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SEGMENTS OF A
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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. 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. Additional information and CommentsThere are no comments yet for this article.
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