REVIEW OF ANALYTIC GEOMETRY


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1 REVIEW OF ANALYTIC GEOMETRY The points in a plane can be identified with ordered pairs of real numbers. We start b drawing two perpendicular coordinate lines that intersect at the origin O on each line. Usuall one line is horizontal with positive direction to the right and is called the ais; the other line is vertical with positive direction upward and is called the ais. An point P in the plane can be located b a unique ordered pair of numbers as follows. Draw lines through P perpendicular to the  and aes. These lines intersect the aes in points with coordinates a and b as shown in Figure. Then the point P is assigned the ordered pair a, b. The first number a is called the coordinate of P; the second number b is called the coordinate of P. We sa that P is the point with coordinates a, b, and we denote the point b the smbol Pa, b. Several points are labeled with their coordinates in Figure. II _3 O _ III b FIGURE _4 I IV a P(a, b) (_, ) _ _ (_3, _) FIGURE B reversing the preceding process we can start with an ordered pair a, b and arrive at the corresponding point P. Often we identif the point P with the ordered pair a, b and refer to the point a, b. [Although the notation used for an open interval a, b is the same as the notation used for a point a, b, ou will be able to tell from the contet which meaning is intended.] This coordinate sstem is called the rectangular coordinate sstem or the Cartesian coordinate sstem in honor of the French mathematician René Descartes (596 65), even though another Frenchman, Pierre Fermat (6 665), invented the principles of analtic geometr at about the same time as Descartes. The plane supplied with this coordinate sstem is called the coordinate plane or the Cartesian plane and is denoted b. The  and aes are called the coordinate aes and divide the Cartesian plane into four quadrants, which are labeled I, II, III, and IV in Figure. Notice that the first quadrant consists of those points whose  and coordinates are both positive. EXAMPLE Describe and sketch the regions given b the following sets. (a), (b), (c ) {, } SOLUTION (a) The points whose coordinates are or positive lie on the ais or to the right of it as indicated b the shaded region in Figure 3(a) _4 (, 3) (, _4) (5, ) = = Thomson BrooksCole copright 7 FIGURE 3 (a) (b) = (c) < =_
2 REVIEW OF ANALYTIC GEOMETRY fi P (, fi) fi P (, )  P (, ) (b) The set of all points with coordinate is a horizontal line one unit above the ais [see Figure 3(b)]. (c) Recall from Review of Algebra that if and onl if The given region consists of those points in the plane whose coordinates lie between and. Thus, the region consists of all points that lie between (but not on) the horizontal lines and. [These lines are shown as dashed lines in Figure 3(c) to indicate that the points on these lines don t lie in the set.] Recall from Review of Algebra that the distance between points a and b on a number line is a b b a. Thus, the distance between points P, and P 3, on a horizontal line must be and the distance between and on a vertical line must be P, P 3,. (See Figure 4.) To find the distance P P between an two points P, and P,, we note that triangle P P P 3 in Figure 4 is a right triangle, and so b the Pthagorean Theorem we have FIGURE 4 P P s P P 3 P P 3 s s Distance Formula The distance between the points P, and P, is P P s For instance, the distance between, and 5, 3 is s5 3 s4 5 s4 CIRCLES r C(h, k) P(, ) An equation of a curve is an equation satisfied b the coordinates of the points on the curve and b no other points. Let s use the distance formula to find the equation of a circle with radius r and center h, k. B definition, the circle is the set of all points P, whose distance from the center Ch, k is r. (See Figure 5.) Thus, P is on the circle if and onl if. From the distance formula, we have PC r s h k r FIGURE 5 or equivalentl, squaring both sides, we get h ( k r This is the desired equation. Equation of a Circle An equation of the circle with center h, k and radius r is h ( k r In particular, if the center is the origin,, the equation is r Thomson BrooksCole copright 7 For instance, an equation of the circle with radius 3 and center, 5 is ( 5 9
3 REVIEW OF ANALYTIC GEOMETRY 3 EXAMPLE Sketch the graph of the equation 6 7 b first showing that it represents a circle and then finding its center and radius. SOLUTION We first group the terms and terms as follows: ( 6 7 (_, 3) 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: FIGURE = ( or ( 3 3 Comparing this equation with the standard equation of a circle, we see that h, k 3, and r s3, so the given equation represents a circle with center, 3 and radius s3. It is sketched in Figure 6. LINES To find the equation of a line L we use its slope, which is a measure of the steepness of the line. Definition The slope of a nonvertical line that passes through the points P, and P, is P (, ) L m P (, ) Î=fi =rise The slope of a vertical line is not defined. FIGURE 7 Î=  =run m=5 m= m= m= m= Thus the slope of a line is the ratio of the change in,, to the change in,. (See Figure 7.) The slope is therefore the rate of change of with respect to. The fact that the line is straight means that the rate of change is constant. Figure 8 shows several lines labeled with their slopes. Notice that lines with positive slope slant upward to the right, whereas lines with negative slope slant downward to the right. Notice also that the steepest lines are the ones for which the absolute value of the slope is largest, and a horizontal line has slope. Now let s find an equation of the line that passes through a given point P, and has slope m. A point P, with lies on this line if and onl if the slope of the line through P and P is equal to m; that is, m FIGURE 8 m=_ m=_ m=_5 m=_ This equation can be rewritten in the form m and we observe that this equation is also satisfied when and. Therefore, it is an equation of the given line. Thomson BrooksCole copright 7 PointSlope Form of the Equation of a Line An equation of the line passing through the point P, and having slope m is m
4 4 REVIEW OF ANALYTIC GEOMETRY EXAMPLE 3 Find an equation of the line through the points, and 3, 4. SOLUTION The slope of the line is m Using the pointslope form with and, we obtain 3 which simplifies to 3 b =m+b Suppose a nonvertical line has slope m and intercept b. (See Figure 9.) This means it intersects the ais at the point, b, so the pointslope form of the equation of the line, with and b, becomes b m FIGURE 9 This simplifies as follows. SlopeIntercept Form of the Equation of a Line An equation of the line with slope m and intercept b is m b b =a =b In particular, if a line is horizontal, its slope is m, so its equation is b, where b is the intercept (see Figure ). A vertical line does not have a slope, but we can write its equation as a, where a is the intercept, because the coordinate of ever point on the line is a. FIGURE a EXAMPLE 4 Graph the inequalit 5. SOLUTION We are asked to sketch the graph of the set, 5 and we begin b solving the inequalit for : =_ + FIGURE 5 5 Compare this inequalit with the equation, which represents a line with 5 slope 5 and intercept. We see that the given graph consists of points whose coor dinates are larger than those on the line 5. Thus, the graph is the region that lies above the line, as illustrated in Figure. PARALLEL AND PERPENDICULAR LINES Thomson BrooksCole copright 7 Slopes can be used to show that lines are parallel or perpendicular. The following facts are proved, for instance, in Precalculus: Mathematics for Calculus, Fifth Edition b Stewart, Redlin, and Watson (Thomson Brooks/Cole, Belmont, CA, 6).
5 REVIEW OF ANALYTIC GEOMETRY 5 Parallel and Perpendicular Lines. Two nonvertical lines are parallel if and onl if the have the same slope.. Two lines with slopes m and m are perpendicular if and onl if m m ; that is, their slopes are negative reciprocals: m m EXAMPLE 5 Find an equation of the line through the point 5, that is parallel to the line SOLUTION The given line can be written in the form which is in slopeintercept form with m 3. Parallel lines have the same slope, so the required line has slope and its equation in pointslope form is We can write this equation as 3 6. EXAMPLE 6 Show that the lines 3 and 6 4 are perpendicular. SOLUTION The equations can be written as 3 3 from which we see that the slopes are m 3 and and 3 4 m 3 Since m m, the lines are perpendicular. Thomson BrooksCole copright 7 EXERCISES A Click here for answers. S Click here for solutions. Find the distance between the points..,, 4, 5., 3, 5, Find the slope of the line through P and Q. 3. P3, 3, Q, 6 4. P, 4, Q6, 5. Show that the points, 9, 4, 6,,, and 5, 3 are the vertices of a square. 6. (a) Show that the points A, 3, B3,, and C5, 5 are collinear (lie on the same line) b showing that AB BC AC. (b) Use slopes to show that A, B, and C are collinear. 7 Sketch the graph of the equation Find an equation of the line that satisfies the given conditions.. Through, 3, slope 6. Through 3, 5, slope 7 3. Through, and, 6 4. Through, and 4, 3 5. Slope 3, intercept 6. Slope 5, intercept intercept, intercept intercept 8, intercept 6 9. Through 4, 5, parallel to the ais. Through 4, 5, parallel to the ais. Through, 6, parallel to the line 6. intercept 6, parallel to the line Through,, perpendicular to the line Through, perpendicular to the line 4 8 (, 3 )
6 6 REVIEW OF ANALYTIC GEOMETRY 5 8 Find the slope and intercept of the line and draw its graph Sketch the region in the plane. 9., {, } , and 3 {, 3 and }, 4 and,, {, 3} Find an equation of a circle that satisfies the given conditions. 37. Center 3,, radius Center, 5, passes through 4, Show that the equation represents a circle and find the center and radius Show that the lines 4 and are not parallel and find their point of intersection. 4. Show that the lines and are perpendicular and find their point of intersection. 43. Show that the midpoint of the line segment from P, to P, is, 44. Find the midpoint of the line segment joining the points, 3 and 7, Find an equation of the perpendicular bisector of the line segment joining the points A, 4 and B7,. 46. (a) Show that if the  and intercepts of a line are nonzero numbers a and b, then the equation of the line can be put in the form a b This equation is called the twointercept form of an equation of a line. (b) Use part (a) to find an equation of the line whose intercept is 6 and whose intercept is 8. Thomson BrooksCole copright 7
7 REVIEW OF ANALYTIC GEOMETRY 7 ANSWERS S Click here for solutions. 8. m 4 5, b. 5. s = = m 3, b _ =4 = 6. m 3, b =+, = 7. m 3 4, b 3 _ , 5, 4 4., 3, s7 4., 4., , (b) Thomson BrooksCole copright 7
8 8 REVIEW OF ANALYTIC GEOMETRY SOLUTIONS. Use the distance formula with P (, )=(, ) and P (, )=(4, 5) to get P P = (4 ) +(5 ) = 3 +4 = 5 = 5. The distance from (, 3) to (5, 7) is (5 ) +[7 ( 3)] = 4 + = 6 = With P ( 3, 3) and Q(, 6), the slope m of the line through P and Q is m = 6 3 ( 3) = m = ( 4) 6 ( ) = Using A(, 9), B(4, 6), C(, ), andd( 5, 3), wehave AB = [4 ( )] +(6 9) = 6 +( 3) = 45 = 9 5=3 5, BC = ( 4) +( 6) = ( 3) +( 6) = 45 = 9 5=3 5, CD = ( 5 ) +(3 ) = ( 6) +3 = 45 = 9 5=3 5,and DA = [ ( 5)] +(9 3) = 3 +6 = 45 = 9 5=3 5. So all sides are of equal length and we have a rhombus. Moreover, m AB = ( ) =, m BC = 6 4 =, m CD = 3 5 =,and m DA = 9 3 =, so the sides are perpendicular. Thus, A, B, C,andD are vertices of a square. ( 5) 6. (a) Using A(, 3), B(3, ), andc(5, 5), wehave AB = [3 ( )] +( 3) = 4 +8 = 8 = 4 5, BC = (5 3) +(5 ) = +4 = = 5,and AC = [5 ( )] +(5 3) = 6 + = 8 = 6 5.Thus, AC = AB + BC. (b) m AB = 3 3 ( ) = =and mac = 4 5 ( ) = =. Since the segments AB and AC have the 6 same slope, A, B and C must be collinear. 7. The graph of the equation =3is a vertical line with intercept 3. The line does not have a slope. 8. The graph of the equation = is a horizontal line with intercept. The line has slope. Thomson BrooksCole copright 7
9 REVIEW OF ANALYTIC GEOMETRY 9 9. = =or =.Thegraph consists of the coordinate aes.. = =or =. B the pointslope form of the equation of a line, an equation of the line through (, 3) with slope 6 is ( 3) = 6( ) or =6 5.. ( 5) = 7 [ ( 3)] or = The slope of the line through (, ) and (, 6) is m = 6 = 5, so an equation of the line is = 5( ) or = For (, ) and (4, 3), m = 3 ( ) =. An equation of the line is 3=( 4) or =. 4 ( ) 5. B the slopeintercept form of the equation of a line, an equation of the line is =3. 6. B the slopeintercept form of the equation of a line, an equation of the line is = Since the line passes through (, ) and (, 3), its slope is m = 3 =3,soanequationis =3 3. Another method: From Eercise 6, + = 3 + = 3 = For ( 8, ) and (, 6), m = 6 ( 8) = 3 4.Soanequationis = Another method: From Eercise 6, = 3 +4 =4 = The line is parallel to the ais, so it is horizontal and must have the form = k. Since it goes through the point (, ) =(4, 5), the equation is =5.. The line is parallel to the ais, so it is vertical and must have the form = k. Since it goes through the point (, ) =(4, 5), the equation is =4.. Putting the line + =6into its slopeintercept form gives us = +3, so we see that this line has slope. Thus, we want the line of slope that passes through the point (, 6): ( 6) = ( ) = = = 3 4 3,som = 3 and the required line is = = = Since this line has slope 5, a line perpendicular to it would have slope 5, so the required line is ( ) = 5 [ ( )] = 5 +. Thomson BrooksCole copright = =. Since this line has slope, a line perpendicular to it would have slope,sothe 8 required line is 3 = = +. 3
10 REVIEW OF ANALYTIC GEOMETRY = =, 3 so the slope is and the 3 intercept is = = 3 +, so the slope is 3 and the intercept is = = 3 4 3, so the slope is 3 4 and the intercept is = = 4 +, so the slope is and the intercept is. 9. {(, ) <} 3. {(, ) and <3} 3. (, ) = {(, ) } 3. (, ) < 3 and < 33. {(, ) 4, } 34. {(, ) > } 35. {(, ) + } 36. (, ) < ( +3) 37. An equation of the circle with center (3, ) and radius 5 is ( 3) +( +) =5 = The equation has the form ( +) +( 5) = r.since( 4, 6) lies on the circle, we have r =( 4+) +( 6 5) =3.Soanequationis( +) +( 5) = 3. Thomson BrooksCole copright = = = 3+4+5=6 ( ) +( +5) =4. Thus, we have a circle with center (, 5) and radius 4.
11 REVIEW OF ANALYTIC GEOMETRY = = +9 +( +3) =7. Thus, we have a circle with center (, 3) and radius =4 = 4 m =and 6 = =6 =3 5 m =3.Sincem 6=m, the two lines are not parallel. To find the point of intersection: 4=3 5 = =. Thus, the point of intersection is (, ) = 5 =3 +9 = m = 3 5 and +6 5 = 6 = +5 = m = 5 3.Sincemm = =, the two lines are perpendicular. To find the point of intersection: = = =68 = = Let M be the point, MP = + MP = = 5 5.Then + + =5. Thus, the point of intersection is (, 5). = + and. = + Hence, MP = MP ;that is, M is equidistant from P and P. 44. Using the midpoint formula from Eercise 43 with (, 3) and (7, 5),weget +7, 3+5 =(4, 9). 45. With A(, 4) and B(7, ), the slope of segment AB is 4 =, so its perpendicular bisector has slope. The 7 +7 midpoint of AB is, 4+( ) =(4, ), so an equation of the perpendicular bisector is =( 4) or = (a) Since the intercept is a, the point (a, ) is on the line, and similarl since the intercept is b, (,b) is on the line. Hence, the slope of the line is m = b a = b a. Substituting into = m + b gives = b a + b b a + = b a + b =. (b) Letting a =6and b = 8 gives 6 + = 8 +6 = 48 [multipl b 48] 8 6 = =4 4 = Thomson BrooksCole copright 7
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