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(2) If B is equal to the perpendicular, the circle would meet DH at the foot of the perpendicular, and there would be one triangle, right-angled, which fulfils the given conditions.

(3) If B is greater than the perpendicular but less than DE, then the circle will meet DH in two points I, I' as in the figure, on the same side of D, and there will be two triangles EDI, EDI' which fulfil the given conditions.

(4) If B is equal to DE, the point I will coincide with D, and one of the two triangles disappears, and the other is isosceles.

(5) If B is greater than DE, the circle will meet DH in two points on the opposite sides of D, but one only of the triangles made by joining EI, EI' will be found to have the angle D, and the other will have the supplementary angle: that is, there will be only one solution.

The cases of the given angle being a right angle or an obtuse angle are left to the ingenuity of the student.

SECTION V.

Loci.

: When a point has to be found to fulfil one given geometrical condition, the problem is indeterminate : that is, an infinite number of points can be found to fulfil the given condition. For example, if the problem is to find a point at a given distance from a given point, it is plain that all the points in the circumference of a circle, described with that point as centre and the given distance as radius, fulfil this condition. Or again, if a point has to be found at a given distance from a given straight line of indefinite length, it may lie anywhere on either of two straight lines parallel to the given line, and at the given distance from it on either side.

All the points which satisfy a single given geometrical condition lie in general in a line or lines : and this line, or these lines, are called the locus of the point under the given condition. Hence we get the following definition of a locus :

Def. If any and every point on a line or group of lines (straight or curved), and no other point, satisfies an assigned condition, that line or group of lines is called the locus of the point satisfying that condition.

In order that a line or group of lines A may be properly termed the locus of a point satisfying an assigned condition X, it is necessary and sufficient to demonstrate the two following associated Theorems :

If a point is on A, it satisfies X.

If a point is not on A, it does not satisfy X. It may sometimes be more convenient to demonstrate the contrapositive of either of these Theorems.

The following examples of loci are important.

i. The locus of a point at a given distance from a given point is the circumference of a circle having a radius equal to the given distance and having its centre at the given point.

ii. The locus of a point at a given distance from a given straight line is the pair of straight lines parallel to the given line, at the given distance from it, and on opposite sides of it.

The proofs of these two Theorems are obvious.

iii. The locus of a point equidistant from two given points is the straight line that bisects, at right angles, the line joining the given points.

Part. En. Let A, B be the two given points; P a point equidistant from A and B, so that. PA = PB; . it is required to find the locus of P.

Constr. Join AB; and bisect it in 0, and join PO.

Then PO produced is the locus, required.

and

and

Proof. In the triangles AOP, BOP, because

AO=OB, (Constr.)
PO is common,

AP=BP, . (Hyp.)) therefore the angle AOP= the angle BOP, (1. 15.) and therefore PO is at right angles to AB; that is, a point equidistant from A and B lies on the line which bisects AB at right angles. Further, every point not on the bisector, is at unequal distances from A and B, as may be proved by Theorem 14, and therefore the line which bisects AB at right angles is the locus of points equidistant from A and B.

iv. The locus of a point equidistant from two intersecting straight lines is the pair of lines, at right angles to one another, which bisect the angles made by the given lines.

Let AB, DC intersect in 0; it is required to find a point equally distant from AB and DC.

Bisect the angle COB; and in the bisector take any point P. Let fall PN, PM perpendicular to DC, AB.

In the triangles PON, POM, because the angles PON, PNO are respectively equal to the angles POM, PMO,

(Constr.)

and the hypotenuse PO is common,
therefore the triangles are equal in all respects,

(Theorem 17). and therefore PN= PM.

In the same manner every point in the bisector of any one of the four angles at O is equally distant from AB and CD; that is, the locus of points equally distant from two straight lines which intersect, is the bisectors of the angles between the lines.

It may further be proved that no point not in a bisector is equally distant from these lines; that is, the bisectors are the complete locus.

EXERCISES.

Find the following loci :-
(1) Of a point at a given distance from a given point.
(2) Of a point at a given distance from a given line.
(3) Of a point at a given distance from a given circle.

(4) A horse is tethered by a chain fastened to a ring which slides on a rod bent into the form of a rectangle. Find the outline of the area over which he can graze.

(5) Find the locus of a point equidistant from two given points. Prove that the locus found is complete.

(6) Find the locus of points at which two equal lengths, adjacent or not adjacent, of a straight line subtend equal angles.

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