8.1. Distance Formula. 394 CHAPTER 8 Graphs, Functions, and Systems of Equations and Inequalities

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1 94 CHAPTER 8 Graphs, Functions, and Sstems of Equations and Inequalities The points in Figure 2 are plotted on this calculator screen. Wh is E5, not visible? The Rectangular Coordinate Sstem and Circles Each of the pairs of numbers (1, 2), 1, 5, and (, 7) is an eample of an ordered pair; that is, a pair of numbers written within parentheses in which the order of the numbers is important. The two numbers are the components of the ordered pair. An ordered pair is graphed using two number lines that intersect at right angles at the zero points, as shown in Figure 1. The common zero point is called the origin. The horizontal line, the -ais, represents the first number in an ordered pair, and the vertical line, the -ais, represents the second. The -ais and the -ais make up a rectangular (or Cartesian) coordinate sstem. The aes form four quadrants, numbered I, II, III, and IV as shown in Figure 2. (A point on an ais is not considered to be in an of the four quadrants.) Origin -ais C ( 5, 6) II I E ( 5, ) A (, 1) B (4, 1) -ais III IV D ( 4, 5) FIGURE 1 FIGURE 2 Double Descartes After the French postal service issued the above stamp in honor of René Descartes, sharp ees noticed that the title of Descartes s most famous book was wrong. Thus a second stamp (see facing page) was issued with the correct title. The book in question, Discourse on Method, appeared in 167. In it Descartes rejected traditional Aristotelian philosoph, outlining a universal sstem of knowledge that was to have the certaint of mathematics. For Descartes, method was analsis, going from self-evident truths step-b-step to more distant and more general truths. One of these truths is his famous statement, I think, therefore I am. (Thomas Jefferson, also a rationalist, began the Declaration with the words, We hold these truths to be selfevident. ) We locate, or plot, the point on the graph that corresponds to the ordered pair (, 1) b going three units from zero to the right along the -ais, and then one unit up parallel to the -ais. The point corresponding to the ordered pair (, 1) is labeled A in Figure 2. The phrase the point corresponding to the ordered pair (, 1) often is abbreviated the point (, 1). The numbers in an ordered pair are called the coordinates of the corresponding point. The parentheses used to represent an ordered pair also are used to represent an open interval (introduced in an earlier chapter). In general, there should be no confusion between these smbols because the contet of the discussion tells us whether we are discussing ordered pairs or open intervals. Distance Formula Suppose that we wish to find the distance between two points, sa, 4 and 5,. The Pthagorean theorem allows us to do this. In Figure on the net page, we see that the vertical line through 5, and the horizontal line through, 4 intersect at the point 5, 4. Thus, the point 5, 4 becomes the verte of the right angle in a right triangle. B the Pthagorean Theorem, the square of the length of the hpotenuse, d, of the right triangle in Figure is equal to the sum of the squares of the lengths of the two legs a and b: d 2 a 2 b 2. The length a is the distance between the endpoints of that leg. Since the -coordinate of both points is 5, the side is vertical, and we can find a b finding the difference between the -coordinates. Subtract 4 from to get a positive value of a. a 4 7

2 8.1 The Rectangular Coordinate Sstem and Circles 95 Similarl, find b b subtracting 5 from. b 5 8 Substituting these values into the formula, we have d 2 a 2 b 2 d Let a 7 and b 8. d d 2 11 d 11. Square root propert, d. Therefore, the distance between 5, and, 4 is 11. ( 2, 2 ) ( 5, ) a d d a (, 4) ( 5, 4) b ( 1, 1 ) b ( 2, 1 ) FIGURE FIGURE 4 This result can be generalized. Figure 4 shows the two different points 1, 1 and 2, 2. To find a formula for the distance d between these two points, notice that the distance between 2, 2 and 2, 1 is given b a 2 1, and the distance between 1, 1 and 2, 1 is given b b 2 1. From the Pthagorean Theorem, Descartes wrote his Geometr as an application of his method; it was published as an appendi to the Discourse. His attempts to unif algebra and geometr influenced the creation of what became coordinate geometr and influenced the development of calculus b Newton and Leibniz in the net generation. In 1649 he went to Sweden to tutor Queen Christina. She preferred working in the unheated castle in the earl morning; Descartes was used to staing in bed until noon. The rigors of the Swedish winter proved too much for him, and he died less than a ear later. d , and b using the square root propert, we obtain the distance formula. Distance Formula The distance between the points 1, 1 and 2, 2 is d This result is called the distance formula. The small numbers 1 and 2 in the ordered pairs 1, 1 and 2, 2 are called subscripts. We read 1 as sub 1. Subscripts are used to distinguish between different values of a variable that have a common propert. For eample, in the ordered pairs, 5 and 6, 4, can be designated as 1 and 6 as 2. Their common propert is that the are both components of ordered pairs. This idea is used in the following eample.

3 96 CHAPTER 8 Graphs, Functions, and Sstems of Equations and Inequalities EXAMPLE 1 Find the distance between, 5 and 6, 4. When using the distance formula to find the distance between two points, designating the points as 1, 1 and 2, 2 is arbitrar. Let us choose 1, 1, 5 and 2, 2 6, 4. d , 2 4, 1, A program can be written for the distance formula. This one supports the result in Eample 1, since Midpoint Formula The midpoint of a line segment is the point on the segment that is equidistant from both endpoints. Given the coordinates of the two endpoints of a line segment, it is not difficult to find the coordinates of the midpoint of the segment. Midpoint Formula The coordinates of the midpoint of the segment with endpoints 1, 1 and 2, 2 are 1 2, In words, the coordinates of the midpoint of a line segment are found b calculating the averages of the - and -coordinates of the endpoints. EXAMPLE 2 Find the coordinates of the midpoint of the line segment with endpoints 8, 4 and 9, 6. Using the midpoint formula, we find that the coordinates of the midpoint are , ,1. EXAMPLE Figure 5 on the net page depicts how the purchasing power of the dollar declined from 199 to 2, based on changes in the Consumer Price Inde. In this model, the base period is For eample, it would have cost $1. to purchase in 199 what $.766 would have purchased during the base period. Use the graph to estimate the purchasing power of the dollar in 1995, and compare it to the actual figure of $.656. The ear 1995 lies halfwa between 199 and 2, so we must find the coordinates of the midpoint of the segment that has endpoints (199,.766) and (2,.581). This is given b , ,.675.

4 8.1 The Rectangular Coordinate Sstem and Circles 97 Thus, based on this procedure, the purchasing power of the dollar to the nearest thousandth was $.674 in This is fairl close to the actual figure of $.656. THE DECLINE IN PURCHASING POWER OF THE DOLLAR Purchasing power $.766 $ Year Source: U.S. Bureau of Labor Statistics. FIGURE 5 Circles An application of the distance formula leads to one of the most familiar shapes in geometr, the circle. A circle is the set of all points in a plane that lie a fied distance from a fied point. The fied point is called the center and the fied distance is called the radius To graph 2 2 9, we solve for to get and Then we graph both in a square window. EXAMPLE 4 Find an equation of the circle with radius and center at (, ), and graph the circle. If the point, is on the circle, the distance from, to the center (, ) is, as shown in Figure 6. B the distance formula, d 2 2 1, 1, 2, Square both sides. An equation of this circle is It can be graphed b locating all points three units from the origin. (, ) = 9 FIGURE 6

5 98 CHAPTER 8 Graphs, Functions, and Sstems of Equations and Inequalities 6 5 (4, ) ( 4) 2 + ( + ) 2 = 25 FIGURE A circle can be drawn b using the appropriate command, entering the coordinates of the center and the radius. Compare with Eample 5 and Figure 7. 8 EXAMPLE 5 Find an equation for the circle that has its center at 4, and radius 5, and graph the circle. Again use the distance formula Square both sides. The graph of this circle is shown in Figure 7. Eamples 4 and 5 can be generalized to get an equation of a circle with radius r and center at h, k. If, is a point on the circle, the distance from the center h, k to the point, is r. Then b the distance formula, h 2 k 2 r. Squaring both sides gives the following equation of a circle. Equation of a Circle The equation of a circle of radius r with center at h, k is h 2 k 2 r 2. In particular, a circle of radius r with center at the origin has equation 2 2 r 2. EXAMPLE 6 Find an equation of the circle with center at 1, 2 and radius 4. Let h 1, k 2, and r 4 in the general equation above to get h 2 k 2 r In the equation found in Eample 5, multipling out 4 2 and 2 and then combining like terms gives This result suggests that an equation that has both 2 and 2 terms ma represent a circle. The net eample shows how to tell, using the method of completing the square. 5 ( 1, ) = FIGURE 8 EXAMPLE 7 Graph Since the equation has 2 and 2 terms with equal coefficients, its graph might be that of a circle. To find the center and radius, complete the squares on and as follows. (See the previous chapter, where completing the square is introduced.) Add 15 to both sides Rewrite in anticipation of completing the square Complete the squares on both and Factor on the left and add on the right. The final equation shows that the graph is a circle with center at 1, and radius 5. The graph is shown in Figure 8.

6 8.1 The Rectangular Coordinate Sstem and Circles 99 7 A (1, 2) C 5 5 B FIGURE 9 The final eample in this section shows how equations of circles can be used in locating the epicenter of an earthquake. EXAMPLE 8 Seismologists can locate the epicenter of an earthquake b determining the intersection of three circles. The radii of these circles represent the distances from the epicenter to each of three receiving stations. The centers of the circles represent the receiving stations. Suppose receiving stations A, B, and C are located on a coordinate plane at the points 1, 4,, 1, and 5, 2. Let the distances from the earthquake epicenter to the stations be 2 units, 5 units, and 4 units, respectivel. See Figure 9. Where on the coordinate plane is the epicenter located? Graphicall, it appears that the epicenter is located at 1, 2. To check this algebraicall, determine the equation for each circle and substitute 1 and 2. Station A: Station B: Station C: Thus, we can be sure that the epicenter lies at 1, 2.