REVIEW OF ANALYTIC GEOMETRY - Stewart Calculus
REVIEW OF ANALYTIC GEOMETRYThe points in a plane can be identified with ordered pairs of real numbers. We start bydrawing two perpendicular coordinate lines that intersect at the origin O on each line.Usually one line is horizontal with positive direction to the right and is called thex-axis; the other line is vertical with positive direction upward and is called the y-axis.Any point P in the plane can be located by a unique ordered pair of numbers as follows.Draw lines through P perpendicular to the x- and y-axes. These lines intersect the axes inpoints with coordinates a and b as shown in Figure 1. Then the point P is assigned theordered pair 共a, b兲. The first number a is called the x-coordinate of P ; the second numberb is called the y-coordinate of P. We say that P is the point with coordinates 共a, b兲, andwe denote the point by the symbol P共a, b兲. Several points are labeled with their coordinates in Figure 2.yyP (a, b)4b43II( 2, 2)I2III312345x3 2 1 012( 3, 2))3aIV4(5, 0)14FIGURE 1Stewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved.2113 2 1 O12(1, 3)32345x(2, 4)FIGURE 2By reversing the preceding process we can start with an ordered pair 共a, b兲 and arrive atthe corresponding point P. Often we identify the point P with the ordered pair 共a, b兲 andrefer to “the point 共a, b兲.” [Although the notation used for an open interval 共a, b兲 is thesame as the notation used for a point 共a, b兲, you will be able to tell from the context whichmeaning is intended.]This coordinate system is called the rectangular coordinate system or the Cartesiancoordinate system in honor of the French mathematician René Descartes (1596–1650),even though another Frenchman, Pierre Fermat (1601–1665), invented the principles ofanalytic geometry at about the same time as Descartes. The plane supplied with this coordinate system is called the coordinate plane or the Cartesian plane and is denoted by 2.The x- and y-axes are called the coordinate axes and divide the Cartesian plane intofour quadrants, which are labeled I, II, III, and IV in Figure 1. Notice that the first quadrant consists of those points whose x- and y-coordinates are both positive.EXAMPLE 1 Describe and sketch the regions given by the following sets.ⱍⱍ(a) 兵共x, y兲 x 艌 0其(c ) {共x, y兲(b) 兵共x, y兲 y 苷 1其ⱍ ⱍ y ⱍ 1}SOLUTION(a) The points whose x-coordinates are 0 or positive lie on the y-axis or to the right of itas indicated by the shaded region in Figure 3(a).yyyy 1y 10x0x0xy 1FIGURE 3(a) x 0(b) y 1(c) y 11
2 REVIEW OF ANALYTIC GEOMETRY(b) The set of all points with y-coordinate 1 is a horizontal line one unit above the x-axis[see Figure 3(b)].(c) Recall from Review of Algebra thatⱍyⱍ 1 1 y 1if and only ifThe given region consists of those points in the plane whose y-coordinates lie between 1 and 1. Thus, the region consists of all points that lie between (but not on) the horizontal lines y 苷 1 and y 苷 1. [These lines are shown as dashed lines in Figure 3(c) toindicate that the points on these lines don’t lie in the set.]yfi fi-› ›P¡( , ›)P ( , ›) - 0 Recall from Review of Algebra that the distance between points a and b on a number lineis a b 苷 b a . Thus, the distance between points P1共x 1, y1 兲 and P3 共x 2 , y1 兲 on ahorizontal line must be x 2 x 1 and the distance between P2 共x 2 , y2 兲 and P3 共x 2 , y1 兲 ona vertical line must be y2 y1 . (See Figure 4.)To find the distance P1 P2 between any two points P1共x 1, y1 兲 and P2 共x 2 , y2 兲, we notethat triangle P1P2 P3 in Figure 4 is a right triangle, and so by the PythagoreanTheorem we haveⱍP ( , fi)x ⱍ ⱍⱍⱍⱍⱍⱍⱍⱍⱍ P P ⱍ 苷 sⱍ P P ⱍFIGURE 412132ⱍ P2 P3ⱍ2ⱍ苷 s x2 x1ⱍ2ⱍ y2 y1ⱍ2苷 s共x 2 x 1 兲2 共y2 y1 兲2Distance Formula The distance between the points P1共x 1, y1 兲 and P2 共x 2 , y2 兲 isⱍ P P ⱍ 苷 s共x122 x 1 兲2 共y2 y1 兲2For instance, the distance between 共1, 2兲 and 共5, 3兲 iss共5 1兲 2 关3 共 2兲兴 2 苷 s4 2 5 2 苷 s41An equation of a curve is an equation satisfied by the coordinates of the points on thecurve and by no other points. Let’s use the distance formula to find the equation of a circle with radius r and center 共h, k兲. By definition, the circle is the set of all points P共x, y兲whose distance from the center C共h, k兲 is r. (See Figure 5.) Thus, P is on the circle if andonly if PC 苷 r. From the distance formula, we haveyP (x, y)rⱍ ⱍC (h, k)s共x h兲2 共 y k兲2 苷 ror equivalently, squaring both sides, we get0FIGURE 5共x h兲2 (y k兲2 苷 r 2xThis is the desired equation.Equation of a Circle An equation of the circle with center 共h, k兲 and radius r is共x h兲2 (y k兲2 苷 r 2In particular, if the center is the origin 共0, 0兲, the equation isx2 y2 苷 r2For instance, an equation of the circle with radius 3 and center 共2, 5兲 is共x 2兲2 (y 5兲2 苷 9Stewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved.CIRCLES
REVIEW OF ANALYTIC GEOMETRY 3EXAMPLE 2 Sketch the graph of the equation x 2 y 2 2x 6y 7 苷 0 by first show-ing that it represents a circle and then finding its center and radius.SOLUTION We first group the x-terms and y-terms as follows:共x 2 2x兲 (y 2 6y兲 苷 7yThen we complete the square within each grouping, adding the appropriate constants(the squares of half the coefficients of x and y) to both sides of theequation:( 1, 3)共x 2 2x 1兲 (y 2 6y 9兲 苷 7 1 9共x 1兲2 (y 3兲2 苷 3or0x1Comparing this equation with the standard equation of a circle, we see that h 苷 1,k 苷 3, and r 苷 s3 , so the given equation represents a circle with center 共 1, 3兲 andradius s3. It is sketched in Figure 6.FIGURE 6 2x-6y 7 0LINESTo find the equation of a line L we use its slope, which is a measure of the steepness of theline.Definition The slope of a nonvertical line that passes through the points P1共x 1, y1 兲and P2 共x 2 , y2 兲 isyLm苷P (x , y )Îy fi-› riseP¡(x¡, y¡)The slope of a vertical line is not defined.Îx - runThus the slope of a line is the ratio of the change in y, y, to the change in x, x. (SeeFigure 7.) The slope is therefore the rate of change of y with respect to x. The fact that theline is straight means that the rate of change is constant.Figure 8 shows several lines labeled with their slopes. Notice that lines with positiveslope slant upward to the right, whereas lines with negative slope slant downward to theright. Notice also that the steepest lines are the ones for which the absolute value of theslope is largest, and a horizontal line has slope 0.Now let’s find an equation of the line that passes through a given point P1共x 1, y1 兲 andhas slope m. A point P共x, y兲 with x 苷 x 1 lies on this line if and only if the slope of the linethrough P1 and P is equal to m; that is,x0Stewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved.FIGURE 7ym 5m 2m 1m 21y y1苷mx x1m 01m 20FIGURE 8m 1m 2m 5y2 y1 y苷 xx2 x1xThis equation can be rewritten in the formy y1 苷 m共x x 1 兲and we observe that this equation is also satisfied when x 苷 x 1 and y 苷 y1 . Therefore, it isan equation of the given line.Point-Slope Form of the Equation of a Line An equation of the line passing through thepoint P1共x 1, y1 兲 and having slope m isy y1 苷 m共x x 1 兲
4 REVIEW OF ANALYTIC GEOMETRYEXAMPLE 3 Find an equation of the line through the points 共 1, 2兲 and 共3, 4兲.SOLUTION The slope of the line ism苷 4 23苷 3 共 1兲2Using the point-slope form with x 1 苷 1 and y1 苷 2, we obtainy 2 苷 32 共x 1兲which simplifies to3x 2y 苷 1ybSuppose a nonvertical line has slope m and y-intercept b. (See Figure 9.) This means itintersects the y-axis at the point 共0, b兲, so the point-slope form of the equation of the line,with x 1 苷 0 and y1 苷 b, becomesy mx by b 苷 m共x 0兲x0This simplifies as follows.FIGURE 9Slope-Intercept Form of the Equation of a Line An equation of the line with slope m andy-intercept b isy 苷 mx byIn particular, if a line is horizontal, its slope is m 苷 0, so its equation is y 苷 b, whereb is the y-intercept (see Figure 10). A vertical line does not have a slope, but we can writeits equation as x 苷 a, where a is the x-intercept, because the x-coordinate of every pointon the line is a.y bbx a0xaEXAMPLE 4 Graph the inequality x 2y 5.ⱍFIGURE 10x 2y 5y2y x 5y 12 x 522.5y 12x 520FIGURE 115xCompare this inequality with the equation y 苷 12 x 52 , which represents a line withslope 12 and y-intercept 52 . We see that the given graph consists of points whose y-coordinates are larger than those on the line y 苷 12 x 52 . Thus, the graph is the region thatlies above the line, as illustrated in Figure 11.PARALLEL AND PERPENDICULAR LINESSlopes can be used to show that lines are parallel or perpendicular. The following facts areproved, for instance, in Precalculus: Mathematics for Calculus, Fifth Edition by Stewart,Redlin, and Watson (Thomson Brooks/Cole, Belmont, CA, 2006).Stewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved.SOLUTION We are asked to sketch the graph of the set 兵共x, y兲 x 2y 5其 and we beginby solving the inequality for y :
REVIEW OF ANALYTIC GEOMETRY 5Parallel and Perpendicular Lines1. Two nonvertical lines are parallel if and only if they have the same slope.2. Two lines with slopes m1 and m2 are perpendicular if and only if m1m2 苷 1;that is, their slopes are negative reciprocals:m2 苷 1m1EXAMPLE 5 Find an equation of the line through the point 共5, 2兲 that is parallel to the line4x 6y 5 苷 0.SOLUTION The given line can be written in the formy 苷 23 x 56which is in slope-intercept form with m 苷 3 . Parallel lines have the same slope, so the2required line has slope 3 and its equation in point-slope form is2y 2 苷 23 共x 5兲We can write this equation as 2x 3y 苷 16.EXAMPLE 6 Show that the lines 2x 3y 苷 1 and 6x 4y 1 苷 0 are perpendicular.SOLUTION The equations can be written asy 苷 23 x 13andy 苷 32 x 14andm2 苷 32from which we see that the slopes arem1 苷 23Since m1m2 苷 1, the lines are perpendicular.Stewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved.EXERCISESA Click here for answers.11–24Find an equation of the line that satisfies the givenconditions.1–211. Through 共2, 3兲, slope 6Find the distance between the points.1. 共1, 1兲, 3–4共4, 5兲 2. 共1, 3兲, 12. Through 共 3, 5兲, slope 27共5, 7兲 14. Through 共 1, 2兲 and 共4, 3兲Find the slope of the line through P and Q.3. P共 3, 3兲, Q共 1, 6兲 4. P共 1, 4兲, 15. Slope 3, y-intercept 2Q共6, 0兲 5. Show that the points 共 2, 9兲, 共4, 6兲, 共1, 0兲, and 共 5, 3兲 are18. x-intercept 8, y-intercept 66. (a) Show that the points A共 1, 3兲, B共3, 11兲, and C共5, 15兲19. Through 共4, 5兲, parallel to the x-axisare collinear (lie on the same line) by showing thatAB BC 苷 AC .(b) Use slopes to show that A, B, and C are collinear.7–10ⱍ ⱍⱍ21. Through 共1, 6兲, parallel to the line x 2y 苷 622. y-intercept 6, parallel to the line 2x 3y 4 苷 0 23. Through 共 1, 2兲, perpendicular to the line 2x 5y 8 苷 08. y 苷 29. xy 苷 0 20. Through 共4, 5兲, parallel to the y-axisSketch the graph of the equation.7. x 苷 3 ⱍyⱍ 苷 110. 216. Slope 5, y-intercept 417. x-intercept 1, y-intercept 3the vertices of a square.ⱍ ⱍ ⱍ13. Through 共2, 1兲 and 共1, 6兲 24. Through ( 2 , 3 ), perpendicular to the line 4x 8y 苷 11 2
6 REVIEW OF ANALYTIC GEOMETRY41. Show that the linesFind the slope and y-intercept of the line and draw25. x 3y 苷 026. 2x 3y 6 苷 027. 3x 4y 苷 1228. 4x 5y 苷 10 29–36 ⱍ35.36. ⱍ{共x, y兲 ⱍ ⱍ x ⱍ 艋 2}32.ⱍ兵共x, y兲 ⱍ y 2x 1其兵共x, y兲 ⱍ 1 x 艋 y 艋 1 2x其{共x, y兲 ⱍ x 艋 y 共x 3兲} and10x 6y 50 苷 0are perpendicular and find their point of intersection.{共x, y兲 ⱍ ⱍ x ⱍ 3 and ⱍ y ⱍ 2}43. Show that the midpoint of the line segment from P1共x 1, y1 兲 toP2 共x 2 , y2 兲 is冉x 1 x 2 y1 y2,22冊44. Find the midpoint of the line segment joining the points 共1, 3兲12 6x 2y 苷 1042. Show that the lines3x 5y 19 苷 030. 兵共x, y兲 x 艌 1 and y 3其33. 兵共x, y兲 0 艋 y 艋 4 and x 艋 2其34. andare not parallel and find their point of intersection.Sketch the region in the xy-plane.29. 兵共x, y兲 x 0其31. 2x y 苷 4 and 共7, 15兲.37–38Find an equation of a circle that satisfies the givenconditions.45. Find an equation of the perpendicular bisector of the line seg-37. Center 共3, 1兲, radius 546. (a) Show that if the x- and y-intercepts of a line are nonzeroment joining the points A共1, 4兲 and B共7, 2兲.38. Center 共 1, 5兲, passes through 共 4, 6兲 39–40Show that the equation represents a circle and find thecenter and radius.39. x 2 y 2 4x 10y 13 苷 040. x 2 y 2 6y 2 苷 0 numbers a and b, then the equation of the line can be put inthe formxy 苷1abThis equation is called the two-intercept form of an equation of a line.(b) Use part (a) to find an equation of the line whosex-intercept is 6 and whose y-intercept is 8.Stewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved.25–28its graph.
REVIEW OF ANALYTIC GEOMETRY 7ANSWERS1. 57.3. 28.92. 2 s29y4.28. m 苷 5 ,474b苷2x 30x39.29.10.y30.yxy 00x11. y 苷 6x 1514. y 苷 x 112. y 苷 2 x 715. y 苷 3x 217. y 苷 3x 320. x 苷 418. y 苷 4 x 6313. 5x y 苷 1131225. m 苷 3 ,131.32.y16. y 苷 5 x 4219. y 苷 521. x 2y 11 苷 023. 5x 2y 1 苷 0x020222. y 苷 3 x 6224. y 苷 2x b苷013y033.34.yy 4xx 2026. m 苷 3 ,2xb苷235.x36.yy 1 x共0, 1兲0xStewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved.y 1-2x27. m 苷 4 ,3b 苷 3y0x39. 共2, 5兲, 4338. 共x 1兲2 共 y 5兲2 苷 13037. 共x 3兲2 共 y 1兲2 苷 2544. 共4, 9兲40. 共0, 3兲, s 745. y 苷 x 341. 共1, 2兲46. (b) y 苷 3 x 8442. 共2, 5兲
8 REVIEW OF ANALYTIC GEOMETRYSOLUTIONS1. Use the distance formula with P1 (x1 , y1 ) (1, 1) and P2 (x2 , y2 ) (4, 5) to gets P1 P2 (4 1)2 (5 1)2 32 42 25 52. The distance from (1, 3) to (5, 7) iss (5 1)2 [7 ( 3)]2 42 102 116 2 29.3. With P ( 3, 3) and Q( 1, 6), the slope m of the line through P and Q is m 4. m 6 39 . 1 ( 3)20 ( 4)4 6 ( 1)75. Using A( 2, 9), B(4, 6), C(1, 0), and D( 5, 3), we havess AB [4 ( 2)]2 (6 9)2 62 ( 3)2 45 9 5 3 5, CD DA ss (1 4)2 (0 6)2 ( 3)2 ( 6)2 45 9 5 3 5,ss ( 5 1)2 (3 0)2 ( 6)2 32 45 9 5 3 5, andt [ 2 ( 5)]2 (9 3)2 32 62 45 9 5 3 5. So all sides are of equal length andwe have a rhombus. Moreover, mAB mDA 116 90 63 0 , mBC 2, mCD , and4 ( 2)21 4 5 129 3 2, so the sides are perpendicular. Thus, A, B, C, and D are vertices of a square. 2 ( 5)6. (a) Using A( 1, 3), B(3, 11), and C(5, 15), we haves AB [3 ( 1)]2 (11 3)2 42 82 80 4 5,s BC (5 3)2 (15 11)2 22 42 20 2 5, ands AC [5 ( 1)]2 (15 3)2 62 122 180 6 5. Thus, AC AB BC .81211 315 3 2 and mAC 2. Since the segments AB and AC have the3 ( 1)45 ( 1)6same slope, A, B and C must be collinear.(b) mAB 7. The graph of the equation x 3 is a vertical linewith x-intercept 3. The line does not have a slope.8. The graph of the equation y 2 is a horizontalline with y-intercept 2. The line has slope 0.Stewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved. BC
REVIEW OF ANALYTIC GEOMETRY 99. xy 0 10. y 1 y 1 or y 1x 0 or y 0. The graphconsists of the coordinate axes.11. By the point-slope form of the equation of a line, an equation of the line through (2, 3) with slope 6 isy ( 3) 6(x 2) or y 6x 15.12. y ( 5) 72 [x ( 3)] or y 72 x 31213. The slope of the line through (2, 1) and (1, 6) is m y 1 5(x 2) or y 5x 11.14. For ( 1, 2) and (4, 3), m 6 1 5, so an equation of the line is1 23 ( 2) 1. An equation of the line is y 3 1(x 4) or y x 1.4 ( 1)15. By the slope-intercept form of the equation of a line, an equation of the line is y 3x 2.16. By the slope-intercept form of the equation of a line, an equation of the line is y 25 x 4.17. Since the line passes through (1, 0) and (0, 3), its slope is m Another method: From Exercise 61, 3 0 3, so an equation is y 3x 3.0 1xy 1 3x y 3 y 3x 3.1 336 0 . So an equation is y 34 x 6.0 ( 8)4yx 1 3x 4y 24 y 34 x 6.Another method: From Exercise 61, 8618. For ( 8, 0) and (0, 6), m 19. The line is parallel to the x-axis, so it is horizontal and must have the form y k. Since it goes through the pointStewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved.(x, y) (4, 5), the equation is y 5.20. The line is parallel to the y-axis, so it is vertical and must have the form x k. Since it goes through the point(x, y) (4, 5), the equation is x 4.21. Putting the line x 2y 6 into its slope-intercept form gives us y 12 x 3, so we see that this line hasslope 12 . Thus, we want the line of slope 12 that passes through the point (1, 6): y ( 6) 12 (x 1) y 12 x 11.222. 2x 3y 4 0 y 23 x 43 , so m 23 and the required line is y 23 x 6.23. 2x 5y 8 0 y 25 x 85 . Since this line has slope 25 , a line perpendicular to it would have slope 52 ,so the required line is y ( 2) 52[x ( 1)] y 52 x 12 .24. 4x 8y 1 y 12 x 18 . Since this line has slope 12 , a line perpendicular to it would have slope 2, so the y 2x 13 .required line is y 23 2 x 12
10 REVIEW OF ANALYTIC GEOMETRY25. x 3y 0 y 13 x,so the slope is 13and the26. 2x 3y 6 0 y y 45 x 2, so the slope is 2, so the slope is23and the y-intercept is 2.y-intercept is 0.28. 4x 5y 10 2x329. {(x, y) x 0}27. 3x 4y 12 y 34 x 3, so the slope is34and the y-intercept is 3.30. {(x, y) x 1 and y 3} 45 and the y-intercept is 2. rq (x, y) x 2 32.{(x, y) 2 x 2}34. {(x, y) y 2x 1} rq (x, y) x 3 and y 235. {(x, y) 1 x y 1 2x}33. {(x, y) 0 y 4, x 2}36. (x, y) x y 12 (x 3)37. An equation of the circle with center (3, 1) and radius 5 is (x 3)2 (y 1)2 52 25.38. The equation has the form (x 1)2 (y 5)2 r2 . Since ( 4, 6) lies on the circle, we haver2 ( 4 1)2 ( 6 5)2 130. So an equation is (x 1)2 (y 5)2 130.39. x2 y 2 4x 10y 13 0 x2 4x y 2 10y 13 2 x 4x 4 y 2 10y 25 13 4 25 16 (x 2)2 (y 5)2 42 . Thus, we have acircle with center (2, 5) and radius 4.Stewart: Calculus, Sixth Edition. ISBN: 0495011606. 2008 Brooks/Cole. All rights reserved.31.
REVIEW OF ANALYTIC GEOMETRY 11 x2 y 2 6y 9 2 9 with center (0, 3) and radius 7.40. x2 y 2 6y 2 0 x2 (y 3)2 7. Thus, we have a circle41. 2x y 4 y 2x 4 m1 2 and 6x 2y 10 2y 6x 10 y 3x 5 m2 3. Since m1 6 m2 , the two lines are not parallel. To find the point of intersection: 2x 4 3x 5 x 1 y 2. Thus, the point of intersection is (1, 2).42. 3x 5y 19 0 5y 3x 19 y 35 x 6y 10x 50 y 53 x 253and 10x 6y 50 0 m2 53 . Since m1 m2 35 53 1, the two lines areperpendicular. To find the point of intersection: 35 x 195195 m1 53 x 25335 9x 57 25x 125 25 5. Thus, the point of intersection is (2, 5).34x 68 x 2 y 35 · 2 1955 x x y y 1212,43. Let M be the point. Then22 x1 x2 2 y1 y2 2 x1 x2 2 y1 y2 2 y1 and MP1 2 x1 2222 x1 x2 2 y1 y2 2 x2 x1 2 y2 y1 2 MP2 2 x2 y2 . Hence, MP1 MP2 ; that2222is, M is equidistant from P1 and P2 . 7 3 15 44. Using the midpoint formula from Exercise 43 with (1, 3) and (7, 15), we get 1 (4, 9)., 22 4 1, so its perpendicular bisector has slope 1. The45. With A(1, 4) and B(7, 2), the slope of
REVIEW OF ANALYTIC GEOMETRY The points in a plane can be identified with ordered pairs of real numbers. We start by drawing two perpendicular coordinate lines that intersect at the origin on each line. Usually one line is horizontal with positive direction to the right and is called the -axis; the other line is vertical with positive direction upward and is called the -axis. Any point in the .
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
www.ck12.orgChapter 1. Basics of Geometry, Answer Key CHAPTER 1 Basics of Geometry, Answer Key Chapter Outline 1.1 GEOMETRY - SECOND EDITION, POINTS, LINES, AND PLANES, REVIEW AN- SWERS 1.2 GEOMETRY - SECOND EDITION, SEGMENTS AND DISTANCE, REVIEW ANSWERS 1.3 GEOMETRY - SECOND EDITION, ANGLES AND MEASUREMENT, REVIEW AN- SWERS 1.4 GEOMETRY - SECOND EDITION, MIDPOINTS AND BISECTORS, REVIEW AN-
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
