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SEGMENTS OF A

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Originally appearing in Volume V11, Page 690 of the 1911 Encyclopedia Britannica.
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SEGMENTS OF A  See also:

LINE § 8 . Any two points A and B in space determine on the line through them a finite See also:part, which may be considered as being described by a point moving from A to B . This we shall denote by AB, and distinguish it from BA, which is supposed as being described by a point moving from B to A, and hence in a direction or in a " sense " opposite to AB . Such a finite line, which has a definite sense, we shall See also:call a " segment," so that AB and BA denote different segments, which are said to be equal in length but of opposite sense . The one sense is often called See also:positive and the other negative . P In introducing the word " sense " for direction in a line, we have the word direction reserved for direction of the line itself, so that different lines have different directions, unless they be parallel, whilst in each line we have a positive and negative sense . We may also say, with See also:Clifford, that AB denotes the " step " of going from A to B . § q . If we have three points A, B, C in a line (fig . 2), the step AB will bring us from A to B, and the step A B BC from B to C . Hence both steps are —*— See also:equivalent to the one step AC . This is expressed by saying that AC is the " sum " of AB and BC ; in symbols A B AB+BC=AC, where See also:account is to be taken of the A G B sense .

This See also:

equation is true whatever be the position of ' the three points on the line . As a See also:special See also:case we have AB+BA=o, (I) AB-FBC+CA=o, (2) which again is true for any three points in a line . We further write AB =—BA, where — denotes negative sense . We can then, just as in See also:algebra, See also:change subtraction of segments into addition by changing the sense, so that AB—CB is the same as AB+(—CB) or AB+BC . A figure will at once show the truth of this . The sense is, in fact, in every respect equivalent to the " sign " of a number in algebra . § Io . Of the many formulae which exist between points in a line we shall have to use only one more, which connects the segments between any four points A, B, C, D in a line . We have BC=BD+DC, CA=CD+DA, AB=AD+DB; or multiplying these by AD, BD, CD respectively, we get BC . AD =BD . AD+DC . AD =BD' .

AD—CD . AD CA . BD=CD . BD+DA . BD =CD . BD—AD . BD AB . CD=AD . CD-I-DB . CD=AD . CD—BD . CD .

It will be seen that the sum of the right-See also:

hand sides vanishes, hence that BC AD+CA . BD+AB : CD =o (3) for any four points on a line . § ii . If C is any point in the line AB, then we say that C divides the segment AB in the ratio AC/CB, account being taken of the sense of the two segments AC and CB . If C lies between A and B the ratio is positive, as AC and CB have the same sense . But if C lies without the segment AB, i.e. if C divides AB externally, then the ratio is negative . Q A M B P To see how the value of this ratio changes with the whole line (fig . 3), whilst A and B remain fixed . If C lies at the point A, then AC =o, hence the ratio AC:CB vanishes . As C moves towards B, AC increases and CB decreases, so that our ratio increases . At the See also:middle point M of AB it assumes the value +I, and then increases till it reaches an infinitely large value, when C arrives at B . On passing beyond B the ratio becomes negative .

Phoenix-squares

If C is at P we have AC=AP=AB+BP, hence AC AB BP 'AB CB=PB+PB= BP—I' In the last expression the ratio AB:BP is positive, has its greatest value co when C coincides with B, and vanishes when BC becomes See also:

infinite . Hence, as C moves from B to the right to the point at infinity, the ratio AC:CB varies from — to—I . If, on the other hand, C is to the See also:left of A, say at Q, we have AC=AQ=AB+BQ=AB—QB, hence CB=Q--I . Here AB <QB, hence the ratio AB :QB is positive and always ' less than one, so that the whole is negative and <1 . If C is at the point at infinity it is—I, and then increases as C moves to the right, till for C at A we get the ratio = o . Hence " As C moves along the line from an infinite distance to the left to an infinite distance at the right, the ratio always increases; it starts with the value—1, reaches o at A, +1 at M, at B, now changes sign to —co , and increases till at an infinite distance it reaches again the value—1 . It assumes therefore all possible values from —m to+ xo , and each value only once, so that not only does every position of C determine a definite value of the ratio AC :CB, but also, conversely, to every positive or negative value of this ratio belongs onefrom algebraic identities is very See also:simple . For example, if a, b, c, x be any four quantities, then a (a—b)(a ¢ c)(x—a) +(b—c) (b ba)(x—b)+ x (c—a)(c—b)(x-c) (x—a)b)(x—c)' this may be proved, cumbrously, by multiplying up, or, simply, by decomposing the right-hand member of the identity into partial fractions . Now take a line ABCDX, and let AB = a, AC = b, AD =c, AX =x . Then obviously (a—b) =AB—AC =—BC, paying regard to signs; (a—c)=AB—AD=DB, and so on . Substituting these values in the identity we obtain the following relation connecting the segments formed by five points on a line : AB AC AD AX BC.BD.BX+CD.CB.CX+DB.DC.DX=BX.CX.DX' Conversely, if a metrical relation be given, its validity may be tested by reducing to an algebraic equation, which is an identity if the relation be true . For example, if ABCDX be five collinear points, prove AD.AX BD.BX CD.CX AB .

AC'+BC . BA+CA . CB'=I' Clearing of fractions by multiplying throughout by AB . BC . CA, we have to prove —AD . AX . BC — BD . BX . CA—CD . CX . AB =AB . BC .

CA . Take A as origin and let AB = a, AC = b, AD =c, AX =x . Substituting for the segments in terms of a, b, c, x, we obtain on simplification a2b—ab2 =—ab2+a2b, an obvious identity . An alternative method of testing a relation is illustrated in the following example:—If A, B, C, D, E, F be six collinear points, then AE.AF BE . BF CE.CF DE.DF AB.AC.AD+BC . BD.BA+CD.CA.CB+DA . DB . DC=0' Clearing of fractions by multiplying throughout by AB . BC . CD: DA, and reducing to a See also:

common origin 0 (calling OA=a, OB=b, &c.), an equation containing the second and See also:lower rowers of OA (=a), &c., is obtained . Calling OA=x, it is found that x=b, x=c,'x'=d are solutions . Hence the quadratic has three roots; consequently it is an identity .

End of Article: SEGMENTS OF A
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