# 3 Rectangular Coordinate System and Graphs

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1 060_CH03_ QXP 10/9/10 10:56 AM Page 13 3 Rectangular Coordinate Sstem and Graphs In This Chapter 3.1 The Rectangular Coordinate Sstem 3. Circles and Graphs 3.3 Equations of Lines 3.4 Variation Chapter 3 Review Eercises In Section 3.3 we will see that parallel lines have the same slope. A Bit of Histor Ever student of mathematics pas the French mathematician René Descartes ( ) homage whenever he or she sketches a graph. Descartes is considered the inventor of analtic geometr, which is the melding of algebra and geometr at the time thought to be completel unrelated fields of mathematics. In analtic geometr an equation involving two variables could be interpreted as a graph in a two-dimensional coordinate sstem embedded in a plane. The rectangular or Cartesian coordinate sstem is named in his honor. The basic tenets of analtic geometr were set forth in La Géométrie, published in The invention of the Cartesian plane and rectangular coordinates contributed significantl to the subsequent development of calculus b its co-inventors Isaac Newton ( ) and Gottfried Wilhelm Leibniz ( ). René Descartes was also a scientist and wrote on optics, astronom, and meteorolog. But beond his contributions to mathematics and science, Descartes is also remembered for his impact on philosoph. Indeed, he is often called the father of modern philosoph and his book Meditations on First Philosoph continues to be required reading to this da at some universities. His famous phrase cogito ergo sum (I think, therefore I am) appears in his Discourse on the Method and Principles of Philosoph. Although he claimed to be a fervent Catholic, the Church was suspicious of Descartes philosoph and writings on the soul, and placed all his works on the Inde of Prohibited Books in

2 060_CH03_ QXP 10/9/10 10:56 AM Page The Rectangular Coordinate Sstem Introduction In Section 1. we saw that each real number can be associated with eactl one point on the number, or coordinate, line. We now eamine a correspondence between points in a plane and ordered pairs of real numbers. The Coordinate Plane A rectangular coordinate sstem is formed b two perpendicular number lines that intersect at the point corresponding to the number 0 on each line. This point of intersection is called the origin and is denoted b the smbol O. The horizontal and vertical number lines are called the -ais and the -ais, respectivel. These aes divide the plane into four regions, called quadrants, which are numbered as shown in FIGURE 3.1.1(a). As we can see in Figure 3.1.1(b), the scales on the - and -aes need not be the same. Throughout this tet, if tick marks are not labeled on the coordinates aes, as in Figure 3.1.1(a), then ou ma assume that one tick corresponds to one unit. A plane containing a rectangular coordinate sstem is called an -plane, a coordinate plane, or simpl -space. II Second quadrant III Third quadrant O I First quadrant Origin IV Fourth quadrant (a) Four quadrants FIGURE Coordinate plane (b) Different scales on - and -aes The rectangular coordinate sstem and the coordinate plane are also called the Cartesian coordinate sstem and the Cartesian plane after the famous French mathematician and philosopher René Descartes ( ). Coordinates of a Point Let P represent a point in the coordinate plane. We associate an ordered pair of real numbers with P b drawing a vertical line from P to the -ais and a horizontal line from P to the -ais. If the vertical line intersects the -ais at the number a and the horizontal line intersects the -ais at the number b, we associate the ordered pair of real numbers (a, b) with the point. Conversel, to each ordered pair (a, b) of real numbers, there corresponds a point P in the plane. This point lies at the intersection of the vertical line through a on the -ais and the horizontal line passing through b on the -ais. Hereafter, we will refer to an ordered pair as a point and denote it b either P(a, b) or (a, b).* The number a is the -coordinate of the point and the number b is the -coordinate of the point and we sa that P has coordinates (a, b). For eample, the coordinates of the origin are (0, 0). See FIGURE *This is the same notation used to denote an open interval. It should be clear from the contet of the discussion whether we are considering a point (a, b) or an open interval (a, b). 14 CHAPTER 3 RECTANGULAR COORDINATE SYSTEM AND GRAPHS

4 060_CH03_ QXP 10/9/10 10:56 AM Page 16 We see from Figure that, 0 for all points (, ) in the second and fourth quadrants. Hence we can represent the set of points for which, 0 b the shaded regions in FIGURE The coordinate aes are shown as dashed lines to indicate that the points on these aes are not included in the solution set. (b) In Section.6 we saw that 00 \$ means that either \$ or #. Since is not restricted in an wa it can be an real number, and so the points (, ) for which \$ and `,,` or # and `,,` can be represented b the two shaded regions in FIGURE We use solid lines to represent the boundaries 5and 5 of the region to indicate that the points on these boundaries are included in the solution set. and < < < 0 < 0 FIGURE Region in the -plane satisfing condition in (a) of Eample 3 and < < FIGURE Region in the -plane satisfing condition in (b) of Eample 3 1 P 1 ( 1, 1 ) P 3 ( 1, ) 1 P (, ) FIGURE Distance between points P 1 and P d Distance Formula Suppose P 1 ( 1, 1 ) and P (, ) are two distinct points in the -plane that are not on a vertical line or on a horizontal line. As a consequence, P 1, P, and P 3 ( 1, ) are vertices of a right triangle as shown in FIGURE The length of the side P 3 P is 0 1 0, and the length of the side P 1 P 3 is If we denote the length of P 1 P b d, then b the Pthagorean theorem. Since the square of an real number is equal to the square of its absolute values, we can replace the absolute-value signs in (1) with parentheses. The distance formula given net follows immediatel from (1). THEOREM d Distance Formula (1) The distance between an two points P 1 ( 1, 1 ) and P (, ) in the -plane is given b d(p 1, P ) 5 "( 1 ) 1 ( 1 ). () Although we derived this equation for two points not on a vertical or horizontal line, () holds in these cases as well. Also, because ( 1 ) 5 ( 1 ), it makes no difference which point is used first in the distance formula, that is, d(p 1, P ) 5 d(p, P 1 ). 16 CHAPTER 3 RECTANGULAR COORDINATE SYSTEM AND GRAPHS

5 060_CH03_ QXP 10/9/10 10:56 AM Page 17 EXAMPLE 4 Distance Between Two Points Find the distance between the points A(8, 5) and B(3, 7). Solution From () with A and B plaing the parts of P 1 and P : B(3, 7) The distance d is illustrated in FIGURE d(a, B) 5 "(3 8) 1 (7 (5)) 5 "(5) 1 (1) 5! d(a, B) Three Points Form a Triangle Determine whether the points P 1 (7, 1), P (4, 1), and P 3 (4, 5) are the vertices of a right triangle. Solution From plane geometr we know that a triangle is a right triangle if and onl if the sum of the squares of the lengths of two of its sides is equal to the square of the length of the remaining side. Now, from the distance formula (), we have Since EXAMPLE 5 d(p 1, P ) 5 "(4 7) 1 (1 1) 5! !15, d(p, P 3 ) 5 "(4 (4)) 1 (5 (1)) 5! ! , d(p 3, P 1 ) 5 "(7 4) 1 (1 5) 5! ! [d(p 3, P 1 )] 1 [d(p, P 3 )] [d(p 1, P )], we conclude that P 1, P, and P 3 are the vertices of a right triangle with the right angle at P 3. See FIGURE P ( 4, 1) A(8, 5) FIGURE Distance between two points in Eample 4 P 3 (4, 5) P 1 (7, 1) FIGURE Triangle in Eample 5 Midpoint Formula In Section 1. we saw that the midpoint of a line segment between two numbers a and b on the number line is the average, (a 1 b)/. In the plane, each coordinate of the midpoint M of a line segment joining two points P 1 ( 1, 1 ) and P (, ), as shown in FIGURE , is the average of the corresponding coordinates of the endpoints of the intervals [ 1, ] and [ 1, ]. To prove this, we note in Figure that triangles P 1 CM and MDP are congruent since corresponding angles are equal and d(p 1, M) 5 d(m, P ). Hence, d(p 1, or Solving the last equation for gives 5 C) 5 d(m, D) Similarl, d(c, M) 5 d(d, P ) so that 1 5 and therefore We summarize the result. C P 1 ( 1, 1 ) D M(, ) P (, ) FIGURE M is the midpoint of the line segment joining P 1 and P THEOREM 3.1. Midpoint Formula The coordinates of the midpoint of the line segment joining the points P 1 ( 1, 1 ) and P (, ) in the -plane are given b a 1 1, 1 1 b. (3) 3.1 The Rectangular Coordinate Sstem 17

6 060_CH03_ QXP 10/9/10 10:56 AM Page 18 A(, 5) Midpoint (1, 3) B(4, 1) FIGURE Midpoint of line segment in Eample 6 EXAMPLE 6 Midpoint of a Line Segment Find the coordinates of the midpoint of the line segment joining A(, 5) and B(4, 1). Solution From formula (3) the coordinates of the midpoint of the line segment joining the points A and B are given b a 1 4, b or This point is indicated in red in FIGURE (1, 3). 3.1 Eercises Answers to selected odd-numbered problems begin on page ANS-5. In Problems 1 4, plot the given points. 1. (, 3), (4, 5), (0, ), (1, 3). 3. ( 1, ), (0, 0), (1, 4 3), (3, 3) 4. (1, 4), (3, 0), (4, ), (1, 1) (0, 0.8), (, 0), (1., 1.), (, ) In Problems 5 16, determine the quadrant in which the given point lies if (a, b) is in quadrant I. 5. (a, b) 6. (a, b) 7. (a, b) 8. (b, a) 9. (b, a) 10. (b, a) 11. (a, a) 1. (b, b) 13. (a, a) 14. (a, a) 15. (b, a) 16. (b, b) 17. Plot the points given in Problems 5 16 if (a, b) is the point shown in FIGURE Give the coordinates of the points shown in FIGURE (a, b) FIGURE Point (a, b) in Problem CHAPTER 3 RECTANGULAR COORDINATE SYSTEM AND GRAPHS a b A B D C E F FIGURE Points in Problem The points (, 0), (, 6), and (3, 0) are vertices of a rectangle. Find the fourth verte. 0. Describe the set of all points (, ) in the coordinate plane. The set of all points (, ). In Problems 1 6, sketch the set of points (, ) in the -plane that satisf the given conditions # 1 and 00 # 4. # and \$ # 1 In Problems 7 3, find the distance between the given points. 7. A(1, ), B(3, 4) 8. A(1, 3), B(5, 0) 9. A(, 4), B(4, 4) 30. A(1, 3), B(5, 7) 31. A( 3, 1), B( 5, ) 3. A( 5 3, 4), B( 3, 1) 1 1 G

7 060_CH03_ QXP 10/9/10 10:56 AM Page 19 In Problems 33 36, determine whether the points A, B, and C are vertices of a right triangle. 33. A(8, 1), B(3, 1), C(10, 5) 34. A(, 1), B(8, ), C(1, 11) 35. A(, 8), B(0, 3), C(6, 5) 36. A(4, 0), B(1, 1), C(, 3) 37. Determine whether the points A(0, 0), B(3, 4), and C(7, 7) are vertices of an isosceles triangle. 38. Find all points on the -ais that are 5 units from the point (4, 4). 39. Consider the line segment joining A(1, ) and B(3, 4). (a) Find an equation that epresses the fact that a point P(, ) is equidistant from A and from B. (b) Describe geometricall the set of points described b the equation in part (a). 40. Use the distance formula to determine whether the points A(1, 5), B(, 4), and C(4, 10) lie on a straight line. 41. Find all points with -coordinate 6 such that the distance from each point to (1, ) is! Which point, or (0.5, 0.97), is closer to the origin? (1/!, 1/! ) In Problems 43 48, find the midpoint of the line segment joining the points A and B. 43. A(4, 1), B(, 4) 44. A( 3, 1), B( 7 3, 3) 45. A(1, 0), B(8, 5) 46. A( 1, 3 ), B( 5, 1) 47. A(a, 3b), B(4a, 6b) 48. A(, ), B(, 1 ) In Problems 49 5, find the point B if M is the midpoint of the line segment joining points A and B. 49. A(, 1), M( 3, 0) 50. A(4, 1 ), M(7, 5 ) 51. A(5, 8), M(1, 1) 5. A(10, ), M(5, 1) 53. Find the distance from the midpoint of the line segment joining A(1, 3) and B(3, 5) to the midpoint of the line segment joining C(4, 6) and D(, 10). 54. Find all points on the -ais that are 3 units from the midpoint of the line segment joining (5, ) and (5, 6). 55. The -ais is the perpendicular bisector of the line segment through A(, 5) and B(, ). Find and. 56. Consider the line segment joining the points A(0, 0) and B(6, 0). Find a point C(, ) in the first quadrant such that A, B, and C are vertices of an equilateral triangle. 57. Find points P 1 ( 1, 1 ), P (, ), and P 3 ( 3, 3 ) on the line segment joining A(3, 6) and B(5, 8) that divide the line segment into four equal parts. Miscellaneous Applications 58. Going to Chicago Kansas Cit and Chicago are not directl connected b an interstate highwa, but each cit is connected to St. Louis and Des Moines. See FIGURE Des Moines is approimatel 40 mi east and 180 mi north of Kansas Cit, St. Louis is approimatel 30 mi east and 40 mi south of Kansas Cit, and Chicago is approimatel 360 mi east and 00 mi north of Kansas Cit. Assume that this part of the Midwest is a flat plane and that the connecting highwas are straight lines. Which route from Kansas Cit to Chicago, through St. Louis or through Des Moines, is shorter? Kansas Cit Des Moines St. Louis Chicago FIGURE Map for Problem The Rectangular Coordinate Sstem 19

8 060_CH03_ QXP 10/9/10 10:56 AM Page 130 For Discussion 59. The points A(1, 0), B(5, 0), C(4, 6), and D(8, 6) are vertices of a parallelogram. Discuss: How can it be shown that the diagonals of the parallelogram bisect each other? Carr out our ideas. 60. The points A(0, 0), B(a, 0), and C(a, b) are vertices of a right triangle. Discuss: How can it be shown that the midpoint of the hpotenuse is equidistant from the vertices? Carr out our ideas. 3. Circles and Graphs Introduction In Chapter we studied equations as an equalit of two algebraic quantities involving one variable. Our goal then was to find the solution set of the equation. In this and subsequent sections that follow we stud equations in two variables, sa and. Such an equation is simpl a mathematical statement that asserts two quantities involving these variables are equal. In the fields of the phsical sciences, engineering, and business, equations in two (or more) variables are a means of communication. For eample, if a phsicist wants to tell someone how far a rock dropped from a great height travels in a certain time t, he or she will write s 5 16t. A mathematician will look at s 5 16t and immediatel classif it as a certain tpe of equation. The classification of an equation carries with it information about properties shared b all equations of that kind. The remainder of this tet is devoted to eamining different kinds of equations involving two or more variables and studing their properties. Here is a sample of some of the equations in two variables that ou will see: 5 1, 1 5 1, 5, 5!, 5 5 1, 5 3 3, 5, 5 ln, 5 1, , (1) (1, 3) (1, 0) = 1 FIGURE 3..1 Graph of equation 5 1 Terminolog A solution of an equation in two variables and is an ordered pair of numbers (a, b) that ields a true statement when 5 a and 5 b are substituted into the equation. For eample, (, 4) is a solution of the equation 5 because 5 4 T T () is a true statement. We also sa that the coordinates (, 4) satisf the equation. As in Chapter, the set of all solutions of an equation is called its solution set. Two equations are said to be equivalent if the have the same solution set. For eample, we will see in Eample 4 of this section that the equation is equivalent to ( 1 5) 1 ( 1) 5 3. In the list given in (1), ou might object that the first equation 5 1 does not involve two variables. It is a matter of interpretation! Because there is no eplicit dependence in the equation, the solution set of 5 1 can be interpreted to mean the set 5(, ) 0 5 1, where is an real number6. The solutions of 5 1 are then ordered pairs (1, ), where ou are free to choose arbitraril so long as it is a real number. For eample, (1, 0) and (1, 3) are solutions of the equation 5 1. The graph of an equation is the visual representation in the rectangular coordinate sstem of the set of points whose coordinates (a, b) satisf the equation. The graph of 5 1 is the vertical line shown in FIGURE CHAPTER 3 RECTANGULAR COORDINATE SYSTEM AND GRAPHS

9 060_CH03_ QXP 10/9/10 10:56 AM Page 131 Circles The distance formula discussed in the preceding section can be used to define a set of points in the coordinate plane. One such important set is defined as follows. DEFINITION 3..1 Circle A circle is the set of all points P(, ) in the -plane that are a given distance r, called the radius, from a given fied point C, called the center. If the center has coordinates C(h, k), then from the preceding definition a point P(, ) lies on a circle of radius r if and onl if Since ( h) 1 ( k) is alwas nonnegative, we obtain an equivalent equation when both sides are squared. We conclude that a circle of radius r and center C(h, k) has the equation ( h) 1 ( k) 5 r. () In FIGURE 3.. we have sketched a tpical graph of an equation of the form given in (). Equation () is called the standard form of the equation of a circle. We note that the smbols h and k in () represent real numbers and as such can be positive, zero, or negative. When h 5 0, k 5 0, we see that the standard form of the equation of a circle with center at the origin is See FIGURE When r 5 1, we sa that () or (3) is an equation of a unit circle. For eample, is an equation of a unit circle centered at the origin. EXAMPLE 1 d(p,c) 5 r or "( h) 1 ( k) 5 r. 1 5 r. Center and Radius Find the center and radius of the circle whose equation is ( 8) 1 ( 1 ) Solution To obtain the standard form of the equation, we rewrite (4) as ( 8) 1 ( ()) 5 7. Comparing this last form with () we identif h 5 8, k 5, and r 5 7. Thus the circle is centered at (8, ) and has radius 7. (3) (4) C(h, k) r P(, ) FIGURE 3.. Circle with radius r and center (h, k) (0, r) ( r, 0) (r, 0) (0, r) FIGURE 3..3 Circle with radius r and center (0, 0) EXAMPLE Equation of a Circle Find an equation of the circle with center C(5, 4) with radius!. Solution Substituting h 55, k 5 4, and r 5! in () we obtain ( (5)) 1 ( 4) 5 (!) or ( 1 5) 1 ( 4) 5. EXAMPLE 3 Equation of a Circle Find an equation of the circle with center C(4, 3) and passing through P(1, 4). Solution With h 5 4 and k 5 3, we have from () ( 4) 1 ( 3) 5 r. (5) 3. Circles and Graphs 131

11 060_CH03_ QXP 10/9/10 10:56 AM Page 133 It is possible that an epression for which we must complete the square has a leading coefficient other than 1. For eample, Note c c is an equation of a circle. As in Eample 4 we start b rearranging the equation: (3 18 ) 1 (3 1 6 ) 5. Now, however, we must do one etra step before attempting completion of the square; that is, we must divide both sides of the equation b 3 so that the coefficients of and are each 1: ( 6 ) 1 ( 1 ) 5. 3 At this point we can now add the appropriate numbers within each set of parentheses and to the right-hand side of the equalit. You should verif that the resulting standard form is ( 3) 1 ( 1 1) Semicircles If we solve (3) for we get 5 r or 5 ; "r. This last epression is equivalent to the two equations 5 "r and 5"r. In like manner if we solve (3) for we obtain 5 "r and 5"r. B convention, the smbol! denotes a nonnegative quantit, thus the -values defined b an equation such as 5 "r are nonnegative. The graphs of the four equations highlighted in color are, in turn, the upper half, lower half, right half, and the left half of the circle shown in Figure Each graph in FIGURE 3..6 is called a semicircle. (0, r) ( r, 0) (r, 0) = r = r ( r, 0) (r, 0) (0, r) (a) Upper half (b) Lower half (0, r) (0, r) = r (r, 0) ( r, 0) = r (0, r) (c) Right half FIGURE 3..6 Semicircles (0, r) (d) Left half 3. Circles and Graphs 133

12 060_CH03_ QXP 10/9/10 10:56 AM Page 134 Inequalities One last point about circles. On occasion we encounter problems where we must sketch the set of points in the -plane whose coordinates satisf inequalities such as 1, r or 1 \$ r. The equation 1 5 r describes the set of points (, ) whose distance to the origin (0, 0) is eactl r. Therefore, the inequalit 1, r describes the set of points (, ) whose distance to the origin is less than r. In other words, the points (, ) whose coordinates satisf the inequalit 1, r are in the interior of the circle. Similarl, the points (, ) whose coordinates satisf 1 \$ r lie either on the circle or are eterior to it. Graphs It is difficult to read a newspaper, read a science or business tet, surf the Internet, or even watch the news on TV without seeing graphical representations of data. It ma even be impossible to get past the first page in a mathematics tet without seeing some kind of graph. So man diverse quantities are connected b means of equations and so man questions about the behavior of the quantities linked b the equation can be answered b means of a graph, that the abilit to graph equations quickl and accuratel like the abilit to do algebra quickl and accuratel is high on the list of skills essential to our success in a course in calculus. For the rest of this section we will talk about graphs in general, and more specificall, about two important aspects of graphs of equations. Intercepts Locating the points at which the graph of an equation crosses the coordinates aes can be helpful when sketching a graph b hand. The -intercepts of a graph of an equation are the points at which the graph crosses the -ais. Since ever point on the -ais has -coordinate 0, the -coordinates of these points (if there are an) can be found from the given equation b setting 5 0 and solving for. In turn, the -intercepts of the graph of an equation are the points at which its graph crosses the -ais. The -coordinates of these points can be found b setting 5 0 in the equation and solving for. See FIGURE intercepts -intercepts (a) Five intercepts (b) Two -intercepts (c) Graph has no intercepts FIGURE 3..7 Intercepts of a graph EXAMPLE 5 Intercepts Find the intercepts of the graphs of the equations (a) 5 9 (b) Solution (a) To find the -intercepts we set 5 0 and solve the resulting equation 5 9 for : or ( 1 3)( 3) 5 0 gives 53 and 5 3. The -intercepts of the graph are the points (3, 0) and (3, 0). To find the -intercepts we set 5 0 and solve 5 9 or 59for. Because there are no real numbers whose square is negative we conclude the graph of the equation does not cross the -ais. 134 CHAPTER 3 RECTANGULAR COORDINATE SYSTEM AND GRAPHS

13 060_CH03_ QXP 10/9/10 10:56 AM Page 135 (b) Setting 5 0 ields This is a quadratic equation and can be solved either b factoring or b the quadratic formula. Factoring gives ( 1 4)( 3) 5 0 and so 54and 5 3. The -intercepts of the graph are the points (4, 0) and ( 3, 0). Now, setting 5 0 in the equation immediatel gives 51. The -intercept of the graph is the point (0, 1). EXAMPLE 6 Eample 4 Revisited Let s return to the circle in Eample 4 and determine its intercepts from equation (7). Setting 0 in and using the quadratic formula to solve shows the -intercepts of this circle are (5!, 0) and (5 1!, 0). If we let 5 0, then the quadratic formula shows that the roots of the equation are comple numbers. As seen in Figure 3..5, the circle does not cross the -ais. Smmetr A graph can also possess smmetr. You ma alread know that the graph of the equation 5 is called a parabola. FIGURE 3..8 shows that the graph of 5 is smmetric with respect to the -ais since the portion of the graph that lies in the second quadrant is the mirror image or reflection of that portion of the graph in the first quadrant. In general, a graph is smmetric with respect to the -ais if whenever (, ) is a point on the graph, (, ) is also a point on the graph. Note in Figure 3..8 that the points (1, 1) and (, 4) are on the graph. Because the graph possesses -ais smmetr, the points (1, 1) and (, 4) must also be on the graph. A graph is said to be smmetric with respect to the -ais if whenever (, ) is a point on the graph, (, ) is also a point on the graph. Finall, a graph is smmetric with respect to the origin if whenever (, ) is on the graph, (, ) is also a point on the graph. FIGURE 3..9 illustrates these three tpes of smmetries. = (, 4) (, 4) ( 1, 1) (1, 1) FIGURE 3..8 Graph with -ais smmetr (, ) (, ) (, ) (, ) (, ) (, ) (a) Smmetr with respect to the -ais FIGURE 3..9 Smmetries of a graph (b) Smmetr with respect to the -ais (c) Smmetr with respect to the origin Observe that the graph of the circle given in Figure 3..3 possesses all three of these smmetries. As a practical matter we would like to know whether a graph possesses an smmetr in advance of plotting its graph. This can be done b appling the following tests to the equation that defines the graph. 3. Circles and Graphs 135

14 060_CH03_ QXP 10/9/10 10:56 AM Page 136 THEOREM 3..1 Tests for Smmetr The graph of an equation is smmetric with respect to the (i) -ais if replacing b results in an equivalent equation; (ii) -ais if replacing b results in an equivalent equation; (iii) origin if replacing and b and results in an equivalent equation. The advantage of using smmetr in graphing should be apparent: If sa, the graph of an equation is smmetric with respect to the -ais, then we need onl produce the graph for \$ 0 since points on the graph for, 0 are obtained b taking the mirror images, through the -ais, of the points in the first and second quadrants. EXAMPLE 7 Test for Smmetr B replacing b in the equation 5 and using () 5, we see that 5 () is equivalent to 5. This proves what is apparent in Figure 3..8; that is, the graph of 5 is smmetric with respect to the -ais. (0, 10) (0, 10) (10, 0) FIGURE Graph of equation in Eample 8 EXAMPLE 8 Intercepts and Smmetr Determine the intercepts and an smmetr for the graph of Solution Intercepts: Setting 5 0 in equation (8) immediatel gives The graph of the equation has a single -intercept, (10, 0). When 5 0, we get 5 10, which implies that 5!10 or 5!10. Thus there are two -intercepts, (0,!10) and (0,!10 ). Smmetr: If we replace b in the equation , we get This is not equivalent to equation (8). You should also verif that replacing and b and in (8) does not ield an equivalent equation. However, if we replace b, we find that 1 () 5 10 is equivalent to Thus, the graph of the equation is smmetric with respect to the -ais. Graph: In the graph of the equation given in FIGURE the intercepts are indicated and the -ais smmetr should be apparent. (8) 3. Eercises Answers to selected odd-numbered problems begin on page ANS-5. In Problems 1 6, find the center and the radius of the given circle. Sketch its graph ( 3) ( 1 ) ( 1 3) 1 ( 5) 5 5 ( 1 ) 1 ( 3 ) CHAPTER 3 RECTANGULAR COORDINATE SYSTEM AND GRAPHS

15 060_CH03_ QXP 10/9/10 10:57 AM Page 137 In Problems 7 14, complete the square in and to find the center and the radius of the given circle In Problems 15 4, find an equation of the circle that satisfies the given conditions. 15. center (0, 0), radius center (1, 3), radius center (0, ), radius! 18. center (9, 4), radius 19. endpoints of a diameter at (1, 4) and (3, 8) 0. endpoints of a diameter at (4, ) and (3, 5) 1. center (0, 0), graph passes through (1, ). center (4, 5), graph passes through (7, 3) 3. center (5, 6), graph tangent to the -ais 4. center (4, 3), graph tangent to the -ais In Problems 5 8, sketch the semicircle defined b the given equation " " "1 ( 1) 8. 5"9 ( 3) 9. Find an equation for the upper half of the circle 1 ( 3) 5 4. The right half of the circle. 30. Find an equation for the lower half of the circle ( 5) 1 ( 1) 5 9. The left half of the circle. In Problems 31 34, sketch the set of points in the -plane whose coordinates satisf the given inequalit \$ 9 3. ( 1) 1 ( 1 5) # # 1 # In Problems 35 and 36, find the - and -intercepts of the given circle. 35. The circle with center (3, 6) and radius The circle In Problems 37 6, find an intercepts of the graph of the given equation. Determine whether the graph of the equation possesses smmetr with respect to the -ais, -ais, or origin. Do not graph ( 1 4) ( 3) ( ) ( 1 ) 49. 5" Circles and Graphs 137

16 060_CH03_ QXP 10/9/10 10:57 AM Page ! ( 1 )( 8) ! In Problems 63 66, state all the smmetries of the given graph FIGURE Graph for Problem 63 FIGURE 3..1 Graph for Problem FIGURE Graph for Problem 65 FIGURE Graph for Problem 66 In Problems 67 7, use smmetr to complete the given graph. 67. The graph is smmetric 68. The graph is smmetric with respect to the -ais. with respect to the -ais. FIGURE Graph for Problem 67 FIGURE Graph for Problem The graph is smmetric 70. The graph is smmetric with respect to the origin. with respect to the -ais. FIGURE Graph for Problem 69 FIGURE Graph for Problem CHAPTER 3 RECTANGULAR COORDINATE SYSTEM AND GRAPHS

17 060_CH03_ QXP 10/9/10 10:57 AM Page The graph is smmetric 7. The graph is smmetric with respect to the - and -aes. with respect to the origin. FIGURE Graph for Problem 71 For Discussion FIGURE 3..0 Graph for Problem Determine whether the following statement is true or false. Defend our answer. If a graph has two of the three smmetries defined on page 135, then the graph must necessaril possess the third smmetr. 74. (a) The radius of the circle in FIGURE 3..1(a) is r. What is its equation in standard form? (b) The center of the circle in Figure 3..1(b) is (h, k). What is its equation in standard form? 75. Discuss whether the following statement is true or false. Ever equation of the form 1 1 a 1 b 1 c 5 0 is a circle. (a) (b) FIGURE 3..1 Circles in Problem Equations of Lines Introduction An pair of distinct points in the -plane determines a unique straight line. Our goal in this section is to find equations of lines. Fundamental to finding equations of lines is the concept of slope of a line. (, ) Rise ( 1, 1 ) Run 1 Slope If P 1 ( 1, 1 ) and P (, ) are two points such that 1, then the number m 5 1 (1) 1 is called the slope of the line determined b these two points. It is customar to call 1 the change in or rise of the line; 1 is the change in or the run of the line. Therefore, the slope (1) of a line is m 5 rise () run. See FIGURE 3.3.1(a). An pair of distinct points on a line will determine the same slope. To see wh this is so, consider the two similar right triangles in Figure 3.3.1(b). Since we know that the ratios of corresponding sides in similar triangles are equal we have Hence the slope of a line is independent of the choice of points on the line. 1 1 (a) Rise and run ( 1, 1 ) ( 4, 4 ) ( 3, 3 ) Rise 4 3 (, ) Run 4 3 Rise 1 Run 1 (b) Similar triangles FIGURE Slope of a line 3.3 Equations of Lines 139

18 060_CH03_ QXP 10/9/10 10:57 AM Page 140 In FIGURE 3.3. we compare the graphs of lines with positive, negative, zero, and undefined slopes. In Figure 3.3.(a) we see, reading the graph left to right, that a line with positive slope (m. 0) rises as increases. Figure 3.3.(b) shows that a line with negative slope (m, 0) falls as increases. If P 1 ( 1, 1 ) and P (, ) are points on a horizontal line, then 1 5 and so its rise is Hence from (1) the slope is zero (m 5 0). See Figure 3.3.(c). If P 1 ( 1, 1 ) and P (, ) are points on a vertical line, then 1 5 and so its run is In this case we sa that the slope of the line is undefined or that the line has no slope. See Figure 3.3.(d). ( 1, 1 ) (, ) Run > 0 Rise > 0 ( 1, 1 ) Run > 0 ( 1, 1 ) (, 1 ) Rise < 0 Rise = 0 (, ) ( 1, ) Run = 0 ( 1, 1 ) (a) m > 0 (b) m < 0 (c) m = 0 (d) m undefined FIGURE 3.3. Lines with slope (a) (c); line with no slope (d) Run = 5 P 1 (, 6) Rise = 10 P (3, 4) FIGURE Line in Eample 1 In general, since it does not matter which of the two points is called P 1 ( 1, 1 ) and which is called P (, ) in (1). EXAMPLE ( 1 ) 1 ( 1 ) 5 1, 1 Slope and Graph Find the slope of the line through the points (, 6) and (3, 4). Graph the line. Solution Let (, 6) be the point P 1 ( 1, 1 ) and (3, 4) be the point P (, ). The slope of the straight line through these points is m () Thus the slope is, and the line through P 1 and P is shown in FIGURE Point-Slope Equation We are now in a position to find an equation of a line L. To begin, suppose L has slope m and that P 1 ( 1, 1 ) is on the line. If P(, ) represents an other point on L, then (1) gives m Multipling both sides of the last equalit b 1 gives an important equation. THEOREM Point-Slope Equation The point-slope equation of the line through P 1 ( 1, 1 ) with slope m is 1 5 m( 1 ). (3) 140 CHAPTER 3 RECTANGULAR COORDINATE SYSTEM AND GRAPHS

19 060_CH03_ QXP 10/9/10 10:57 AM Page 141 EXAMPLE Point-Slope Equation Find an equation of the line with slope 6 and passing through ( 1, ). Solution Letting m 5 6, 1 5 1, and 1 5 we obtain from (3) 5 6[ ( 1 )]. Simplifing gives 5 6( 1 1 ) or EXAMPLE 3 Point-Slope Equation Find an equation of the line passing through the points (4, 3) and (, 5). Solution First we compute the slope of the line through the points. From (1), The point-slope equation (3) then gives m the distributive law T T ( 4) or Slope-Intercept Equation An line with slope (that is, an line that is not vertical) must cross the -ais. If this -intercept is (0, b), then with 1 5 0, 1 5 b, the point-slope form (3) gives b 5 m( 0). The last equation simplifies to the net result. THEOREM 3.3. Slope-Intercept Equation The slope-intercept equation of the line with slope m and -intercept (0, b) is 5 m 1 b. (4) The distributive law a(b 1 c) 5 ab 1 ac is the source of man errors on students papers. A common error goes something like this: ( 3) 5 3. The correct result is: ( 3) 5 (1)( 3) 5 (1) (1) = = 4 = 1 = = 4 = = 1 = When b 5 0 in (4), the equation 5 m represents a famil of lines that pass through the origin (0, 0). In FIGURE we have drawn a few of the members of that famil. FIGURE Lines through the origin are 5 m EXAMPLE 4 Eample 3 Revisited We can also use the slope-intercept form (4) to obtain an equation of the line through the two points in Eample 3. As in that eample, we start b finding the slope m The equation of the line is then b. B substituting the coordinates of either point (4, 3) or (, 5) into the last equation enables us to determine b. If we use and, then # and so b The equation of the line is b Horizontal and Vertical Lines We saw in Figure 3.3.(c) that a horizontal line has slope m 5 0. An equation of a horizontal line passing through a point (a, b) can be obtained from (3), that is, b 5 0( a) or 5 b. 3.3 Equations of Lines 141

20 060_CH03_ QXP 10/9/10 10:57 AM Page 14 THEOREM Equation of Horizontal Line The equation of a horizontal line with -intercept (0, b) is 5 b. (5) A vertical line through (a, b) has undefined slope and all points on the line have the same -coordinate. The equation of a vertical line is then THEOREM Equation of Vertical Line The equation of a vertical line with -intercept (a, 0) is 5 a. (6) = 1 (3, 1) = 3 FIGURE Horizontal and vertical lines in Eample 5 EXAMPLE 5 Vertical and Horizontal Lines Find equations for the vertical and horizontal lines through (3, 1). Graph these lines. Solution An point on the vertical line through (3, 1) has -coordinate 3. The equation of this line is then 5 3. Similarl, an point on the horizontal line through (3, 1) has -coordinate 1. The equation of this line is 51. Both lines are graphed in FIGURE Linear Equation The equations (3), (4), (5), and (6) are special cases of the general linear equation in two variables and a 1 b 1 c 5 0, where a and b are real constants and not both zero. The characteristic that gives (7) its name linear is that the variables and appear onl to the first power. Observe that (7) a 5 0, b 0, gives 5 c b, d horizontal line a 0, b 5 0, gives 5 c a, d vertical line a 0, b 0, gives 5 a b c b. d line with nonzero slope EXAMPLE 6 Slope and -intercept Find the slope and the -intercept of the line Solution We solve the linear equation for : Comparing the last equation with (4) we see that the slope of the line is m 5 3 and the -intercept is (0, 5 7 7). If the - and -intercepts are distinct, the graph of the line can be drawn through the corresponding points on the - and -aes. 14 CHAPTER 3 RECTANGULAR COORDINATE SYSTEM AND GRAPHS

21 060_CH03_ QXP 10/9/10 10:57 AM Page 143 EXAMPLE 7 Graph the linear equation Graph of a Linear Equation Solution There is no need to rewrite the linear equation in the form 5 m 1 b. We simpl find the intercepts. -intercept: Setting 5 0 gives or 5 4. The -intercept is (0, 4). -intercept: Setting 5 0 gives or The -intercept is ( 8 3, 0). As shown in FIGURE 3.3.6, the line is drawn through the two intercepts (0, 4) and ( 8 3, 0). Parallel and Perpendicular Lines Suppose L 1 and L are two distinct lines with slope. This assumption means that both L 1 and L are nonvertical lines. Then necessaril L 1 and L are either parallel or the intersect. If the lines intersect at a right angle the are said to be perpendicular. We can determine whether two lines are parallel or are perpendicular b eamining their slopes. (0, 4) ( 8, 0) = 0 FIGURE Line in Eample 7 THEOREM Parallel and Perpendicular Lines If L 1 and L are lines with slopes m 1 and m, respectivel, then (i) L 1 is parallel to L if and onl if m 1 5 m, and (ii) L 1 is perpendicular to if and onl if m 1 m 51. L There are several was of proving the two parts of Theorem The proof of part (i) can be obtained using similar right triangles, as in FIGURE 3.3.7, and the fact that the ratios of corresponding sides in such triangles are equal. We leave the justification of part (ii) as an eercise. See Problems 49 and 50 in Eercises 3.3. Note that the condition m 1 m 51implies that m 51/m 1, that is, the slopes are negative reciprocals of each other. A horizontal line 5 b and a vertical line 5 a are perpendicular, but the latter is a line with no slope. EXAMPLE 8 Parallel Lines The linear equations and can be rewritten in the slopeintercept forms and , respectivel. As noted in color in the preceding line the slope of each line is 3. Therefore the lines are parallel. The graphs of these equations are shown in FIGURE EXAMPLE 9 Perpendicular Lines Find an equation of the line through (0, 3) that is perpendicular to the graph of Solution We epress the given linear equation in slope-intercept form: implies Dividing b 3 gives This line, whose graph is given in blue in FIGURE , has slope 3. The slope of an line perpendicular to it is the negative reciprocal of 4 3, namel, 3 4. Since (0, 3) is the -intercept of the required line, it follows from (4) that its equation is The graph of the last equation is the red line in Figure parallel lines L 1 Run L Rise Run Rise FIGURE Parallel lines = 3 + = FIGURE Parallel lines in Eample 8 3 = 3 4 FIGURE Perpendicular lines in Eample 9 4 = + 3 (0, 3) 3.3 Equations of Lines 143

22 060_CH03_ QXP 10/9/10 10:57 AM Page Eercises Answers to selected odd-numbered problems begin on page ANS-6. In Problems 1 6, find the slope of the line through the given points. Graph the line through the points. 1. (3, 7), (1, 0). (4, 1), (1, 1) 3. (5, ), (4, 3) 4. (1, 4), (6, ) 5. (1, ), (3, ) 6. (8, 1 ), (, 5 ) In Problems 7 and 8, use the graph of the given line to estimate its slope FIGURE Graph for Problem 7 5 FIGURE Graph for Problem 8 5 In Problems 9 16, find the slope and the - and -intercepts of the given line. Graph the line In Problems 17, find an equation of the line through (1, ) with the indicated slope undefined In Problems 3 36, find an equation of the line that satisfies the given conditions. 3. Through (, 3) and (6, 5) 4. Through (5, 6) and (4, 0) 5. Through (8, 1) and (3, 1) 6. Through (, ) and (, ) 7. Through (, 0) and (, 6) 8. Through (0, 0) and (a, b) 9. Through (, 4) parallel to Through (1, 3) parallel to Through (5, 7) parallel to the -ais 3. Through the origin parallel to the line through (1, 0) and (, 6) 33. Through (, 3) perpendicular to Through (0, ) perpendicular to Through (5, 4) perpendicular to the line through (1, 1) and (3, 11) 36. Through the origin perpendicular to ever line with slope In Problems 37 40, determine which of the given lines are parallel to each other and which are perpendicular to each other. 37. (a) (b) 5 53 (c) (d) (e) (f) CHAPTER 3 RECTANGULAR COORDINATE SYSTEM AND GRAPHS

23 060_CH03_ QXP 10/9/10 10:57 AM Page (a) (b) 5 (c) (d) 5 4 (e) (f) (a) (b) (c) (d) (e) (f) (a) (b) 5 7 (c) (d) (e) (f) Find an equation of the line L shown in FIGURE if an equation of the blue curve is A tangent to a circle is defined to be a straight line that touches the circle at onl one point P. Find an equation of the tangent line L shown in FIGURE L P L 1 FIGURE Graphs in Problem 41 3 FIGURE Circle and tangent line in Problem 4 For Discussion 43. How would ou find an equation of the line that is the perpendicular bisector of the line segment through ( 1, 10) and ( 3, 4)? 44. Using onl the concepts of this section, how would ou prove or disprove that the triangle with vertices (, 3), (1, 3), and (4, ) is a right triangle? 45. Using onl the concepts of this section, how would ou prove or disprove that the quadrilateral with vertices (0, 4), (1, 3), (, 8), and (3, 7) is a parallelogram? 46. If C is an arbitrar real constant, an equation such as 3 5 C is said to define a famil of lines. Choose four different values of C and plot the corresponding lines on the same coordinate aes. What is true about the lines that are members of this famil? 47. Find the equations of the lines through (0, 4) that are tangent to the circle For the line a 1 b 1 c 5 0, what can be said about a, b, and c if (a) the line passes through the origin, (b) the slope of the line is 0, (c) the slope of the line is undefined? In Problems 49 and 50, to prove part (ii) of Theorem ou have to prove two things, the onl if part (Problem 49) and then the if part (Problem 50) of the theorem. 49. In FIGURE , without loss of generalit, we have assumed that two perpendicular lines 5 m 1, m 1. 0, and 5 m, m, 0, intersect at the origin. Use the information in the figure to prove the onl if part: If L 1 and L are perpendicular lines with slopes m 1 and m, then m 1 m Reverse our argument in Problem 49 to prove the if part: If L 1 and L are lines with slopes m 1 and m such that m 1 m 51, then L 1 and L are perpendicular. L L 1 (1, m 1 ) 1 (1, m ) = m 1 FIGURE Lines through origin in Problems 49 and 50 = m m 1 m 3.3 Equations of Lines 145

24 060_CH03_ QXP 10/9/10 10:57 AM Page Variation Introduction In man disciplines, a mathematical description b means of an equation, or mathematical model, of a real-life problem can be constructed using the notion of proportionalit. For eample, in one model of a growing population (sa, bacteria) it is assumed that the rate of growth at time t is directl proportional to the population at that time. If we let R represent the rate of growth, P the population, then the preceding sentence translates into R ~ P, where the smbol ~ is read proportional to. The mathematical statement in (1) is an eample of variation. In this section we eamine four tpes of variation: direct, inverse, joint, and combined. Each of these tpes of variation produce an equation in two or more variables. Direct Variation We begin with the formal definition of direct variation. (1) DEFINITION Direct Variation A quantit varies directl, or is directl proportional to, a quantit if there eists a nonzero number k such that 5 k. () In () we sa that the number k is the constant of proportionalit. Comparing () with Figure we know that the graph of an equation of the form given in () is a line through the origin with slope k. FIGURE illustrates the graph of () in the case of when k. 0 and \$ 0. Of course, smbols other than and are often used in (). In the stud of springs in phsics, the force F required to keep a spring stretched units beond its natural, or unstretched, length is assumed to be directl proportional to the elongation, that is, F 5 k. See FIGURE The result in (3) is called Hooke s law after the irascible English phsicist Robert Hooke ( ). (3) Natural length = k, k > 0 Stretched spring Force is F = k FIGURE Graph of 5 k, k. 0, \$ 0. 0 Elongation FIGURE 3.4. Stretched spring 146 CHAPTER 3 RECTANGULAR COORDINATE SYSTEM AND GRAPHS

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