that this also be written as: -1 (1,m ) m 1 = m = -1 m -1 m 1

19 Slide 9 / 41 If one line has a slope of -1/, what must be the slope of an line perpendicular to it? Slide 1 / 4 If one line passes through the points (, ) and (4, ) what must be the slope of an line perpendicular to that first line? Slide 111 / 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 11 / 44 If one line passes through the points (1, ) and (, 6) and a perpendicular line passes through the point (1, ) which of these points could lie on that second line? A (, ) B (4, 4) C (, 4) D (-1, 3) Slide 113 / 4 If one line passes through the points (-3, 4) and (, ) and a perpendicular line passes through the point (-1, -4) which of these points could lie on that second line? A (, -) B (, -) C (3, 1) D (-3, -) Slide 114 / 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? A (, ) B (, ) C (4, 8) D (-4, -4)

20 Slide 11 / Equations of Parallel & Perpendicular Lines Return to Table of Contents Slide 116 / Writing the Equation of a Line m = -1-1 Let's start with the above definition of slope, multipl both sides b ( - 1), and rearrange to get: ( - 1) = m( - 1) Now, if I enter one point ( 1, 1) this defines the infinite locus of other points on the line. - 1 = m(- 1) Point-slope form Then, I can add 1 to both sides to isolate the variable. = m(- 1)+ 1 Slide 117 / Writing the Equation of a Line Slide 118 / Slope Intercept Equation of a Line = m(- 1)+ 1 = m(- 1)+ 1 The term in red is just our steps to the right along the -ais, our "run". A common simplification of this is to use the -intercept for ( 1, 1). Multipling b slope tells ou how man steps up ou must take along the -ais, our "rise". Those steps are added to the term in green, our original position on the -ais to find our final -coordinate. That tells ou the -coordinate on the line for the given - coordinate. The -intercept is the point where the line crosses the - ais. Its smbol is "b" so the coordinates of that point are (,b), since = on the -ais. Just substitute (,b) in the above equation for ( 1, 1). Slide 119 / Slope Intercept Equation of a Line = m(- 1)+ 1 = m(-) + b = m + b This is ver useful since both the slope and the -intercepts often have important meaning when we are solving real problems. In the graph to the left, b=, so the equation is =. Slide 1 / Writing Equations of Parallel Lines 1. Find an equation of the line passing through the point (-4, ) and parallel to the line whose equation is -3 + = -1. Step 1: Identif the information given in the problem. Contains the point (-4, ) & is parallel to the line -3 + = -1 Step : 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 -3 + = -1 Click Click = 3-1 = 3/ - 1/ Therefore m = 3/ click

21 Step 3: Create the equation. Slide 11 / Writing Equations of Parallel Lines - 1= m( - 1) - = 3/ ( +4) Click - = 3/ + 6 Click = 3/ + 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. Slide 1 / Writing Equations of Perpendicular Lines. Write the equation of the line passing through the point (-, ) and perpendicular to the line = 1/ + 3 Step 1: Identif the slope according to the given equation. Given equation: m = 1/ Perpendicular Line: m = - Step : Use the given point and the point-slope formula to write the equation of the perpendicular line. - = - ( + ) Click - = Click = Click click click Point-Slope Form Slope-Intercept Form Slide 13 / Writing Equations of Perpendicular Lines 3. Write the equation of the line passing through the point (4, 7) and perpendicular to the line - = Slide 14 / 47 What is the equation of the line passing through (6, -) and parallel to the line whose equation is = - 3? A = + B = - + C = 1/ - D = - 14 Slide 1 / 48 Which is the equation of a line parallel to the line represented b: = - -? A - = B - = C + = -17 D + = - Slide 16 / 49 Two lines are represented b the equation: -3 = 1-14 and = k + 14 For which value of k will the lines be parallel? A 1 B -14 C 3 D -4

22 Slide 17 / Which equation represents a line parallel to the line whose equation is: = 1 A = 1 B 3-4 = C 3 = D = 1 Slide 18 / 1 What is the equation of a line that passes through (9, 3) and is perpendicular to the line whose equation is 4 - =? A - 3 = -/4( - 9) B - 3 = 4/( - 9) C - 3 = 4( - 9) D - 3 = -( - 9) Slide 19 / Slide 13 / What is an equation of the line that passes through point (6, -) and is parallel to the line whose equation is A = + B = + C = - + = +? 3 What is an equation of the line that passes through the point (, -) and is parallel to the line: 9-3 = 1 A = 3-17 B = C = D = D = - + E = Slide 131 / Slide 13 / 4 What is an equation of the line that contains the point (-4, 1) and is perpendicular to the line whose equation is = - - 3? A = + 1 B = 1/ + 3 C = D = - 1/ + 3 Two lines are represented b the given equations. What would be the best statement to describe these two lines? A B C + =1 ( + 1) = - + The lines are parallel. The lines are the same line. The lines are perpendicular. D The lines intersect at an angle other than 9.

23 Slide 133 / Slide 134 / Triangle Coordinate Proof Triangle Coordinate Proofs 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 etra first step in conducting the proof. We'll provide a few eamples and then have ou do some proofs. Return to Table of Contents Slide 13 / Triangle Coordinate Proof Given: The coordinates A(, 4); B(3, ); C(-3, ) and Q(, ) 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. A(,4) C(-3,) Q(,) B(3,) Slide 136 / Eample Given: The coordinates A(, 4); B(3, ); C(-3, ) and Q(, ) are the vertices of ABC and AQB Prove: QA bisects CAB Does this sketch look like ours? 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. A(,4) C(-3,) Q(,) B(3,) Slide 137 / Eample Given: The coordinates A(, 4); B(3, ); C(-3, ) and Q(, ) are the vertices of ABC and AQB Prove: QA bisects CAB 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: AB = ((3-) + (-4) ) 1/ = () 1/ = AC = ((-(-3)) + (4-) ) 1/ = () 1/ = Slide 138 / Eample Given: The coordinates A(, 4); B(3, ); C(-3, ) and Q(, ) are the vertices of ABC and AQB Prove: QA bisects CAB C 3 A Q 4 3 B 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.

24 Slide 139 / Slide 14 / Given: Coordinates of vertices of ABC Coordinates of vertices AQB A Triangle Coordinate Proof Prove: QA bisects CAB 4 Given: The points A(4, -1), B(, 6), and C(1, 3) Prove: ABC is an isosceles right triangle Statements Reasons C 3 Q 3 B Make a sketch and think of a strateg for this proof. 1. CQ = 3 and BQ = 3 1. Given in graph. AC = and AB =. Distance Formula 3. QC QB 3. segments have equal measure 4. AQ QA 4. Refleive Propert of. AC AB. segments have equal measure 6. ΔAQC ΔAQB 6. SSS triangle congruence 7. CAQ BAQ 7. CPCTC 8. QA bisects CAB 8. Definition of angle bisector Slide 141 / Slide 14 / Triangle Coordinate Proof Triangle Coordinate Proof B Given: The points A(4, -1), B(, 6), and C(1, 3) Prove: ABC is an isosceles right triangle B Let's use the distance formula to find the lengths: d = (( - 1) + ( - 1) ) 1/ C A 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. Once we do two sides we ma as well do three and then use Pthagorean Theorem to prove it both isosceles and right. C A AC = ((4-1) +(-1-3) ) 1/ = (9+16) 1/ = () 1/ = BC = ((-1) + (6-3) ) 1/ = (16+9) 1/ = () 1/ = AB = ((4-) + (-1-6) ) 1/ = (1+49) 1/ + () 1/ = # Slide 143 / 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 = # Slide 144 / Triangle Coordinate Proof Given: The points A(1, 1), B(4, 4), and C(6, ) Prove: ABC is a right triangle Make a sketch and think of a strateg. C Since AC = BC this is an isosceles triangle. Since (#) = + = this is a right triangle A So, we have proven this to be a right isosceles triangle

25 Slide 14 / Slide 146 / Triangle Coordinate Proof Triangle Coordinate Proof Given: Points A(1, 1), B(4, 4) & C(6, ) Prove: ABC is a right triangle Given: Points A(1, 1), B(4, 4) & C(6, ) Prove: ABC is a right triangle Now we just have to prove that two sides of this triangle are B perpendicular. A C 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. A B C m = Δ -1 Δ = - 1 m AB = m BC = = = = = = = Since -1 is the negative (or opposite) reciprocal of 1, AB is perpendicular to BC and ABC is a right triangle Slide 147 / Slide 148 / The Circle as a Locus of Points Equation of a Circle & Completing the Square 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." Return to Table of Contents r Slide 149 / (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 (,) this particular point is located at about (4,8). (, ) Let's use the distance formula to find the lengths: (,) r Slide / The Circle as a Locus of Points We can also solve in general for (,) an point (,) which lies on the circumference of the circle whose center is located at the origin (,). 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. d = (( - 1) + ( - 1) ) 1/ r = ((4-) + (8-) ) 1/ click = (16+64) 1/ = (8) 1/ = 4# click If we start with the distance formula d = and square both sides, what will be the resulting equation? d = ( - 1) + ( - 1) click

26 (,) r Slide 11 / The Circle as a Locus of Points (,) d = ( - 1) + ( - 1) Substituting r for d (, ) for (, ) (, ) for ( 1, 1) What will be resulting equation after the substitution? Slide 1 / 6 What is the radius of the circle whose equation is + =? r = ((-) + (-) ) = + click The sides are usuall swapped to ield: click + = r The equation of a circle whose center is at the origin. Slide 13 / Slide 14 / 7 If the coordinate of a point on the circle + = is, what is the coordinate? 8 How man points on the circle + = have an -coordinate of 3? Slide 1 / 9 How man points on the circle + = have an -coordinate of 6? Slide 16 / The Circle as a Locus of Points r In general, an circle centered on the origin will have an equation + = r (,) If a point is on the circle, it must satisf this equation. How about circles whose center is not on the origin (, ).

27 Slide 17 / Slide 18 / (,) r (4,8) The Circle as a Locus of Points If a circle is not centered on the origin, the equation has to be shifted b the amount it is awa from the origin. For eample, let's shift the center of this circle to (, 3). (,3) r (6,11) The Circle as a Locus of Points Shifting the center of this circle from (, ) to (, 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 and up 3 will add that amount to each and coordinate. Slide 19 / Slide 16 / (,3) 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. The new equation will be ( - ) + ( - 3) = r We can check to see if we still get the same radius. (,3) r (6,11) The Circle as a Locus of Points r = (6 - ) + (11-3) r = (4) + (8) r = = 8 These are the same values we had before when the circle was centered on (, ) and that point was located at (4, 8). So, this translation of the center did not change the circle or its radius. r Slide 161 / The Circle as a Locus of Points In general, if the center of a circle is located at (h, k) and its radius is r, the equation for the circle is Slide 16 / Eample Write the equation of a circle with center (-, 3) & radius 3.8. (h,k) ( - h) + ( - k) = r Keep in mind that ou are subtracting the or coordinate of the center of the circle. So, if the center is at (3, ) and the radius is 4, the equation becomes ( - 3) + ( - ) = 16

28 Slide 163 / 6 What is the radius of the circle whose equation is ( - ) + ( - 3) = 36? Slide 164 / 61 What is the radius of the circle whose equation is ( + 3) + ( - 4) = 67? Slide 16 / 6 What is the -coordinate of the center of the circle whose equation is ( - ) + ( - 3) = 47? Slide 166 / 63 What is the center and radius of the circle whose equation is ( + 3) + ( - 4) = 3? Slide 167 / 64 What is the center and the radius of the circle whose equation is ( - ) + ( - 3) = 7? Slide 168 / 6 What is the center and the radius of the circle whose equation is ( + 3) + ( - 4) = 6.36?

29 Slide 169 / Eample The point (-, 6) is on a circle with center (-1, 3). Write the standard equation of the circle. Slide 17 / Eample The equation of a circle is ( - 4) + ( + ) = 36. Graph the circle. Center: (4, -) Click to reveal Radius = 36 = 6 Click to reveal (4, -) Click on the center of the circle to reveal it in the graph. Slide 171 / Slide 17 / 66 Which is the standard equation of the circle below? 67 Which is the standard equation of the circle? A + = 4 A ( - 4) + ( - 3) = 81 B C D ( - ) + ( - ) = 4 ( + ) + ( - ) = 4 ( - ) + ( + ) = 4 1 B ( - 4) + ( - 3) = 9 C ( + 4) + ( + 3) = 81 D ( + 4) + ( + 3) = Slide 173 / Slide 174 / 68 What is the center of ( - 4) + ( - ) = 64? A (, ) B (4, ) C (-4, -) D (4, -) 69 What is the radius of ( - 4) + ( - ) = 89?

30 Slide 17 / 7 What is the diameter of a circle whose equation is ( - ) + ( + 1) = 16? Slide 176 / 71 Which point does not lie on the circle described b the equation ( + ) + ( - 4) =? A (-, -1) A B 4 C 8 D 16 B (1, 8) C (3, 4) D (, ) Slide 177 / 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 b this equation = Slide 178 / Completing the Square = To find the radius and the coordinates of the center, we need to transform this into the form ( - h) + ( - k) = r The first step is to separate groups of terms which have, which have, and are constants. Just moving those around makes this equation: [ - 4] + = 1 Take a moment to confirm that this is true. [ - 4] + = 1 Slide 179 / Completing the Square = We alread see that the -coordinate of the center is (k = ), since is b itself. But what to do with the epression ( - 4)? We have to convert that into the form ( - h) to find the - coordinate of the center...and then the radius. Slide 18 / Completing the Square If ou recall, when ou square a binomial, ou get a trinomial. (-h) = - h + h Our problem starts with an epression in the form of - h, so let's solve for that so we can see what can replace it: - h = (-h) - h So The coefficient (-h) of is -h. The constant of the trinomial (-h ) is -(h). So, to get h, divide the coefficient of b - To make the epressions equivalent, subtract h from the binomial

31 Slide 181 / Completing the Square [ - 4] + = 1 Dividing the coefficient -4 b - ields, so h = Then -h = -4 [ - 4] + = 1 [( - ) - 4] + = 1 ( - ) + = 16 The center is at (, ) and the radius is 4. The same steps are used to find k, when needed, as in the net eample. Slide 183 / 7 What is the radius of the circle described b this equation? = Slide 18 / Eample of Completing the Square Determine the radius and center of this circle = [ - ] + [ + 6] = -6 [ - ] [ + 6] h = -/(-) = 1 k = +6/(-) = -3 - = ( - h) - 1 [ + 6] = ( -(-3)) - (3) - = ( - 1) - 1 [ + 6] = ( + 3) - 9 ( - 1) ( + 3) - 9 = -6 ( - 1) + ( + 3) = 4 The center is (1,-3) and the radius is Slide 184 / 73 What is the -coordinate of the center of the circle described b this equation? = Slide 18 / 74 What is the -coordinate of the center of the circle described b this equation? = Slide 186 / 7 What is the radius of the circle described b this equation? =

32 Slide 187 / 76 What is the radius of the circle described b this equation? = Slide 188 / 77 What is the -coordinate of the center of the circle described b this equation? = Slide 189 / 78 Part A The equation = b describes a circle. Determine the -coordinate of the center of the circle. 79 Part B Slide 19 / The equation = 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) PARCC Released Question (EOY) Slide 191 / Slide 19 / 8 The equation = 9 defines a circle in the coordinate plane. To find the radius of the circle, the equation can be rewritten as ( ) + =. (Select two answers.) A + 4 B - 4 C + 16 E F 13 G 9 PARCC Sample Questions D - 16 H PARCC Released Question (EOY) Return to Table of Contents

33 Slide 193 / Slide 194 / Question /7 Topic: Partitions of a Line Segment Question 6/7 Topic: Equation of a Circle PARCC Released Question (EOY) Slide 19 / PARCC Released Question (EOY) Slide 196 / Question 7/ Topic: Equation of a Circle Question /7 Topic: Partitions of a Line Segment a. + 4 b. - 4 c d e. f. 13 g. 9 h. PARCC Released Question (EOY) PARCC Released Question (PBA) Slide 197 / Slide 198 / Write the equation of a line, in slope-intercept form, which has a point of tangenc at (3,6) with a circle whose center is at the origin. General Problems Return to Table of Contents

34 Slide 199 / Write the equation of a line, in slope-intercept form, which has a point of tangenc at (3,6) with a circle whose center is at the origin. Slide / Write an equation of a line which has a point of tangenc at (3,6) with a circle whose center is at the origin. Strateg The slope of the radius of that circle to that point can be determined. Then the slope of the line tangent at that point will be the negative reciprocal of the slope of the radius since the are perpendicular. Given a point and the slope, the equation of the line can be written. Solution m radius = (6-)/(3-) = m tangent = -1/ (- 1) = m(- 1) (-6) = (1/)(-3) = = b =.() + 4. b = 4. Slide 1 / 81 What is the slope of a line tangent at (7,) to a circle whose center is at (,3)? Slide / 8 What is the -intercept of the line in the prior problem which was tangent at (7,) to a circle whose center is at (,3)?