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from AB and AC. So also any point in AK which bisects the angle supplementary to _BAC is equidistant from AB and AC. And no point out of AD
or AK can be equidistant from AB and AC. The above are examples of problems where we are required to find a point satisfying a certain condition; and when the problem is solved it is found that not one point, but any point in a certain line or group of lines satisfies the given condition, and no other point does so: then the line, or group of lines, is called the locus of the point satisfying the given condition.
DEF.-The locus of a point satisfying a given condition is the line, or group of lines, any point in which, and no other point, satisfies the given condition.
The method of loci may be applied in the solution of certain problems which are determinate, or which give a limited number of solutions, e.g.
6. To find a point in a plane equidistant from three given points in the plane. Let A, B, and G be
B the three given
EF bisecting AB A-
A and B.
of points equidistant from B and G.
Since all points equidistant from A and B are in EF,
and all points equidistant from B and G are in HK, : a point, to be equidistant from A, B, and G, must be in EF and also in HK, and .. can only be the point P, the point of intersection of
the two loci. This problem is impossible of solution when A, B, and G are in the same straight line, for then EF becomes || HK.
If the locus of a point satisfying one condition is found, and the locus of a point satisfying another condition, then if it is possible for a point to exist which satisfies both conditions simultaneously, it must be common to the two loci, and so must be their point, or points, of intersection.
7. To find a point at a given distance from a given point, and also at a given distance from a given line.
The locus of a point at a given distance from a given point is a circle (Loci 2).
The locus of a point at a given distance from a given line is a pair of lines || the given line at the given distance from it (Loci 3).
If the circle and either or both of these lines intersect, their points of intersection satisfy the given conditions.
If either of the lines touches the circle, there is one point of contact which satisfies the given conditions.
If the circle and the lines do not meet at all, then the problem is impossible with the given conditions.
Additional Examples of Loci. 8. Find the locus of the vertices of equal triangles on equal bases which are in the same straight line.
9. Find the locus of the middle point of any straight line drawn from a given point to meet a given straight line.
10. Show how the problems (1) To construct an equilateral triangle on a given straight line, (2) To construct a triangle with sides equal respectively to three given lines, are examples of intersection of loci.
11. To find a point in a line at a given distance from (1) a given point, (2) a given line, (3) a given circle.
12. Investigate all the different cases of Loci 1, viz., To find a point in a given line equidistant from two given points.
13. To find a point equally distant from two given intersecting lines, and at a given distance from a given point.
14. AB and CD are two equal parts of a given straight line, find the locus of the point at which they subtend equal angles.
15. To find a point equally distant from three given straight lines which intersect so as to form a triangle. How many such points are there?
16. A house is described as being 200 yards from a road and 100 yards from a railway. The road and the railway are both straight, and they intersect. Find the position of the house. How many different positions would the given conditions admit of its occupying?
17. Show how the description of the position of a place on a map by its latitude and longitude is an example of intersection of loci.
18. If any straight line be drawn to intersect two parallel lines, find the locus of the middle point of the part intercepted between the parallel lines.
19. If a point be taken in the circumference of a circle and any line be drawn from it to the circumference, find the locus of its middle point (see Exercises 91, 92, and 100).
20. The position of a boulder in a field is described as 60 ft. from the western boundary, and 75 ft. from a particular tree. Does this description fix the position exactly?
APPENDIX TO BOOK I.
EU. I. 2. PROB.
From a given point to draw a straight line equal to a given straight line. Let A be the given point, and BC the
given st. line.
A ADB (1. 1).
describe O CGH. From D as centre, with DG as radius, describe O GKE. AE is the line required. :: D is centre of O GKE, DG = DE (Def. 11), and parts of
these DB = DA (Def. 17), .. remainder BG AE. But ::: B is centre of O CHG, BG BC (Def. 11),
:. AE = BC. Q.E.F. Note.-It is evident that there can be drawn from A an innumerable number of lines each equal to BC, and yet this proposition gives
us the means of drawing only eight of these, (1.) By joining A to either B or C; (2.) By drawing the equil. A on either side of its base; (3.) By producing DB in either direction.
If A is on BC, there are four cases; but if A is B or C, any number of lines can be drawn equal to BC.
This proposition and the next may be superseded in two ways :
(1.) By reading Post. 3 thus,-Let it be granted that a circle may be described from any point as centre, and with radius equal to any finite distance.
(2.) By retaining Eu. Post. 3 with the restriction usually read into it, but allowing that a circle or a straight line may be taken up and placed down anywhere without alteration of its magnitude.
EU. I. 3. PROB. From the greater of two given straight lines to cut off a part equal to the less.
Let AB and C be the given st. lines.
a O with A as centre and AD as radius,
cutting AB at E. ::: A is centre of O DEF, AE
AD (Def. 11), and AD = C (cons.), .: AE
EU. I. 5. THEOR. The angles at the base of an isosceles triangle are equal to one another; and if the equal sides be produced, the angles on the other side of the base shall also be equal to one another. Given AB AC, and let AB and AC be A
produced To prove LABC
= L ACB, and DBC =
=- ECB. Take F any point in BD, cut off in AE, AG = AF (I. 3). Join FC and GB.
G :: FA = AG (cons.), AC = AB (given), and
LA is com., .: AFAC = AGAB in every respect (I. 4). .: BG = FC, L ABG = L ACF, and LAFC = L AGB. Again :: AF AG, and AB AC, :: BF
CG (Ax. 3), and FC BG (proved), and 2 BFC LCGB (proved). .: ABFC = ACGB in every respect (1. 4). .: LFBC
= LGCB, these are the angles on the other side of the base, and _BCF = L CBG. But / ACF = L ABG (proved), and LBCF
LCBG (proved), .: LACB LABC (Ax. 3), and these are the angles at the