Analytic Geometry - Content.njctl
Slide 1 / 202Slide 2 / 202GeometryAnalytic Geometry2015-10-02www.njctl.orgSlide 3 / 202Table of ContentsOrigin of Analytic GeometryThe Distance FormulaThe Midpoint FormulaPartitions of a Line SegmentSlopes of Parallel & Perpendicular LinesEquations of Parallel & Perpendicular LinesTriangle Coordinate ProofsEquation of a Circle & Completing the SquarePARCC Sample QuestionsGeneral Problemsclick on the topic to goto that section
Throughout this unit, the Standards for Mathematical Practiceare used.Slide 4 / 202MP1: Making sense of problems & persevere in solving them.MP2: Reason abstractly & quantitatively.MP3: Construct viable arguments and critique the reasoning ofothers.MP4: Model with mathematics.MP5: Use appropriate tools strategically.MP6: Attend to precision.MP7: Look for & make use of structure.MP8: Look for & express regularity in repeated reasoning.Additional questions are included on the slides using the "MathPractice" Pull-tabs (e.g. a blank one is shown to the right onthis slide) with a reference to the standards used.If questions already exist on a slide, then the specific MPs thatthe questions address are listed in the Pull-tab.Slide 5 / 202Origin of AnalyticGeometryReturn to Tableof ContentsThe Origin of Analytic GeometryAnalytic Geometry is a powerful combination ofgeometry and algebra.Many jobs that are looking for employees now, and willbe in the future, rely on the process or results ofanalytic geometry.This includes jobs in medicine, veterinary science,biology, chemistry, physics, mathematics, engineering,financial analysis, economics, technology,biotechnology, etc.Slide 6 / 202
The Origin of Analytic GeometrySlide 7 / 202Euclidean Geometry· Was developed in Greece about2500 years ago.· Was lost to Europe for more than athousand years.· Was maintained and refined duringthat time in the Islamic world.· Its rediscovery was a critical partof the European oismilliet-5541389-details.aspxThe Origin of Analytic GeometrySlide 8 / 202Algebra· Started by Diophantus in Alexandriaabout 1700 years ago.· Ongoing contributions fromBabylon, Syria, Greece and Indians.· Named from the Arabic word al-jabrwhich was used by al-Khwarizmi inthe title of his 7th century book.http://en.wikipedia.org/wiki/Mathematics in medieval IslamThe Origin of Analytic GeometryAnalytic Geometry· A powerful combination ofalgebra and geometry.· Independently developed, andpublished in 1637, by ReneDescartes and Pierre deFermat in France.· The Cartesian Plane is namedfor Descartes.Slide 9 / 202
The Origin of Analytic GeometrySlide 10 / 202How would youdescribe to someonethe location of thesefive points so theycould draw them onanother piece ofpaper without seeingyour drawing?Discuss.Slide 11 / 202The Origin ofAnalytic GeometryAdding thisCartesiancoordinate planemakes that tasksimple since thelocation of eachpoint can be givenby just twonumbers: an xand y-coordinate,written as theordered pair (x,y).1050-55-5-10Slide 12 / 202The Origin ofAnalytic GeometryWith the CartesianPlane providing anumerical descriptionof locations on theplane, geometricfigures can beanalyzed usingalgebra.105-50-5-105
Slide 13 / 202The Distance FormulaLab: Derivation of theDistance FormulaReturn to Tableof ContentsyThe DistanceFormula(x2, y2)10Let's derive the formulato find the distancebetween any twopoints: call the points(x1, y1) and (x2, y2).5(x1, y1)0-5Slide 14 / 202x5First, let's zoom in sowe have more room towork.-5-10The Distance FormulayNow, we'll construct two pathsbetween the points.(x2, y2)10The first path will be directlybetween them.5(x1, y1)05xSlide 15 / 202
The Distance FormulayThe second path will have asegment parallel the x-axisand a segment parallel to they-axis.(x2, y2)10Slide 16 / 202That will be a big help sinceit's easy to read the length ofeach of those from the axis.5(x1, y1)0x5The Distance FormulaWe now have a right triangle.yFor convenience, let's labelthe two legs "a" and "b" andthe hypotenuse "c."(x2, y2)10cLet's also label the pointwhere the two legs meet byits coordinates: (x2, y1).b5a(x1, y1)0(x2, y1)x5Take a moment to see thatthose are the coordinates ofthat vertex of the triangle.Which formula relates thelengths of the sides of a righttriangle?The Distance FormulaDid you get this?yc2 a2 b2(x2, y2)10cThe next step is to writeexpressions for the lengths of sidesa and b based on the x and ysubscript coordinates.bUse the distance along each axisto find those.5a(x1, y1)0Slide 17 / 2025(x2, y1)xThe coordinates that we just addedfor the third vertex should help.Slide 18 / 202
Slide 19 / 202The Distance FormulaDid you get these?ya lx2 - x1l(x2, y2)10ANDb ly2 - y1lWe could equally well writeca lx1 - x2lb5a(x1, y1)0(x2, y1)x5ANDb ly1 - y2lWe use absolute values since weare just concerned with lengths,which are always positive. That'swhy the order doesn't matter andall of the above are OK.Substitute one pair of these intoc2 a2 b2The Distance FormulaSlide 20 / 202We'll use the first pair:yc2 (lx2 - x1l)2 (ly2 - y1l)2(x2, y2)10cSince the quantities are squared,we don't need to indicateabsolute value, the result willalways be positive anyway.b5c2 (x2 - x1)2 (y2 - y1)2a(x1, y1)0(x2, y1)x5Since "c" is the distancebetween the points, we can use"d" for distance and say thatd2 (x2 - x1)2 (y2 - y1)2If we solve this equation for d,what will be the final formula?The Distance Formulayd ((x2 - x1)2 (y2-y1)2)1/2(x2, y2)10cORd b5a(x1, y1)05(x2, y1)xSlide 21 / 202
1 What is the distance between the points: (4, 8) and (7, 3)?Round your answer to the nearest hundredth.2 What is the distance between the points: (-4, 8) and(7, -3)? Round your answer to the nearest hundredth.3 What is the distance between the points: (-2, -5) and(-7, 3)? Round your answer to the nearest hundredth.Slide 22 / 202Slide 23 / 202Slide 24 / 202
4 What is the distancebetween the indicatedpoints? Round youranswer to the nearesthundredth.1050-5Slide 25 / 2025-5-105 What is the distancebetween the indicatedpoints? Round youranswer to the nearesthundredth.105-50Slide 26 / 2025-5-106 What is the distancebetween the indicatedpoints? Round youranswer to the nearesthundredth.105-50-5-105Slide 27 / 202
Slide 28 / 202The Midpoint FormulaLab - Midpoint FormulaReturn to Tableof ContentsThe Midpoint FormulaySlide 29 / 202Another question which is easilysolved using analytic geometryis to find the midpoint of a line.10Once again, we make use of thefact that it's easy to determinedistance parallel to an axis, solet's add those lines.505xThe Midpoint FormulaThe x-coordinate for themidpoint between (x1, y1) and(x2, y2) is halfway between x1and x2.y10Similarly, the y-coordinate ofthat midpoint will be halfwaybetween y1 and y2.5If you're provided a graph of aline, and asked to mark themidpoint, you can often do thatwithout much calculating.05xSlide 30 / 202
The Midpoint FormulaySlide 31 / 202In this graph, x1 1 and x2 9.They are 8 units apart, so justgo 4 units along the x-axis andgo up until you intersect the line.10That will give you the midpoint.In this case, that is at (5,7),which can be read from thegraph.50x5We would get the same answerif we had done this along the yaxis, as we do on the next slide.The Midpoint FormulaThe y-coordinates of the twopoints are 11 and 3, so they arealso 8 units apart.y10So, just go up 4 from the lowery-coordinate and then across tothe line to also get (5,7) to bethe midpoint.5So, the order doesn't matter andif you have the graph and thenumbers are easy to read, youmay as well use it.0x5If you are not given a graph orthe lines don't fall so that thevalues are easy to read, youcan do a quick calculation to getthe same result.The Midpoint FormulaWe can calculate the x-coordinatemidway between that of the twogiven points by finding theiraverage.y(9,11)10The same for the y-coordinate.Just add the two values anddivide by two.5The x-coordinate of the midpointis (1 9)/2 5(1,3)0Slide 32 / 2025xThe y-coordinate of the midpointis (3 11)/2 7That's the same answer: (5,7)Slide 33 / 202
Slide 34 / 202The Midpoint FormulaThe more general solution isgiven below for any points:(x1, y1) and (x2, y2).y(x2, y2)10The x-coordinate of themidpoint is given byxM (x1 x2) / 2The y-coordinate of themidpoint is given byyM (y1 y2) / 25(x1, y1)0So, the coordinates of themidpoint are:5xSlide 35 / 2027 What is themidpoint betweenthe indicatedpoints?105A (4, 9)B (-5, -4)0-5C (5, 6)5-5D (5, 7)-10Slide 36 / 2028 What is themidpoint betweenthe indicatedpoints?105A (0, 0)B (5, 10)C (5, 5)-50-5D (10, 10)-105
Slide 37 / 2029 What is themidpoint betweenthe indicatedpoints?105A (3, 3)B (3, 4)C (4, 3)-505-5D (5, 3)-1010 What is the midpoint between the points: (4, 8)and (7, 3)?Slide 38 / 202A (8, 2)B (4, 7)C (5.5, 5.5)D (6, 5)11 What is the midpoint between the points: (-4, 8)and (4, -8)?A (8, 2)B (0, 0)C (12, 12)D (4, 4)Slide 39 / 202
12 What is the midpoint between the points: (-4, -8)and (-6, -4)?Slide 40 / 202A (-5, -6)B (-10, -12)C (2, 4)D (5, 6)Finding the Coordinates of an E ndpointof a SegmentThe midpoint of AB is M(1, 2).One endpoint is A(4, 6). Findthe coordinates of the otherendpoint B(x, y).You can use the midpointformula to write equationsusing x and y. Then, set up2 equations and solve eachone.A(4, 6)M(1, 2) (4 2 x, 6 2 y )B(x, y)Finding the Coordinates of an E ndpointof a SegmentM(1, 2) (4 2 x, 6 2 y )1 4 x22 4 x-2 xA(4, 6)2 6 y2M(1, 2) 4 6 y-2 yTherefore, the coordinatesof B(-2, -2).Slide 41 / 202(4 2 x, 6 2 y )B(x, y)Can you find a shortcut to solve this problem? How would yourshortcut make the problem easier?Slide 42 / 202
Finding the Coordinates of an E ndpointof a SegmentAnother way of approaching thisproblem is to look for the pattern thatoccurs between the endpointA (4, 6) and midpoint M (1, 2).Looking only at our points, we candetermine that we traveled left 3 unitsand down 4 units to get from A to M.If we travel the same units in thesame direction starting at M, we willget to B(-2, -2).Slide 43 / 202left 3down4left 3down4A(4, 6)M(1, 2)B(x, y)Finding the Coordinates of an E ndpointof a SegmentSlide 44 / 202Similarly, if we line up the points vertically and determine the pattern ofthe numbers, without a graph, we can calculate the coordinates for ourmissing endpoint.A (4, 6)-3-4M (1, 2)-3B (-2, -2)If you use this method, always determinethe operation required to get from the givenendpoint to the midpoint. The reverse willnot work.-413 Find the other endpoint of the segment withthe endpoint (7, 2) and midpoint (3, 0)A(-1, -2)B(-2, -1)C(4, 2)D(2, 4)Slide 45 / 202
14 Find the other endpoint of the segment withthe endpoint (1, 4) and midpoint (5, -2)A(11, -8)B(9, 0)C(9, -8)D(3, 1)Slide 46 / 202Slide 47 / 20215 Find the other endpoint of the segment withthe endpoint (-4, -1) and midpoint (-2, 3).A(-6, -5)B(-3, -2)C(0, 7)D(1, 9)Slide 48 / 20216 Find the other endpoint of the segment withthe endpoint (-2, 5) and midpoint (0, 2).A(-1, -3.5)B(-4, 8)C(1, 0.5)D(2, -1)
Slide 49 / 202Partitions of a LineSegmentReturn to Tableof ContentsPartitions of a Line SegmentBSlide 50 / 202Partitioning a line segmentsimply means to divide thesegment into 2 or more parts,based on a given ratio.The midpoint partitions asegment into 2 congruentsegments, forming a ratio of 1:1.But if you need to partition asegment so that its ratio issomething different, for example3:1, how can it be done?APartitions of a Line SegmentBIn order to divide the segment inthe ratio of 3:1, think of dividingthe segment into 3 1, or 4congruent pieces.Plot the points that would divideAB into 4 congruent pieces.- Click on one of the points inthe grid to show them all.ASlide 51 / 202
Slide 52 / 202Partitions of a Line SegmentyB10If we add a coordinate plane toour segment, could we alsodetermine the coordinates ofour points?Yes, we can. We can alsoeliminate one of our points thatdoes not divide our segmentinto the ratio 3:1. In our case,the midpoint.- Click on midpoint to hide it5A0x5Let's say that the ratio of the twosegments that we're looking forfrom left to right is 3:1. Whichother point should beeliminated?- Click on that point to hide itSlide 53 / 202Partitions of a Line SegmentyB10Could we also determine thecoordinates of our point if theratio was reversed (1:3 insteadof 3:1)?Yes, we can. Again, ourmidpoint can be eliminated.- Click on midpoint to hide it5A0x5Partitions of a LineSegmentyB (9,11)105A(1,3)05xNow, the ratio of the twosegments that we're looking forfrom left to right is 1:3. Whichother point should beeliminated?- Click on that point to hide itWe can also calculate both the xand y coordinates between thetwo given points by using aformula that is similar to themidpoint formula.But instead of having a commonratio (1:1) and dividing by 2, whatthe midpoint formula has us do,we need to multiply the one set ofcoordinates by the first number inthe ratio and the other set ofcoordinates by the secondnumber in the ratio and divide bythe number of segments that arerequired for our ratio 3:1, or3 1 4.3(9) 1(1) 3(11) 1(3)3 13 127 133 3 (7, 9)44((,,))Slide 54 / 202
Slide 55 / 202Partitions of a Line SegmentIf we use the same formula, butswitch to the other ratio 1:3, wewill get the other point thatpartitions AB.yB (9,11)10(1(9)3 13(1) , 1(11)3 13(3))(9 4 3 , 11 4 9) (3, 5)5A (1,3)05xSlide 56 / 202Partitions of a Line SegmentThe more general solution isgiven below for any points:(x1, y1) and (x2, y2).y(x2, y2)10If a point P partitions asegment in a ratio of m:n, thenthe coordinates of P areP 5y521Partitions of a Line SegmentExample with a graph:Line segment CD in the coordinateplane has endpoints withcoordinates (-2, 10) and (8, -5).Graph CD and find two possiblelocations for a point P that dividesCD into two parts with lengths in aratio of 2:3.5-51x1002Remember to also use theformula twice switching the m:nratio to n:m.(x1, y1)0my ny(mxm nxn , m n )5xSlide 57 / 202
yC (-2, 10)10Partitions of a Line SegmentStep #1: Graph the points. C(-2, 10)and D(8, -5)- Click on the points to show.(0, 7)5Step #2: Connect them w/ thesegment CD.- Click on the segment to show.(2, 4)(4, 1)05(6, -2)-5yC (-2, 10)10xStep #3: Determine the totalnumber of ways required to dividethe segment using the ratio 2:3.2 3 5ClickStep #4: Plot the points on thegraph that divide CD into thedesired number of sections.- Click on the points to show.D (8, -5)Partitions of a Line SegmentSlide 59 / 202Step #5: Since the ratio is 2:3,determine which points to eliminate(3 of them) leaving only 1 pointremaining so that, from left to right,the ratio is 2:3.(0, 7)5Slide 58 / 202(2, 4)(4, 1)05(6, -2)D (8, -5)-5yC (-2, 10)10xPartitions of a Line SegmentStep #6: Because the question asksfor 2 points, we also need toconsider the ratio 3:2. Similar toStep #5, determine which points toeliminate (3 of them) leaving only 1point remaining so that, from left toright, the ratio is 3:2.(0, 7)5(2, 4)(4, 1)0-55(6, -2)xD (8, -5)Slide 60 / 202
Partitions of a Line SegmentSlide 61 / 202Example without a graph:Line segment EF in the coordinate plane has endpoints withcoordinates (10, -11) and (-4, 10).Find two possible locations for a point P that divides EF into twoparts with lengths in a ratio of 5:2.Partitions of a Line SegmentSlide 62 / 202(10, -11) and (-4, 10)Step #1: Determine the total number of ways required to divide thesegment using the ratio 5:2.5 2 7ClickStep #2: Use the formula to determine the coordinates of our firstpoint P.my ny(mxm nxn , m n )5(-4) 2(10) 5(10) 2(-11), 5 2 )P (5 2clickP 212P -20 207150 (-22)7P (0, 4)Partitions of a Line Segment(10, -11) and (-4, 10)Step #3: Reverse the ratio 2:5 and reuse the formula to determine thecoordinates of our second point P.my ny(mxm nxn , m n )( 2(-4) 5(10) , 2(10) 5(-11))P P 2121click2 52 5-8 5020 (-55)P 77P (6, 5)Slide 63 / 202
Slide 64 / 20217 Line segment GH inthe coordinate planehas endpoints withcoordinates (-7, -9) &(8, 3), shown in thegraph below. Find 2possible locations fora point P that dividesGH into two partswith lengths in a ratioof 2:1.ABCD(0.5, -3)(3, -1)(6, 1)(-2, -5)105H(8, 3)0-55-5G (-7, -9)-10E (-5, -7)18 Line segment JK inthe coordinate planehas endpoints withcoordinates (10, 12)& (-10, -13), shownin the graph below.Find 2 possiblelocations for a pointP that divides JK intotwo parts withlengths in a ratio of4:1.A (-6, -8)B (-2, -3)C (0, -0.5)D (2, 2)E (6, 7)(-6, 10)(-4, 7)(-2, 4)(0, 1)E (2, -3)Slide 65 / 202105-505-5-10K(-10, -13)19 Line segment LM inL(-8, 13)the coordinate planehas endpoints withcoordinates (-8, 13)& (6, -8), shown inthe graph below.Find 2 possiblelocations for a pointP that divides LMinto two parts withlengths in a ratio of3:4.ABCDJ(10, 12)Slide 66 / 202105-505-5M(6, -8)-10
20 Line segment LM in the coordinate plane has endpointswith coordinates (-8, 13) & (6, -8), shown in the graphbelow. Find 2 possible locations for a point P that dividesLM into two parts with lengths in a ratio of 6:1.Slide 67 / 202A (-6, 10)B (-4, 7)C (-2, 4)D (0, 1)E (4, -6)21 Line segment NO in the coordinate plane has endpointswith coordinates (10, 12) & (-10, -13), shown in the graphbelow. Find 2 possible locations for a point P that dividesNO into two parts with lengths in a ratio of 3:2.Slide 68 / 202A (-6, -8)B (-2, -3)C (0, -0.5)D (2, 2)E (6, 7)22 Line segment QR in the coordinate plane has endpointswith coordinates (-12, 11) & (12, -13), shown in the graphbelow. Find 2 possible locations for a point P that dividesQR into two parts with lengths in a ratio of 5:3.A (3, -4)B (0, -1)C (-3, 2)D (-6, 5)E (-9, 8)Slide 69 / 202
Line segment JK in the coordinate plane has endpoints with coordinates(-4, 11) & (8, -1). Graph JK and find two possible locations for point Mso that M divides JKJ(-4, 11)into two parts withlengths in a ratio of 1:3.Slide 70 / 202To graph a linesegment, click thelocations for the twopoints. Then click inbetween the two pointsto make the segment.K(8, -1)23 Find two possible locations for point M so that M dividesJK into two parts with lengths in a ratio of 1:3.A (-2, 9)Slide 71 / 202J (-4, 11)B (-1, 8)C (0, 7)D (4, 3)E (5, 2)K (8, -1)F (6, 1)PARCC Released Question (EOY) - Part 2 - Response Format24 Point Q lie on ST, where point S is located at (-2, -6) andpoint T is located at (5, 8). If SQ:QT 5:2, where is pointQ on ST?A (-1, -4)B (0, -2)C (1, 0)D (2, 2)E (3, 4)F (4, 6)PARCC Released Question (PBA)Slide 72 / 202
Slide 73 / 202Slopes of Parallel &Perpendicular LinesReturn to Tableof ContentsSlopeySlide 74 / 202The slope of a line indicates theangle it makes with the x-axis.10The symbol for slope is "m".505xSlopeyA horizontal line has a slope ofzero.10A vertical line has an undefinedslope.A line which rises as you movefrom left to right has a positiveslope.505xA line which falls as you movefrom left to right has a negativeslope.Slide 75 / 202
Slide 76 / 20225 The slope of the indicated line is:A negativeyB positiveC zeroD undefinedxSlide 77 / 20226 The slope of the indicated line is:A negativeyB positiveC zeroD undefinedxSlide 78 / 20227 The slope of the indicated line is:A negativeyB positiveC zeroD undefinedx
Slide 79 / 20228 The slope of the indicated line is:A negativeyB positiveC zeroD undefinedxSlide 80 / 20229 The slope of the indicated line is:A negativeyB positiveC zeroD undefinedxSlide 81 / 20230 The slope of the indicated line is:A negativeyB positiveC zeroD undefinedx
Slide 82 / 202SlopeyThe slope of a line is notgiven in degrees.Rather, it is given as theratio of "rise" over "run".10The slope of a line is thesame anywhere along theline, so any two points onthe line can be used tocalculate the slope.505xSlide 83 / 202Slopey x(x1, y2) "run"10The "rise" is the change in thevalue of the y-coordinatewhile the "run" is the change in(x2, y2) the value of the x-coordinate.The symbol for change is theGreek letter delta, " ", whichjust means "change in"."rise" ySo the slope is equal to thechange in y divided by thechange in x, or y divided by x.delta y over delta x.5(x1, y1)05Δy y2-y1m Δx x -x2 1xSlide 84 / 202Slope105 x(2, 11)"run"Δy y2-y1m Δx x -x2 1(8,11) yThe rise is from 4 to 11 y 11 - 4 7And the run is from 2 to 8,(2,4)0In this case:"rise"y x 8 - 2 65xSo the slope ism y 7 x6
Slide 85 / 202Slopey x10 x yAny points on the line canbe used to calculate its(x2, y2) slope, since the slope of aline is the sameeverywhere.The values of y and xmay be different for otherpoints, but their ratio will bethe same. y5(x1, y1)05xYou can check that with thered and green trianglesshown here.Δy y2-y1m Δx x -x2 1Using Slope to Draw a LineySlide 86 / 202The slope also allows us toquickly graph a line, givenone point on the line.10For instance, if I know onepoint on a line is (1, 1) andthat the slope of the line is2, I can find a second point,and then draw the line.505xΔy y2-y1m Δx x -x2 1Using Slope to Draw a LineyI do this by recognizing thatthe slope of 2 means that ifI go up 2 units on the y-axisI have to go 1 unit to theright on the x-axis .105 x y05xΔy y2-y1m Δx x -x2 1Slide 87 / 202
Using Slope to Draw a LineyI do this by recognizing thatthe slope of 2 means that ifI go up 2 units on the y-axisI have to go 1 unit to theright on the x-axis . x10Or if I go up 10, I have togo over 5 units, etc. y505xΔy y2-y1m Δx x -x2 1Using Slope to Draw a LineyThis method is the easiestto use if you just have todraw a line given a pointand slope. y5Slide 89 / 202Then I draw the linethrough any two of thosepoints. x10Slide 88 / 202The same approach worksfor writing the equation of aline.05xΔy y2-y1m Δx x -x2 1Slopes of Parallel LinesParallel lines have the same slope.With what we'velearned about parallellines this is easy tounderstand.The slope of a line isrelated to the angle itmakes with the x-axis.Now, think of the x-axisas a transversal.Slide 90 / 202
31 What is the name of this pair of angles, formed by atransversal intersecting two lines.A Alternate Interior AnglesC Alternate Exterior AnglesB Corresponding AnglesD Same Side Interior Angles1232 We know those angles must then be:A SupplementaryC EqualB ComplementaryD Adjacent1Slide 92 / 202233 That means that the slopes of parallel lines must be:A ReciprocalsC EqualB InversesD Nothing special1Slide 91 / 2022Slide 93 / 202
34 If one line has a slope of 4, what must be the slope ofany line parallel to it?35 If one line passes through the points (0, 0) and (2, 2)what must be the slope of any line parallel to that firstline?36 If one line passes through the points (-5, 9) and (5, 8)what must be the slope of any line parallel to that firstline?Slide 94 / 202Slide 95 / 202Slide 96 / 202
37 If one line passes through the points (2, 2) and (5, 5) anda parallel line passes through the point (1, 5) which ofthese points could lie on that second line?Slide 97 / 202A (2, 2)B (4, 4)C (5, 6)D (-1, 3)38 If one line passes through the points (-3, 4) and (0, 10)and a parallel line passes through the point (-1, -4) whichof these points could lie on that second line?Slide 98 / 202A (0, -2)B (2, -5)C (3, 1)D (-1, 3)39 If one line passes through the points (3, 5) and (5, -1)and a parallel line passes through the point (-1, -1) whichof these points could lie on that second line?A (0, 1)B (-2, 2)C (4, 8)D (-4, -4)Slide 99 / 202
Slopes of Perpendicular LinesSlide 100 / 202The slopes of perpendicular lines are negative(or opposite) reciprocals.There are three ways ofexpressing thissymbolically:m1m2 -1m1 -1m2m2 -1m1These all have the samemeaning.Slopes of Perpendicular LinesSlide 101 / 202The slopes of perpendicular lines are negative (oropposite) reciprocals.This will be useful in caseswhere you need to provelines perpendicular,including proving thattriangles are right triangles.First, let's prove this is true.Slopes ofPerpendicular Lines10First, let's move our lines tothe origin so our work getseasier and we can focus onthe important parts.5-50-5-105We can move them since wecan just think of drawing newparallel lines through theorigin that have the sameslopes as these lines.We earlier showed thatparallel lines have the sameslope, so the slopes of thesenew lines will be the same asthat of the original lines.Slide 102 / 202
Slopes of Perpendicular Lines1(1,m1)Slide 103 / 202Now, let's zoom in and focuson the lines between x 0 andx 1.When the lines are at x 1,their y-coordinates are m1 forthe first line and m2 for thesecond line.01That's because if the slope ofthe first line is m1, then whenwe move 1 along the x-axis,the y-value must increase bythe amount of the slope, m1.-1The same for the second line,whose slope is m2.(1,m2)Slopes of Perpendicular LinesSlide 104 / 202Perpendicular lines obey thePythagorean Theorem:1(1,m1)ac2 a2 b2Let's find expressions for thosethree terms so we can substitutethem in.01cb-1Side "c" is hypotenuse and is thedistance between the pointsalong the vertical line x 1. It haslength m1-m2 (in this case, thateffectively adds their magnitudessince m2 is negative).We can use the distance formulato find the length of each leg.(1,m2)Slopes of Perpendicular Lines1(1,m1)aThe lengths of legs "a" and "b"can be found using thedistance formula.d2 (x2-x1)2 (y2-y1)20a2 (1-0)2 (m1-0)2 1 m121cb2 (1-0)2 (m2-0)2 1 m22And, we get c2 by squaring ourresult for c from the prior slide.b-1(1,m2)c2 (m1-m2)2 m12 - 2m1m2 m22Slide 105 / 202
Slopes of Perpendicular Lines1(1,m1)aSlide 106 / 202We're going to substitute theexpressions from the prior slide intothe Pythagorean Theorem.We'll color code them so we cankeep track.c2 m12 - 2m1m1 m2201ca2 1 m12b2 1 m22bc2 a2 b2m12 - 2m1m2 m22 1 m12 1 m22-1(1,m2)Slopes of Perpendicular LinesSlide 107 / 202m12 - 2m1m2 m22 1 m12 1 m221Now, let's identify like terms(1,m1)am12 - 2m1m2 m22 1 m12 1 m22Notice that we have m12 and m22 onboth sides so they can be canceled,and that 1 1 2. So,01c-2m1m2 2m1m2 -1which is what we set out to proveNotice that this also be written as:b-1(1,m2)m1 -1m2-1m2 m140 If one line has a slope of 4, what must be the slope ofany line perpendicular to it?Slide 108 / 202
41 If one line has a slope of -1/2, what must be the slope ofany line perpendicular to it?42 If one line passes through the points (0, 0) and (4, 2)what must be the slope of any line perpendicular to thatfirst line?43 If one line passes through the points (-5, 9) and (5, 8)what must be the slope of any line perpendicular to thatfirst line?Slide 109 / 202Slide 110 / 202Slide 111 / 202
44 If one line passes through the points (1, 2) and (5, 6) anda perpendicular line passes through the point (1, 5)which of these points could lie on that second line?Slide 112 / 202A (2, 2)B (4, 4)C (2, 4)D (-1, 3)45 If one line passes through the points (-3, 4) and (0, 10)and a perpendicular line passes through the point (-1, -4)which of these points could lie on that second line?Slide 113 / 202A (0, -2)B (2, -5)C (3, 1)D (-3, -5)46 If one line passes through the points (3, 5) and (5, -1)and a perpendicular line passes through the point (-1, -1)which of these points could lie on that second line?A (5, 0)B (2, 0)C (4, 8)D (-4, -4)Slide 114 / 202
Slide 115 / 202Equations of Parallel &Perpendicular LinesReturn to Tableof ContentsWriting the Equation of a LineSlide 116 / 202y2-y1m x -xy21Let's start with the abovedefinition of slope, multiplyboth sides by (x2-x1), andrearrange to get: x10 y(y2-y1) m(x2-x1)5Now, if I enter one point (x1,y1)this defines the infinite locusof other points on the line.y-y1 m(x-x1)05xPoint-slopeformThen, I can add y1 to bothsides to isolate the variable y.y m(x-x1) y1Writing the Equation of a Lineyy m(x-x1) y1 xThe term in red is just your stepsto the right along the x-axis, your"run".10Multiplying by slope tells youhow many steps up you musttake along the y-axis, your "rise". y505xThose steps are added to theterm in green, your originalposition on the y-axis to find yourfinal y-coordinate.That tells you the y-coordinateon the line for the given xcoordinate.Slide 117 / 202
Slide 118 / 202Slope Intercept Equation of a Lineyy m(x-x1) y1 xA common simplification of thisis to use the y-intercept for(x1,y1).10The y-intercept is the pointwhere the line crosses the yaxis. Its symbol is "b" so thecoordinates of that point are(0,b), since x 0 on the y-axis. y50x5Just substitute (0,b) in theabove equation for (x1,y1).Slide 119 / 202Slope Intercept Equation of a Lineyy m(x-x1) y1 xy m(x-0) b10y mx bThis is very useful since boththe slope and the y-interceptsoften have important meaningwhen we are solving realproblems. y50x5In the graph to the left, b 0, sothe equation is y 2x.Writing Equations of Parallel Lines1. Find an equation of the line passing through the point (-4, 5)and parallel to the line whose equation is -3x 2y -1.Step 1: Identify the information given in the problem.Contains the point (-4, 5) & is parallel to the line -3x 2y -1Step 2: Identify what information you still need to create theequation and choose the method to obtain it.The slope: Use the equation of the parallel line to determine theslope-3x 2y -1Click2y 3x - 1y 3/2 x - 1/2ClickTherefore m 3/2clickSlide 120 / 202
Writing Equations of Parallel LinesSlide 121 / 202Step 3: Create the equation.y - y1 m(x - x1)y - 5 3/2 (x 4)Point-Slope FormClicky - 5 3/2 x 6Clicky 3/2 x 11ClickSlope-Intercept FormThe correct solution to the original problem is either form of theequation. Point-Slope Form and Slope-Intercept Form are twoways to write the same linear equations.Writing Equations of Perpendicular LinesSlide 122 / 2022. Write the equation of the line passing through the point(-2, 5)and perpendicular to the line y 1/2 x
Oct 02, 2015 · Origin of Analytic Geometry Return to Table of Contents Slide 5 / 202 Analytic Geometry is a powerful combination of geometry and algebra. Many jobs that are looking for employees now, and will be in the future, rely on the process or results of analytic geometry. This includes jobs in medicine, veterinary science,
COMPLEX ANALYTIC GEOMETRY AND ANALYTIC-GEOMETRIC CATEGORIES YA’ACOV PETERZIL AND SERGEI STARCHENKO Abstract. The notion of a analytic-geometric category was introduced by v.d. Dries and Miller in [4]. It is a category of subsets of real analytic manifolds which extends the c
Analytic Geometry Geometry is all about shapes and their properties. If you like playing with objects, or like drawing, then geometry is for you! Geometry can be divided into: Plane Geometry is about flat shapes like lines, circles and triangles . shapes that can be drawn on a piece of paper S
Analytic geometry connects algebra and geometry, resulting in powerful methods of analysis and problem solving. The first unit of Analytic Geometry involves similarity, congruence, and proofs. Students will understand similarity in terms of similarity transformations, prove
Analytic geometry connects algebra and geometry, resulting in powerful methods of analysis and problem solving. The first unit of Analytic Geometry involves similarity, congruence, and proofs. Students will understand similarity in terms of similarity transformations, prove
analytic geometry can be done over a countable ordered field. . I give Hilbert’s axioms for geometry and note the essential point for analytic geometry: when an infinite straight line is conceived as an ordered additive group, then this group can be
Analytic Continuation of Functions. 2 We define analytic continuation as the process of continuing a function off of the real axis and into the complex plane such that the resulting function is analytic. More generally, analytic continuation extends the representation of a function
Georgia Department of Education Georgia Standards of Excellence Framework Accelerated GSE Analytic Geometry B/Advanced Algebra Unit 2 Mathematics 2Accelerated GSE Analytic Geometry B/Advanced Algebra Unit : Geometric and Algebraic Connections
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