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than is contained by that figure may be sponding to the first condition. But inclosed with the same perimeter. But again, the point required must be the area inclosed by a given perimeter equidistant from the two given points, cannot exceed a certain limit, which that is, it must be in the straight line limit, being the greatest possible that which bisects the distance of the two can be so inclosed, some figure with the given points at right angles; for this, it given perimeter must be capable of con- is easily seen, (I. 6.) is the locus corfaining. Therefore the circle only con- responding to this second condition. tains the greatest area of all figures Therefore, if this straight line be drawn, having the same perimeter.
and intersect the given line, the point Cor. In the same manner it may be of intersection (or any of those points, shown that if a figure is to be inclosed if there be more than one) will satisfy by a given perimeter, of which part is to both conditions, and will be the point be a given finite straight line, and if it be required. not made a circular segment of which If there be no point of intersection, the the given line is chord, a greater may be problem is impossible. inclosed with the same conditions, and To take another instancetherefore that of all figures so inclosed Let it be required “to find a point in a the circular segment is the greatest. certain plane, which shall be, first, at a
given distance from a given point in the PROP. 43.
plane; and, secondly, at a second given Of all plane figures having the same distance from a second given point in area, the circle has the least perimeter. the same plane."
Let the circle C have the same area Here it is evident that the locus corwith any other plane figure F: C shall responding to the first condition is the be contained by a less perimeter than F. circumference of a circle described about
the given point as a centre with the given distance as radius: and again, that the locus corresponding to the second condition is the circumference of a circle described about the second
given point as a centre with the second Let C' be a second circle, having the given distance as radius. Therefore game perimeter with F; then by the the points which are common to the two last proposition, c' has a greater area circumferences, that is, their points of than F has, that is, than C has. But intersection, if there be any, will either the areas of circles (33.) are as the of them be the point required. squares of their radii ; therefore the ra- If the circles do not intersect one anodius of C' is greater than the radius of ther, the problem is impossible. C; and the radii of circles (33.) are as Such is the use of loci in the solution of their circumferences; therefore the cir- problems. We have seen also in the above cumference of C', or perimeter of F, is example, that they serve to determine in greater than the circumference of C. what cases the solution is possible or imTherefore, &c.
possible. Thus, in the latter example, it Section 6.-Simple and Plane Loci.
will be impossible, if the distances of
the point required from the given points Def. 14. A locus in Plane Geometry is differ by more than the mutual distance a straight line, circle, or plane curve, of those points, or together fall short of every point of which, and none else in that distance: and in the first example that plane, satisfies a certain condition. it will be impossible, if the given line,
The nature and use of loci will be being straight, be perpendicular to the readily apprehended from the following line which passes through the two given example:
points, and does not pass through the “Required a point in a certain plane point which bisects that line; for if it which shall be, first, in a given line in does so pass, the two conditions prothe plane ; and, secondly, equidistant posed are identical, and any point in this from two given points in the same line will answer them. plane."
Every locus is the limit between exHere, as far as the first condition cess and defect. The points upon one only is concerned, any point in the given side of it fail by defect, and those upon line, but none else, will answer. The the other side by excess, of possessing given line is therefore the locus corre- the required property which is possessed
by every point in the locus. Thus, in
PROP. 44. the case of the circle, the distance of a point within the circle falls short of the
Required the locus of all points which distance of the circumference, while that
are equidistant from two given points
A, B. of a point without exceeds it. When a locus is merely a straight line,
Let P be a point in the it is called a simple locus; when the cir: locus, and join PA, PB. cumference of a circle, #t is called a plane
Then, because PAB is an locus ; when any other curve, it is said to isosceles triangle, if the t be of higher dimensions than the circle. base AB be bisected in C, The following propositions afford ex
PC joined will be at right amples of the two first only; and, the angles to AB (I. 6. Cor.3). three concluding propositions excepted,
Therefore the point P lies they will be found the same in substance in the straight line which with theorems which have been stated bisects AB at right anbefore, and which only reappear in this gles; and, it is easily shown, reversely,
that place under a different form.
every point in this straight line is It will be observed that they are in- equidistant from A and B (I.
4.); therevestigated—a species of analytical rea
fore this straight line is the locus resoning which has not hitherto been ex
quired. emplified either in the demonstration of We may observe, that if any point Q a theorem or in the solution of a pro A, Q A will be less than Q B; if upon
lie upon the same side of the locus with blem, but which is the method commonly pursued in arriving both at the one and the other side, QA will be greater than the other, and is known under the name
QB (I. 11.). of “Geometrical Analysis." The nature
PROP. 45. of this, as opposed to the ordinary di Required the locus of all points which dactic method of solution, commonly are equidistant from two given straight called that of synthesis or composition, lines AB, C D. is pointed out by its name, and will be at
If the given once apprehended from any of the fol- straight lines be lowing examples. We need oniy ob- parallel, the reserve that rather than two distinct me- quired locus is evithods of doing the same thing, as might dently a straight at arst appear to be the case, they are line, which is pabut the different parts of one full and rallel to each, and perfect method ; that the use of each is bisects the disessential to a complete solution ; the tances between latter (synthesis) always taking the sub- them. jeet up where the other leaves it; the But if not, let analysis first descending, by geometrical them meet in E, and let P be a point in reasoning, from the thing proposed to the locus. Then if PA be drawn perpenthe minutest particulars of the solution, dicular to AB, and PC perpendicular to and the synthesis ascending back through CD, PA will be equal to PC. Join the same steps from these particulars to PE: then, because the right-angled trithe thing proposed. It is true, that, be- angles PAE, PC E have a common cause the steps in each are for the most hypotenuse P E, and equal sides PA, part the same, occurring only in an in- PC, they are equal to one another in verted order, the same principles are every respect (I. 13.), and the angle PEA developed in each, and, therefore, the is equal to the angle PEC. Therefore detail of either (more especially the syn- the point P is in the straight line which thesis) commonly furnishes a satisfac- bisects the angle A EC: and, reversely, tory view of the question; for which rea- it may be shown that every point in this son, and for the sake of brevity, one is straight line is equidistant from AB and usually given to the exclusion of the CD (I. 13.): therefore, this straight line other. The student will, however, find is the locus required. himself amply repaid by entering into If any point Q lie upon the same side both, and he is recommended, after fol- of the locus with the straight line AB, lowing the analysis of the problems of its distance from A B will be less than the present section, to supply in each its distance from CD; if, upon the case where it is omitted, the synthesis other side, its distance from A B will be necessary to complete the solution, greater than that from CD.
the point P is in the circular arc passing PROP. 46.
through C, and having A B for its Straight lines being drawn from a given point A to a given straight line chord (15. Cor. 3.): and, reversely, it B C, required the focus of all points may be shown that every point in this
arc has the given property (15.); theredividing them in the same given ratio
fore it is the locus required. Let AB be any
If Q be any point upon the same side straight line drawn
of the locus with AB, the angle A QB from A to BC, and
will be greater than A CB; if upon the divided in the given
other side, less. ratio in the point D;
PROP. 49. and let P be a point in the locus. Then, because AP is to Required the locus of the vertices of PC in the same ratio as A D to DB, all triangles upon the same base AB, DP joined is parallel to B C (II. 29.). having the side terminated in A greater Therefore the point P lies in a straight than that terminated in B, and ihe difline drawn through D parallel to BC: ference of the squares of the sides equal and reversely it may be shown that to a given square. every point in this straight line has the Let P be a point given property (II. 29.); therefore it is in the locus, and the locus required.
from P draw P C at If Q be any point on the same side of right angles to A B, the locus with A, it will divide the line or AB produced. A C which passes through it, in a less Then, because the ratio than that of A D to D B: if upon
difference of the the other side, in a greater ratio.
squares of A C,BC
is equal to the difPROP. 47.
ference of the squares of AP, BP (1.38.), Required the locus of the vertices of the difference of the squares of AC, BC all equal triangles, upon the same base is equal to the given square; and the A B, and upon the same side of it. point C may be found (I. 34.) by taking Let ABC be
AD (11.52.) a third proportional to A B any triangle, upon
and the side of the given square, so that the given side of
the rectangle under AB, AD may be the base, and hay
(II. 38. Cor.1.) equalto the given square, ing the given area,
and bisecting BD in C.* And it may and let P be a
be shown, reversely, that if from the point in the locus. Then, because the tri- point C so taken, PC be drawn perangle PAB is equal to CAB, PC joined pendicular to AC, every point in PC is parallel to A B (I. 27.). Therefore will satisfy the given condition; therethe point P lies in a straight line drawn fore P C is the locus required. through C parallel to X B: and, re
If Q be any point upon the same side versely, it may be shown that every A B, the diff
of the locus with the middle point of point in this straight line has the given
nce of the squares of property (I. 27.); therefore it is the 2 A, Q B will be less than the given locus required.
the other side, If Q be any point upon the same side
greater. For, if a perpendicular Q E will be less than CAB; if upon the the squares of Q A, Q B will be equal to of the locus with AB, the triangle QAB be drawn from Q to A B, and E F be
taken equal to EB, the difference of other side, greater.
the rectangle under AB, AF, (1, 39.) Prop. 48.
which is less or greater than the rectanRequired the locus of the vertices of gle under A B, A D, according as the all triangles having equal vertical an. position of Q is one or the other of the gles, upon the same base AB, and upon
two just mentioned. the same side of it.
The figure represents the point in Let A CB be the
A B produced ; if, however, the given given vertical angle,
square be sufficiently small, the point C and let P be any
may lie between A and B. point in the locus.
Prop. 50. Then, because the
Required the locus of the vertices of angle at P is equal
all triangles upon the same base AB, to the angle at c.
having the sum of the squares of their sum of the angles A PB, B PX, that is sides equal to a given square.
to the half of two right angles (I. 2.) or . Let the given square
to one right angle. And because D Pd be the square of C,
is a right angle, the point P lies in the and let P be a point
circumference of a circle described upon in the locus. Bisect
the diameter D d. (15. Cor. 3.) AB in D, and join PD.
And reversely, if P be any point in Then, because the base
the circumference of this circle, PA shall AB of the triangle
be to P B in the given ratio. For, take PAB is bisected in
C the middle point of D d, that is, the D, the sum of the
centre of the circle, and join CP. Then, squares of PA, PB is equal to twice because A D is to DB as Ad to d B, the square of PD, together with twice the line A d is harmonically divided in D the square of DA (1.40.) But it is also and B (II. def. 19, page 68); and because equal to the square of C. Therefore twice the harmonical mean D d is bisected in C, the square of PD is equal to the dif- (II. 46.) C A, C D and CB are proporference between the square of C and tionals : also, CD is equal to CP: twice the square of AD, that is, if twice therefore, in the triangles ACP, PCB, the square of D E be equal to the same A C is to C P as CP to CB; and given difference, to twice the square of consequently (II. 32.) the triangles are DE; and the point P lies in the circum- equiangular. Therefore (II. 31.) PA is ference of a circle described from the to P B as A C to CP, that is, as AC to centre D with the radius D E. And it CD, or (because CA, C D, and CB may be shown, reversely, that every are proportionals) as A D to D B (II. point in the circumference of this circle 22. Cor. 1.). satisfies the given condition (I. 40.); If any point Q be taken within the therefore it is the locus required.
locus, Q A will be to Q B in a greater If Q be any point without the circle, ratio than that of A D to DB; if withthe sum of the squares of Q A, Q B out it, in a less ratio. For, if AB be diwill be greater than the given sum; if vided in E in the ratio of A Q to Q B, within it, less. For QD2 will be greater and if A B produced be divided in the than P D2 in the former case, and less same ratio in e; then, joining Q E and in the latter; and therefore the sum of Qe, the angle E Qe will be a right anthe squares of Q A, Q B will be (I. 40) gle, as is above shown. And if one of greater than twice the sum of the squares the points E, e lie between D and d, the of PD, DA, that is than the given sum, other will also lie between D and d; for if in the former case, and less in the latter. AE is to EB in a greater ratio than AD
to DB, which is the case when E lies PROP. 51.
between D and d, Ae will be to e B in Required the locus of the vertices of a greater ratio than A d to dB, which all triangles upon the sume base AB, is the case (as may easily be shown) only haring the side terminated in A greater when e lies between D and d: and conthan the side terminated in B, and their versely. Therefore, if the point Q be ratio the same with the given ratio of within the locus, and the angle D Qd AD to DB.
(by consequence) greater than a right Let P be a point in the locus. Di- angle (15. Cor. 3.), that is, than E Qe, vide AB produced (11.55.) in the point the point E cannot lie otherwise than PT
between D and d; and consequently the ratio of A E to E B, that is, the ratio of AQ to Q B, must be greater than the ratio of A D to D B. In the same manner it may be shown that, if the point Q lie without the locus, AQ will be to QB
in a less ratio. d, so that A d may be to d B as A D to Cor. If there be taken in the same D B, and join PA, PB, PD, Pd. straight line, and in the same direction
Then, because in the triangle PAB from a common extremity, three harmothe straight lines P D, Pd divide the nical progressionals, and if upon the base and the base produced in the ratio mean progressional for a diameter, a of the sides (II. 50.) they bisect the ver- circle be described, the distances of any tical and exterior vertical angles: there- point in the circumference from the fore the angle D Pd is equal to half the other extremities of the first and third
D E B
shall have to one another always the KPQ: this circle shall be the focus resame ratio, viz. that of the first to the quired. third.
For, let P be any point in the circum
ference of the circle KPQ ; join PA, Scholium.
draw the tangent PT to the circle BCD, The last proposition may be stated and join PE, cutting the circumference thus: “Required the locus of all points BCD in L; join also GP, and draw P, the distances of which from two PM perpendicular to AE. given points A and B, are to one another Then, by Prop. 51, because the circle in a given ratio." And it has been shown KPQ is described from the centre F that the locus is a circle in every case in with the radius FK, which is a mean which the given ratio is not that of equa- proportional between FA and FG, and lity; and in that particular case it is that P is a point in the circumference (44.) a straight line which bisects AB KPQ, PA is to PG as AK to KG, or as at right angles. Under this form it FA to FK (II. 22. Cor. 1.) because FA, readily suggests two other questions of FK and FG are proportionals. Therefore the same kind, which likewise lead to also PA2 is to PG2 as FA2 is to FK , or plane loci, and are at the same time so as FA to FG (II. 37.). And, because elegant and so nearly related to that we PAP : PG :: FA : FG, have been discussing, that they claim PARX EG : PGPX AE::FA XEG : some notice in this place.
FG * AE (Rule 2. Scholium, II. .) First, then, let it be "required to find Therefore, convertendo PA2X ÉG: the locus of ail points P such that the PG’X AE-PA? EG::FA EG : FG distance PA from a given point A, and <AE-FAXEG.(a) But, because PA the tangent PT drawn to a given circle
= PE + AE2 + 2 AEXEM (I. 37.) PCD are to one another in a given PA?X EG = PE: EG+ AE: * EGÉ ratio."
2AEXEM XEG ; and, for the like reaTake E the centre of the given circle; son, join AE; and, if PA is to be greater PG2X AE = PE2X AE+EG2X AEŁ than PT, produce AE to F (II. 55.) so 2AE EMEG; therefore PGP X AEthat AF may be to FE in the duplicate PAX AG=PEP X AE-PEEG, + of the given ratio (fig. 1.); but, if PA is EGP X AE-AEX EG, that is, = PE2X to be less than PT, produce EA to F so
AG-AG XAEXEG. that AF may be to FE in the duplicate of Again, because FG is equal to FEŁ the given ratio (fig. 2.); take EG (II. 52.) EG, FG XAE is equal to FEXAEI a third proportional to EA and ED, and EG XAE, that is, to FEXAG+FEX
EGŁEG XAGŁEG?, because AE is Fig. 1.
equal to AG+EG; and, in like manner, because FA is equal to FEFEG AG, FAX EG is equal to FE REG EGʻIEG AG; therefore, FG X AE -FAXEG is equal to FE x AG.
Therefore, substituting these values instead of the second and fourth terms of the proportion, (a),
PAXẾ G : PE2 x AG-AGXAEX EG::FAX EG : FEXAG, and hence, (Rule 2. Scholium, II.
). Fig. 2.
PA : PE: – AEXEG:: FA : FE, that is, because AEX EG is equal to ED. (II. 38. Cor. 1.), and PEP-ED is equal to PE2-ET? or PT, PA!: PT?:: FA: FE. Therefore PA is to PT in the subduplicate ratio of FA to FE, that is in the given ratio; and the circumference KPQ is the locus required.
If the given ratio be the ratio of equality, the difference of the squares of PA,
PË will be equal to the square of ED; FK a mean proportional (II. 51.) be- and therefore the locus is a straight tween FA and FG ; and, from the centre line (49.) cutting AE at right angles, and F with the radius FK describe a circle may be determined as in Prop. 49.