Flick, T. Michael

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FROM THE MIDPOINT FORMULA TO GENERALIZED DIVISION OF LINE SEGMENTS T. Michael Flick Purcell Marian High School Cincinnati, Ohio To most students of Algebra 2 and above, the division of a line segment in the coordinate plane is limited to what is classically known as "The Midpoint Formula." Many texts barely touch on this interesting relationship and often dismiss its derivation as beyond the scope of the text or leave it to the reader. Often only a few sentences of superficial explanation accompany the statement of the formula. Indeed, this may be all some students can handle, but an advanced or honors group deserves much more. Most students of Algebra 2, and others who might study the midpoint formula, have completed a course in Plane Geometry. This means that they already have begun to develop an understand-ing of a ratio and its applications. The ratio and a little algebra are all that is needed not only to derive the midpoint formula, but to generalize and extend it into what the author calls the division of the line segment. The procedure for dividing the line segment into equal parts will lead to greater and easier use of this technique by the student and will there-fore expand problem solving ability while enhancing a full understanding of underlying concepts. A pattern arises in the generalization that greatly simpli-fies the division of the line segment, just as Pascal's Triangle (or the combination approach) vastly simplifies the expansion of binomials raised to high powers. Figure 1 shows line segment AB which is to be bisected at midpoint (a,b). From the geometry of ratios, it is known that 1 a - x1 x2 - a Since "a" is the x-coordinate of the midpoint, simply solve this ratio to get the standard result -2-x2 - a a - x1 2a = x1 + x2 x1 + x2 a = 2 A similar ratio relates to the y-axis the corresponding result b Yl + Y2 2 Therefore, the point that bisects segment AB is given by (Xl + X2 Yl ; y9· This is a derivation that is easily understood by better students. The derivation easily generalizes to trisecting a line segment. Figure 2 shows a setup similar to Figure 1. This time (a,b) is located so that 1/3 of the length of AB is to the left and 2/3 is to the right (A ratio of 1:2). The dividing ratio now becomes 1 = 2 a - x1 x2 - a or x2 - a = 2a - x1 3a = 2x1 + x2 2x1 + x2 a = J 2n + Y2 b = which gives one 3 Similarly, of the two trisecting points as(2x1 + x2 2n ~). By letting the length to the left of (a,b) equal 2/3 of the total length with the length to the right equalling 1/3 of the length (ratio of 2:1), the dividing ratio 2 a - x1 other trisection point as(x1 + 2x2 ' Yl + 2y2). _1_ yields the x2 - a The process easily generalizes. By applying the same ratios to the general setup in Figures 3 and 4 we get -J-.. FROM THE MIDPOINT FORMULA TO GENERALIZED DIVISION OF LINE SEGMENTS T. Michael Flick Purcell Marian High School Cincinnati, Ohio To most students of Algebra 2 and above, the division of a line segment in the coordinate plane is limited to what is classically known as "The Midpoint Formula." Many texts barely touch on this interesting relationship and often dismiss its derivation as beyond the scope of the text or leave it to the reader. Often only a few sentences of superficial explanation accompany the statement of the formula. Indeed, this may be all some students can handle, but an advanced or honors group deserves much more. Most students of Algebra 2, and others who might study the midpoint formula, have completed a course in Plane Geometry. This means that they already have begun to develop an understand-ing of a ratio and its applications. The ratio and a little algebra are all that is needed not only to derive the midpoint formula, but to generalize and extend it into what the author calls the division of the line segment. The procedure for dividing the line segment into equal parts will lead to greater and easier use of this technique by the student and will there-fore expand problem solving ability while enhancing a full understanding of underlying concepts. A pattern arises in the generalization that greatly simpli-fies the division of the line segment, just as Pascal's Triangle (or the combination approach) vastly simplifies the expansion of binomials raised to high powers. Figure 1 shows line segment AB which is to be bisected at midpoint (a,b). From the geometry of ratios, it is known that 1 a - x1 x2 - a Since "a" is the x-coordinate of the midpoint, simply solve this ratio to get the standard result -2-·~ .. ,, x2 - a= a - x1 2a = x1 + x2 x1 + x2 a= --2 A similar ratio relates to the y-axis the corresponding result b Yl + Y2 2 Therefore, the point that bisects segment AB is given by (Xl + X2 Yl ; y9· This is a derivation that is easily understood by better students. The derivation easily generalizes to trisecting a line segment. Figure 2 shows a setup similar to Figure 1. This time (a,b) is located so that 1/3 of the length of AB is to the left and 2/3 is to the right (A ratio of 1:2). The dividing ratio now becomes 1 = 2 a - x1 x2 - a or x2 - a = 2a - Xl 3a = 2x1 + x2 2x1 + x2 a = J 2n + Y2 b = which gives one 3 Similarly, of the two trisecting points as(2x1 + x2 2n ~). By letting the length to the left of (a,b) equal 2/3 of the total length with the length to the right equalling 1/3 of the length (ratio of 2:1), the dividing ratio __ 2_ = __ 1_ yields the a - x1 x2 - a other trisection point as(x1: 2x2 'Yl: 2y2). The process easily generalizes. By applying the same ratios to the general setup in Figures 3 and 4 we get -J-_____£._ = _.9.._ _____£._ = _.9.._ a - x1 x2 - a b - Yl Y2 - b px2 - pa= qa - qx1 PY2 - pb = bq - qn qx1 + pxz qy1 + PY2 a = p + q b = p + q Now it is possible to simply write down the points that divide segment AB by substituting p and q into the formula qy1 + PY2). (qx1 + px2 p + q ( EQUATION 1) If we maintain a left-to-right orientation from (x1,y1) to (x2,y2), it is possible to establish an easily remembered pattern for dividing a line segment into any desired number of pieces. To divide segment AB of figures 3 and 4 into four equal pieces will require working with the three substitutions: 1) let p 2) let P 3) let P 1 and q 2 and q 3 and q 3 2 1. This maintains a left-to-right orientation for finding the points that divide AB into four equal parts. It is also important to keep A(x1,y1) left of B(x2,y2) for a left to right orientation. The results of substituting 1, 2, and 3 above into EQUATION 1 are shown in Table 1 under N = 4. Note that in Table 1 the values of p and q are shown to the right of each point along the vertical line. The results of the above procedure are also included for N = 5 and N = 6. Additional entries for greater values of N and close inspection of the results show that an interesting pattern has emerged in the triangular shape of Table 1. Note that left-to-right orientation has been maintained throughout Table 1. The pattern is shown mathematically for a line segment that is divided into "N" equal parts. What this shows is that if we maintain left-to-right orientation of (x1,Y1) and (x2,y2), the coefficients of x1 and Yl count down from left-to-right beginning with N-1 and ending with 1, while the coefficients of x2 and Y2 count up from left-to-right beginning with 1 and ending with N-1. The denominator is Nin all cases. -4-The verbal description of this pattern sounds much more complicated than is the case. An inspection of Table 1 gives a much clearer picture of the ease with which this pattern applies. The following specific examples illustrate the process for concrete points: EXAMPLE 1: Find the points that divide the line segment connect-ing (2,1) and (8,4) into 3 equal parts. SOLUTION: To divide into three equal parts we apply the pattern established and shown in Table 1. Let the coeffi-cients of x1 and Yl count down from 2 to 1 and the coefficients of x2 and Y2 count up from 1 to 2 while dividing by 3. The required points are (2Xl + X2 2n + y~ (x1 + 2x2 - and -3 3 Let (2,1) = (x1,Y1), since this point (8,4), and (8,4) = (x2,y2). (2(2) + 8 2(1) + 4) (2 + 2(8) 3 , 3 /and 3 the required points are (4,2), (6,3). Y1 ; ''f is left of 1 +32(4)) EXAMPLE 2: Find the points that divide the line segment connect-ing (2,-8) and (-3,2) into 5 equal parts. SOLUTION: Again the established pattern is applied. The four required points are shown in Table 1 with N = 5. Now since (-3,2) is left of (2,-8), we set (-3,2) (x1,Y1) and (2,-8) = (x2,Y2). Substitution yields the solution (-2,0), (-1,-2), (0,-4), (1,-6), (2,-8). In conclusion, the process of deriving the midpoint formula gives an excellent opportunity to generalize the important procedure of dividing a line segment. In fact, the results are so nice that dividing a line segment into "N" equal parts can be·as easy (and mechanical) as simply bisecting a line segment. The generalization is far more useful than the one specific case and greatly expands the student's understanding and problem solving ability. -5-_____E_ = __g_ _____E_ = __g_ a - x1 x2 - a b - Yl Y2 - b px2 - pa= qa - qx1 PY2 - pb = bq - qy1 qx1 + px2 qn + PY2 a = p + q b = p + q Now it is possible to simply write down the points that divide segment AB by substituting p and q into the formula qy1 + PY2). (qx1 + px2 p + q (EQUATION 1) If we maintain a left-to-right orientation from (x1,y1) to (x2,y2), it is possible to establish an easily remembered pattern for dividing a line segment into any desired number of pieces. To divide segment AB of figures 3 and 4 into four equal pieces will require working with the three substitutions: 1) let p 2) let p 3) let p 1 and q 2 and q 3 and q 3 2 1. This maintains a left-to-right orientation for finding the points that divide AB into four equal parts. It is also important to keep A(x1,y1) left of B(x2,y2) for a left to right orientation. The results of substituting 1, 2, and 3 above into EQUATION 1 are shown in Table 1 under N = 4. Note that in Table 1 the values of p and q are shown to the right of each point along the vertical line. The results of the above procedure are also included for N = 5 and N = 6. Additional entries for greater values of N and close inspection of the results show that an interesting pattern has emerged in the triangular shape of Table 1. Note that left-to-right orientation has been maintained throughout Table 1. The pattern is shown mathematically for a line segment that is divided into "N" equal parts. What this shows is that if we maintain left-to-right orientation of (x1,n) and (x2,y2), the coefficients of x1 and Yl count down from left-to-right beginning with N-1 and ending with 1, while the coefficients of x2 and Y2 count up from left-to-right beginning with 1 and ending with N-1. The denominator is Nin all cases. -4-The verbal description of this pattern sounds much more complicated than is the case. An inspection of Table 1 gives a much clearer picture of the ease with which this pattern applies. The following specific examples illustrate the process for concrete points: EXAMPLE 1: Find the points that divide the line segment connect-ing (2,1) and (8,4) into 3 equal parts. SOLUTION: To divide into three equal parts we apply the pattern established and shown in Table 1. Let the coeffi-cients of x1 and Yl count down from 2 to 1 and the coefficients of x2 and Y2 count up from 1 to 2 while dividing by 3. The required points are (2Xl + X2 2n + y~ (x1 + 2x2 - and· 3 3 Let (2,1) = (x1,n), since this point (8,4), and (8,4) = (x2,y2). (2(2) + 8 2(1) + 4) (2 + 2(8) 3 , 3 Jand 3 the required points are (4,2), (6,3). Y1 • 2y~. is left of 1 + 2(4)) EXAMPLE 2: Find the points that divide the line segment connect-SOLUTION: ing (2,-8) and (-3,2) into 5 equal parts. Again the established pattern is applied. The four required points are shown in Table 1 with N = 5. Now since (-3,2) is left of (2,-8), we set (-3,2) (x1,n) and (2,-8) = (x2,Y2). Substitution yields the solution (-2,0), (-1,-2), (0,-4), (1,-6), (2,-8). In conclusion, the process of deriving the midpoint formula gives an excellent opportunity to generalize the important procedure of dividing a line segment. In fact, the results are so nice that dividing a line segment into "N" equal parts can be·as easy (and mechanical) as simply bisecting a line segment. The generalization is far more useful than the one specific case and greatly expands the student's understanding and problem solving ability. -5-y y FIGURE 1 a - x1 (x1,u) FIGURE 2 I I I 1 x2 - a (a,OJ 'a-x I x -a I 2 I 2 (xl'o) (a,OJ -6-B(xc}yc) X (xc,OJ B(x2,y2) X (x2,0) ,_,,~~~,.,,,,,-.----,..,,_~~~~~~~~~~~~~~~~1 FIGURE 3 t a - xl i x2 - a X (x ,OJ (a,b) 1 (x2,o) FIGURE 4 ''/ (0,y2) t-- -------------7 B(,, ,y,) Y2-b I t0,b) I b-yl (0,Yl) r--· A( X ,y ) 1 1 X -7-y y 111 ''I 111 11 I i I I 11 1'.i FIGURE 1 I I B(xi::'.}yi::'.) a - x1 , x2 - a , x (x1,o) (a,OJ (xi::'. ,OJ FIGURE 2 B(x2,y2) I I I I I I a - X I Xz - a I 2 I I X (x1,o) (a,OJ (x2,0) -6-FIGURE 3 Y.' B (x2 ,_Y z J a - x1 ! x2 - a X (x ,OJ (a,b) 1 (x2,o) I FIGURE 4 ''/ \ 0 ,y 2) i---------------7 B ( x, ,y 2) IYz-b tO,b) r (0 --~ I ,y 1) A(x ,y) f 1 1 I )( I -7-!111 I JI i :11 '11 i 1!1! I I I * :z0 <+ CD <+ :,-CD :, C: 3 O" CD -s V, i <+ :,-ro X V, "' :, 0. 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For that re-keying, for the majority of the data entry to date as wel as for consultation on the initial file design, a local member of the membership commitee, Barb Suszek, has donated hundreds of hours of her time. With the new software it is now possible to send the JOURNAL Editor, Harold Brockman, and the NEWSLETTER Editor, Charles W •Smith, Jr., only one set of labels al in zip code order instead of the previous several sets each of which was in zip code order • DB master also permits easy counting of the members in each district • This information, summarized in the graph below, has been part of the membership report at the past three board meetings: Sept. 1986, Dec. 1986, and Mar. 1987. 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From the Midpoint Formula to Generalized Division of Line Segments

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From the Midpoint Formula to Generalized Division of Line Segments

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