REVIEW OF ANALYTIC GEOMETRY - Stewart Calculus

1 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 on each 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 point in the plane can be located by a unique ordered pair of numbers as lines through perpendicular to the - and -axes. These lines intersect the axes inpoints with coordinates and as shown in Figure 1.

2 Then the point is assigned theordered pair . The first number is called the x-coordinateof ; the second numberis called the y-coordinateof . We say that is the point with coordinates , andwe denote the point by the symbol . Several points are labeled with their coordi-nates in Figure reversing the preceding process we can start with an ordered pair and arriveat the corresponding point . Often we identify the point with the ordered pair andrefer to the point . [Although the notation used for an open interval is thesame as the notation used for a point , you will be able to tell from the context whichmeaning is intended.]

3 ]This coordinate system is called the rectangular coordinate systemor the Cartesiancoordinate systemin honor of the French mathematician Ren 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 coor-dinate system is called thecoordinate planeor theCartesian planeand is denoted - and -axes are called the coordinate axesand divide the Cartesian plane intofour quadrants, which are labeled I, II, III, and IV in Figure 1. Notice that the first quad-rant consists of those points whose - and -coordinates are both 1 Describe and sketch the regions given by the following sets.

4 (a)(b)(c )SOLUTION(a)The points whose -coordinates are 0 or positive lie on the -axis or to the right of itas indicated by the shaded region in Figure 3(a).FIGURE 3x0yx0yy=1x0yy=1y=_1(a) x 0(b) y=1(c) |y|<1yx{ x, y y 1} x, y y 1 x, y x 0 yxyx 2 a, b a, b a, b a, b PP a, b 0x12345_1_2_31234_2_3_1y_4(5, 0)(1, 3)(_2, 2)(_3, _2))(2, _4)FIGURE 2x12345_1_2_3aO24_2_1by13P(a, b)IIIIVIII_3 FIGURE 1_4P a, b a, b PPbPa a, b PbayxPPyxO1 Thomson Brooks-Cole copyright 20072 REVIEW OF ANALYTIC GEOMETRY (b) The set of all points with -coordinate 1 is a horizontal line one unit above the [see Figure 3(b)].

5 (c)Recall from REVIEW of AlgebrathatThe given region consists of those points in the plane whose -coordinates lie betweenand . Thus, the region consists of all points that lie between (but not on) the hori-zontal lines and . [These lines are shown as dashed lines in Figure 3(c) toindicate that the points on these lines don t lie in the set.]Recall from REVIEW of Algebrathat the distance between points and on a number lineis . Thus, the distance between points and on ahorizontal line must be and the distance between and on a vertical line must be . (See Figure 4.)To find the distance between any two points and , we notethat triangle in Figure 4 is a right triangle, and so by the Pythagorean Theorem we haveDistance FormulaThe distance between the points and isFor instance, the distance between and isCIRCLESAn equation of a curveis an equation satisfied by the coordinates of the points on thecurve and by no other points.

6 Let s use the distance formula to find the equation of a cir-cle with radius and center . By definition, the circle is the set of all points whose distance from the center is . (See Figure 5.) Thus,is on the circle if andonly if . From the distance formula, we haveor equivalently, squaring both sides, we getThis is the desired of a CircleAn equation of the circle with center and radius is In particular, if the center is the origin , the equation isFor instance, an equation of the circle with radius 3 and center is x 2 2 (y 5 2 9 2, 5 x2 y2 r2 0, 0 x h 2 (y k 2 r2r h, k x h 2 (y k 2 r2s x h 2 y k 2 r PC rPrC h, k P x, y h, k rs 5 1 2 3 2 2 s42 52 s41 5, 3 1, 2 P1P2 s x2 x1 2 y2 y1 2 P2 x2, y2 P1 x1, y1 s x2 x1 2 y2 y1 2 P1P2 s P1P3 2 P2P3 2 s x2 x1 2 y2 y1 2 P1P2P3P2 x2, y2 P1 x1, y1 P1P2 y2 y1 P3 x2.)))

7 Y1 P2 x2, y2 x2 x1 P3 x2, y1 P1 x1, y1 a b b a bay 1y 11 1y 1 y 1if and only if y 1x-axisyFIGURE 4P ( , )x 0 fiyP ( , fi)P ( , )| - ||fi- |C(h, k)x0yrP(x, y)FIGURE 5 Thomson Brooks-Cole copyright 2007 REVIEW OF ANALYTIC GEOMETRY 3 EXAMPLE 2 Sketch the graph of the equation by first show-ing that it represents a circle and then finding its center and first group the -terms and -terms as follows:Then we complete the square within each grouping, adding the appropriate constants(the squares of half the coefficients of and ) to both sides of the equation:orComparing this equation with the standard equation of a circle, we see that and , so the given equation represents a circle with center andradius.

8 It is sketched in Figure find the equation of a line we use its slope,which is a measure of the steepness of slopeof a nonvertical line that passes through the points and isThe slope of a vertical line is not the slope of a line is the ratio of the change in ,, to the change in ,. (SeeFigure 7.) The slope is therefore the rate of change of ywith respect to x. The fact that theline is straight means that the rate of change is 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.

9 Notice also that the steepest lines are the ones for which the absolute value of theslope is largest, and a horizontal line has slope let s find an equation of the line that passes through a given point andhas slope . A point with lies on this line if and only if the slope of the linethrough and is equal to ; that is,This equation can be rewritten in the formand we observe that this equation is also satisfied when and . Therefore, it isan equation of the given Form of the Equation of a Line An equation of the line passing through thepoint and having slope isy y1 m x x1 mP1 x1, y1 y y1x x1y y1 m x x1 y y1x x1 mmPP1x x1P x, y mP1 x1, y1 xx yym y x y2 y1x2 x1P2 x2, y2 P1 x1, y1 Ls3 1, 3 r s3k 3,h 1, x 1 2 (y 3 2 3 x2 2x 1 (y2 6y 9 7 1 9yx x2 2x (y2 6y 7yxx2 y2 2x 6y 7 0x0y1(_1, 3)FIGURE 6 + +2x-6y+7=0 FIGURE 7P (x , y )P (x , y ))))

10 L y=fi- =rise x= - =runx0yx0ym=1m=0m=_1m=_2m=_5m=2m=5m=12m= _12 FIGURE 8 Thomson Brooks-Cole copyright 20074 REVIEW OF ANALYTIC GEOMETRYEXAMPLE 3 Find an equation of the line through the points and .SOLUTIONThe slope of the line isUsing the point-slope form with and , we obtainwhich simplifies toSuppose a nonvertical line has slope and -intercept . (See Figure 9.) This means itintersects the -axis at the point , so the point-slope form of the equation of the line,with and , becomesThis simplifies as Form of the Equation of a LineAn equation of the line with slope and-intercept isIn particular, if a line is horizontal, its slope is , so its equation is , whereis the -intercept (see Figure 10).