147. Locus. The path of a point that moves in accordance with certain given geometric conditions is called the locus of the point.

X

-Y

d

A

B

X-

---Y'

Thus, considering only figures in a plane, a point at a given distance from a given line of indefinite length is evidently in one of two lines parallel to the given line and at the given distance from it. Thus, if AB is the given line and d the given distance, the locus is evidently the pair of parallel lines XY and X'Y'.

The locus of a point in a plane at a given distance r from a given paint O is evidently the circle described about O as a center with a radius r.

The plural of locus (a Latin word meaning "place") is loci (pronounced lō-si).

r

We may think of the locus as the place of all points that satisfy certain given geometric conditions, and speak of the locus of points. Both expressions, locus of a point and locus of points, are used in mathematics.

EXERCISE 14

State without proof the following loci in a plane:

1. The locus of a point 2 in. from a fixed point 0.

2. The locus of the tip of the minute hand of a watch.

3. The locus of the center of the hub of a carriage wheel moving straight ahead on a level road.

4. The locus of a point 1 in. from each of two parallel lines that are 2 in. apart.

5. The locus of a point on this page and 1 in. from the edge. 6. The locus of the point of a round lead pencil as it rolls along a desk.

7. The locus of the tips of a pair of shears as they open, provided the fulcrum (bolt or screw) remains always fixed in one position.

8. The locus of the center of a circle that rolls around another circle, always just touching it.

148. Proof of a Locus. To prove that a certain line or group of lines is the locus of a point that fulfills a given condition, it is necessary and sufficient to prove two things:

1. That any point in the supposed locus satisfies the condition. 2. That any point outside the supposed locus does not satisfy the given condition.

D

P

A.

B

For example, if we wish to find the locus of a point equidistant from these intersecting lines AB, CD, it is not sufficient to prove that any point on the angle-bisector PQ is equidistant from AB and CD, because this may be only part of the locus. It is necessary to prove that no point outside of PQ satisfies the condition. In fact, in this case there is another line in the locus, the bisector of the BOD, as will be shown in § 152.

149. Perpendicular Bisector. A line that bisects a given line and is perpendicular to it is called the perpendicular bisector of the line.

EXERCISE 15

Draw the following loci, giving no proofs:

1. The locus of a point in. below the base of a given triangle ABC.

2. The locus of a point in. from a given line AB.

3. The locus of a point 1 in. from a given point 0.

4. The locus of a point in. outside the circle described about a given point 0 with a radius 1 in.

5. The locus of a pointin. within the circle described about a given point O with a radius 11⁄2 in.

6. The locus of a point in. from the circle described about a given point O with a radius 1 in.

7. The locus of a point in. from each of two given parallel lines that are 1 in. apart.

PROPOSITION XXXIV. THEOREM

150. The locus of a point equidistant from the extremities of a given line is the perpendicular bisector of that line.

Y

B

Given YO, the perpendicular bisector of the line AB.

To prove

that YO is the locus of a point equidistant from

A and B.

Proof. Let P be any point in YO, and C any point not in YO.

Draw the lines PA, PB, CA, and CB.

151. COROLLARY. Two points each equidistant from the extremities of a line determine the perpendicular bisector of the line.

PROPOSITION XXXV. THEOREM

152. The locus of a point equidistant from two given intersecting lines is a pair of lines bisecting the angle formed by those lines.

B

C

Given XX' and YY' intersecting at O, AC the bisector of angle X'OY, and BD the bisector of angle YOX.

To

prove that the pair of lines AC and BD is the locus of a point equidistant from XX' and YY'.

Proof. Let P be any point on AC or BD, and Q any point not on AC or BD. Let PM and QR be 1 to XX', PN and QS to YY!

Let QS cut AO at P'. Draw P'T to XX', and draw QT.

.. the pair of lines is the required locus, by § 148.

Post. 3

§ 86

Ax. 10

Ax. 9

Q.E.D.

153. The Synthetic Method of Proof. The method of proof in which known truths are put together in order to obtain a new truth is called the synthetic method.

This is the method used in most of the theorems already given. The proposition usually suggests some known propositions already proved, and from these we proceed to the proof required. The exercises on this page and on pages 78 and 79 may be proved by the synthetic method.

154. Concurrent Lines. If two or more lines pass through the same point, they are called concurrent lines.

155. Median. A line from any vertex of a triangle to the mid-point of the opposite side is called a median of the triangle.

EXERCISE 16

1. If two triangles have two sides of the one equal respectively to two sides of the other, and the angles opposite two equal sides equal, the angles opposite the other two equal sides are equal or supplementary, and if equal the triangles are congruent. Let AC A'C', BC = B'C', and ▲ B = ▲ B'.

=

Place A A'B'C' on AABC so that B'C' shall coincide with BC, and ZA' and ZA shall be on the same side of BC.

=

Since B' LB, B'A' will fall along what line? Then A' will fall at A or at some other point in BA, as D. If A' falls at A, what do we know about the congruency of the ▲ A'B'C' and ABC ?

If A'falls at D, what about the congruency of the A A'B'C' and DBC? Since CDC'A′ = CA, what about the relation of A to Z CDA? Then what about the relation of the CDA and BDC?

Then what about the relation of the

A and BDC?

Draw figures and show that the triangles are congruent :

1. If the given angles B and B′ are both right or both obtuse.

2. If the angles A and A' are both acute, both right, or both obtuse. 3. If AC and A'C' are not less than BC and B'C' respectively.