1 D0 APPENDIX D Precalculus Review APPENDIX D. The Cartesian Plane The Cartesian Plane The Distance and Midpoint Formulas Equations of Circles The Cartesian Plane Just as ou can represent real numbers b points on a real number line, ou can represent ordered pairs of real numbers b points in a plane called the rectangular coordinate sstem, or the Cartesian plane, after the French mathematician René Descartes. The Cartesian plane is formed b using two real number lines intersecting at right angles, as shown in Figure D.. The horizontal real number line is usuall called the -ais, and the vertical real number line is usuall called the -ais. The point of intersection of these two aes is the origin. The two aes divide the plane into four parts called quadrants. -ais Quadrant II Origin Quadrant I, ) -ais, ) 3 3, ) 0, 0) 3, 0) 3 3, 3) 3 Quadrant III Quadrant IV The Cartesian plane Figure D. Points represented b ordered pairs Figure D.5 Each point in the plane is identified b an ordered pair, of real numbers and, called coordinates of the point. The number represents the directed distance from the -ais to the point, and the number represents the directed distance from the -ais to the point see Figure D.). For the point,, the first coordinate is the -coordinate or abscissa, and the second coordinate is the -coordinate or ordinate. For eample, Figure D.5 shows the locations of the points,, 3,, 0, 0, 3, 0, and, 3 in the Cartesian plane. NOTE The signs of the coordinates of a point determine the quadrant in which the point lies. For instance, if > 0 and < 0, then the point, lies in Quadrant IV. Note that an ordered pair a, b is used to denote either a point in the plane or an open interval on the real number line. This, however, should not be confusing the nature of the problem should clarif whether a point in the plane or an open interval is being discussed.
2 APPENDIX D. The Cartesian Plane D The Distance and Midpoint Formulas Recall from the Pthagorean Theorem that, in a right triangle, the hpotenuse c and sides a and b are related b a b c. Conversel, if a b c, then the triangle is a right triangle see Figure D.). a c b, ), ), ) The distance between two points Figure D.7 d The Pthagorean Theorem: a b c Figure D. Suppose ou want to determine the distance d between the two points, and, in the plane. If the points lie on a horizontal line, then and the distance between the points is If the points lie on a vertical line, then and the distance between the points is.. If the two points do not lie on a horizontal or vertical line, the can be used to form a right triangle, as shown in Figure D.7. The length of the vertical side of the triangle is and the length of the horizontal side is B the Pthagorean Theorem, it follows that., d d. Replacing and b the equivalent epressions and produces the following result. Distance Formula The distance d between the points, and, in the plane is given b d. EXAMPLE Finding the Distance Between Two Points Find the distance between the points, and 3,. Solution d 3 Distance Formula Substitute for,,, and
3 D APPENDIX D Precalculus Review EXAMPLE Verifing a Right Triangle, ) d d Verifing a right triangle Figure D.8 d3, 0) 5, 7) Verif that the points,,, 0, and 5, 7 form the vertices of a right triangle. Solution Figure D.8 shows the triangle formed b the three points. The lengths of the three sides are as follows. Because and d d 0 5 d d d d 3 50 Sum of squares of sides Square of hpotenuse ou can appl the Pthagorean Theorem to conclude that the triangle is a right triangle. Each point of the form, 3) lies on this horizontal line., 3) 5, 3) EXAMPLE 3 Using the Distance Formula Find such that the distance between, 3 and, is 5. d = 5 Given a distance, find a point. Figure D.9 d = 5 5, ) Solution Using the Distance Formula, ou can write the following Distance Formula Square each side. Write in general form. Factor. Therefore, 5 or, and ou can conclude that there are two solutions. That is, each of the points 5, 3 and, 3 lies five units from the point,, as shown in Figure D.9. Midpoint of a line segment Figure D , 3) 3, 0) 9, 3) The coordinates of the midpoint of the line segment joining two points can be found b averaging the -coordinates of the two points and averaging the -coordinates of the two points. That is, the midpoint of the line segment joining the points, and, in the plane is,. Midpoint Formula For instance, the midpoint of the line segment joining the points 5, 3 and 9, 3 is 5 9, 3 3, 0 as shown in Figure D.0.
4 APPENDIX D. The Cartesian Plane D3 Center: h, k) Radius: r Point on circle:, ) Equations of Circles A circle can be defined as the set of all points in a plane that are equidistant from a fied point. The fied point is the center of the circle, and the distance between the center and a point on the circle is the radius see Figure D.). You can use the Distance Formula to write an equation for the circle with center h, k and radius r. Let, be an point on the circle. Then the distance between, and the center h, k is given b h k r. B squaring each side of this equation, ou obtain the standard form of the equation of a circle. Definition of a circle Figure D. Standard Form of the Equation of a Circle The point, lies on the circle of radius r and center h, k if and onl if h k r. The standard form of the equation of a circle with center at the origin, h, k 0, 0, is r. If r, the circle is called the unit circle. EXAMPLE Writing the Equation of a Circle, ) 8 3, ) The point 3, lies on a circle whose center is at,, as shown in Figure D.. Write the standard form of the equation of this circle. Solution The radius of the circle is the distance between, and 3,. r 3 0 You can write the standard form of the equation of this circle as 0 0. Write in standard form. Figure D. B squaring and simplifing, the equation h k r written in the following general form of the equation of a circle. can be A A D E F 0, A 0 To convert such an equation to the standard form h k p ou can use a process called completing the square. If p > 0, the graph of the equation is a circle. If p 0, the graph is the single point h, k. If p < 0, the equation has no graph.
5 D APPENDIX D Precalculus Review EXAMPLE 5 Completing the Square Sketch the graph of the circle whose general equation is Solution To complete the square, first divide b so that the coefficients of and are both Write original equation Divide b. A circle with a radius of and center at 5, r = Figure D.3 5, ) = 5 37 Group terms. Complete the square b adding 5 5 and to each side half half Write in standard form. Note that ou complete the square b adding the square of half the coefficient of and the square of half the coefficient of to each side of the equation. The circle is centered at and its radius is, as shown in Figure D.3. 5, 5 You have now reviewed some fundamental concepts of analtic geometr. Because these concepts are in common use toda, it is eas to overlook their revolutionar nature. At the time analtic geometr was being developed b Pierre de Fermat and René Descartes, the two major branches of mathematics geometr and algebra were largel independent of each other. Circles belonged to geometr and equations belonged to algebra. The coordination of the points on a circle and the solutions of an equation belongs to what is now called analtic geometr. It is important to become skilled in analtic geometr so that ou can move easil between geometr and algebra. For instance, in Eample, ou were given a geometric description of a circle and were asked to find an algebraic equation for the circle. So, ou were moving from geometr to algebra. Similarl, in Eample 5 ou were given an algebraic equation and asked to sketch a geometric picture. In this case, ou were moving from algebra to geometr. These two eamples illustrate the two most common problems in analtic geometr.. Given a graph, find its equation. Geometr Algebra. Given an equation, find its graph. Algebra Geometr In the net section, ou will review other eamples of these two tpes of problems.
6 APPENDIX D. The Cartesian Plane D5 EXERCISES FOR APPENDIX D. In Eercises, a) plot the points, b) find the distance between the points, and c) find the midpoint of the line segment joining the points ,,, 0,,,, 3, 3, 5, 5.,,, 5. 3.,, 3, 5. 5., 3,,. 3,, 3, 3, 3, 5,, 0, 0, In Eercises and, find such that the distance between the points is 5.. 0, 0,,.,,, In Eercises 7 0, determine the quadrants) in which, is located so that the conditions) is are) satisfied. 7. and > 0 8. < 9. > 0 0., is in Quadrant II. In Eercises, show that the points are the vertices of the polgon. A rhombus is a quadrilateral whose sides are all of the same length.) Vertices., 0,,,, 5 Right triangle., 3, 3,,, Isosceles triangle 3. 0, 0,,,,, 3, 3 Rhombus. 0, ), 3, 7,,,, Parallelogram 5. Number of Stores The table shows the number of Target stores for each ear from 99 through 003. Source: Target Corp.) Select reasonable scales on the coordinate aes and plot the points,.. Conjecture Plot the points,, 3, 5, and 7, 3 on a rectangular coordinate sstem. Then change the sign of the -coordinate of each point and plot the three new points on the same rectangular coordinate sstem. What conjecture can ou make about the location of a point when the sign of the -coordinate is changed? Repeat the eercise for the case in which the sign of the -coordinate is changed. In Eercises 7 0, use the Distance Formula to determine whether the points lie on the same line. 7. 0,,, 0, 3, 8. 0,, 7,, 5, Polgon Year, Number, Year, Number, In Eercises 3 and, find such that the distance between the points is , 0, 3,. 5,, 5, 5. Use the Midpoint Formula to find the three points that divide the line segment joining, and, into four equal parts.. Use the result of Eercise 5 to find the points that divide the line segment joining the given points into four equal parts. a),,, b), 3, 0, 0 In Eercises 7 30, match the equation with its graph. [The graphs are labeled a), b), c), and d).] a) c), 0) 0, 0) In Eercises 3 38, write the general form of the equation of the circle. b) d) 3. Center: 0, 0 3. Center: 0, 0 Radius: 3 Radius: Center:, 3. Center:, 3 Radius: Radius: 5 8, 3), ) 3
7 D APPENDIX D Precalculus Review 35. Center:, Point on circle: 0, 0 3. Center: 3, Point on circle:, 37. Endpoints of diameter:, 5,, 38. Endpoints of diameter:,,, 39. Satellite Communication Write the standard form of the equation for the path of a communications satellite in a circular orbit,000 miles above Earth. Assume that the radius of Earth is 000 miles.) 0. Building Design A circular air duct of diameter D is fit firml into the right-angle corner where a basement wall meets the floor see figure). Find the diameter of the largest water pipe that can be run in the right-angle corner behind the air duct. D In Eercises 8, write the standard form of the equation of the circle and sketch its graph In Eercises 9 and 50, use a graphing utilit to graph the equation. Use a square setting. Hint: It ma be necessar to solve the equation for and graph the resulting two equations.) 53. Prove that, 3 3 is one of the points of trisection of the line segment joining, and,. Find the midpoint of the line segment joining, 3 3 and, to find the second point of trisection. 5. Use the results of Eercise 53 to find the points of trisection of the line segment joining the following points. a),,, b), 3, 0, 0 True or False? In Eercises 55 58, determine whether the statement is true or false. If it is false, eplain wh or give an eample that shows it is false. 55. If ab < 0, the point a, b lies in either Quadrant II or Quadrant IV. 5. The distance between the points a b, a and a b, a is b. 57. If the distance between two points is zero, the two points must coincide. 58. If ab 0, the point a, b lies on the -ais or on the -ais. In Eercises 59, prove the statement. 59. The line segments joining the midpoints of the opposite sides of a quadrilateral bisect each other. 0. The perpendicular bisector of a chord of a circle passes through the center of the circle.. An angle inscribed in a semicircle is a right angle.. The midpoint of the line segment joining the points, and, is, In Eercises 5 and 5, sketch the set of all points satisfing the inequalit. Use a graphing utilit to verif our result >