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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 O, and join PO.

Then PO produced is the locus, required.

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(I. 15.)

therefore the angle AOP= the angle BOP,

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 O; it is required to find a point equally distant from AB and DC.

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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,

and therefore PN= PM.

(Theorem 17).

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.

(7) Find part of the locus of points at which two adjacent sides of a square subtend equal angles.

(8) Find the locus of a point at which two adjacent sides of a rectangle subtend supplementary angles.

INTERSECTION OF LOCI.

When a point has to be found which satisfies two conditions, the problem is generally determinate if it is possible: and the method of loci is very frequently employed in discovering the point. For if the locus of points which satisfy each condition separately is constructed, it is obvious that the points which satisfy both conditions must be the points common to both loci, that is, must be the point or points where the loci intersect.

For example; a triangle is to be constructed on a given base with its sides of given lengths. Let AB be the base. The two conditions are that the

lengths of the two sides are given; the point sought for is the vertex: now the vertex must be at a certain distance from A= one of the given lengths; its locus is therefore a certain circle round A as centre. Similarly it must be at a certain distance from B; its locus is therefore another circle round B as

B

centre. The points of intersection of these circles are therefore the vertices of the two equal triangles which fulfil the given conditions.

It was this reasoning that suggested the construction in Problem 5.

Occasionally it will be found that with certain conditions among the data in the following Exercises the loci do not intersect, or the solution becomes impossible. So in the case given, it will not be difficult to see that the circles would not intersect unless any two of the given sides were greater than the third side. These conditions among the data for the possibility or impossibility of a solution should always be found.

The principle of the intersection of loci may be thus stated:

If A is the locus of a point satisfying the condition X, and B the locus of a point satisfying the condition Y; then the intersections of A and B, and these points only, satisfy both the conditions X and Y.

The following examples of intersection of loci are important, and are at once demonstrated by the aid of the preceding examples of loci.

i. There is one and only one point in a plane which is equidistant from three given points not in the same straight line.

ii. There are four and only four points in a plane each of which is equidistant from three given straight lines that intersect one another but not in the same point.

EXERCISES ON INTERSECTION OF LOCI.

I. Find a point in a given straight line at equal distances from two given points. Construct the figures for all cases.

2. Find a point in a given straight line at a given distance from a given straight line.

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