A ny point equidistant from two points lies on the perpendicular bisector of the line segmentjoining those two points

Given: I' any point such that AI' = BP; CM 1. AB; AM = BM.

Conclusion: Plies on eM.

ProOf

ST ATEMENTS

REASONS

1. Plies on CM or I' does not lie on

+--,)0

CM.

2. Assume I' does not lie on CM.

3. Draw PM.

I. Law of excluded middle.

2. Temporary assumption.

3. Postulate 2.

324

FUNDAMENTALS OF COLLEGE GEOMETRY

4. AP

= BP. 4. Given.

5. AiH = BM. 5. Given.

6. PM = PM. 6. Theorem 4.1.

7. ,6.AJvlP

,6.BiHP. 7. S.S.S.

8. LAMP

LBMP. 8. § 4.28.

9. JWP 1. AB. 9. Theorem 3.14.

10. CM 1. AB. 10. Given.

11. There are two distinct lines pass- 11. Statements 9 and 10.

ing through P and perpendicular toAB.

12. This is impossible.

13. :.P must lie on CM.

12. Theorem 5.2.

13. Either

p

^or ^not-p; [not(not-p)] ~

Theorem 11.5

11.6. The locus of points in the interior of an angle which are equidistant from the sides of the angle is the bisector of the angle minus its endpoint.

PART I: A ny point on the

bisector of the angle

is equidistant

from the

sidesof the .~

angle. ^'

Given: BF b sects LABC; point P # Bon BF; PE 1. 1M; PD 1. EG.

. n: PE = PD.

Proof

ST ATEMENTS

1. BF bisects LABC.

2. La ₌₌ Lf3.

3. PE 1. EA, PD 1. BC.

4. LBEP and LBDP are right L§.

5. BP=BP.

6. Right ,6.BEP

right ,6.BDP.

7. PE = PD.

---

B<f-L, ~

{J -~f ~

I I I

Theorem 11.5.

REASONS

1. Given.

2. § 1.19.

3. Given.

4. § 1.20.

5. Theorem 4.1.

6. §5.27 and A.S.A.

7. § 4.28.

---

..

GEOMETRIC LOCI

325

PART II: Any point equidistant from the sides of an angle is on the bisector of the angle.

Given: BFbisects LABC; PE 1. BA; PD 1. BC; PE

= PD, iJP Conclusion: P lies on lIF.

Proof

STATEMENTS REASONS

1. PE 1.

BA, PD 1. BC.

2. LBEP and LBDP are right L§.

3. PE = PD.

4. BP=BP.

5. ,6.BEP

,6.BDP.

6. La ₌₌ Lf3.

7. :.BPbisects LABC.

1. Given.

2. § 1.20.

3. Given.

4. Theorem 4.1.

5. Theorem 5.20.

6. § 4.28.

7. § 1.19.

11.7. Corollary: The locus of points equidistant from two given intersect- ing lines is the pair of perpendicular lines which bisects the vertical angles formed by the given lines.

Theorem 11.6

I~& Thi: locu~ <If alll"<I;..b "ud. tl.al DArB

i" ii hiiiugle having

Ail

a fixed line segmefilasn-ypotenuse IS a cIrctenavrng ALJ as amamefer;excepfror points A and B themselves.

11.9. Intersection of loci. In our study of loci thus far we have limited our- selvcs to finding points which satisfy only one condition. Sometimes a point must satisfy each of two or more given conditions. In such cases each condition will determine a locus. The required points will then be the intersection of the loci, since only those points will lie on each of the lines which, in turn, represent the given conditions.

Th us to locate the point

(or set of points)

which satisfies

two or more

conditions, determine the locusfor each condition.

The point (or set of points) at

which

theseloci intersect

will be the required point (or set of points).

In solving a problem involving intersecting loci, it is customary to place the given parts in the most general positions in order to determine the most general solution for the problem. Then, in a discussion that follows the general solution, consideration is given to special positions of the given parts and to the solutions for these special cases.

---..---

,-

Discussion:

1. If

~

is a tangent to 0 A, the required solution set ~l be only one point.

2. If RS falls outside OA (i.e., if distance from A to RS is more than d), there are no points which will satisfy the required conditions. The solution set is a null set.

3. There can never be more than two points in the solution set.

11.11. Illustrative Example 2.

Find all the points that are equidistant from

(

two fixed points and equidistant from two intersecting lines

-

Discussion:

1. If l3 I/l2 (or if l3 I/lI), the solution set will consist 2. If l3 coincides with either II or l2, the solution

number of points.

3. If l3 passes through 0, there is one point in the 4. In all other cases there will be two points in the

State and prove the solutions Discuss each.

1. Find all points a given distance from a fixed point two parallel lines.

Exercises

for each of

and equidistant the following locus

solution set.

from problems 3.

][S, the .1

locus of I Band C.

Circle A with center at A and radius equal to d is the locus of all points a distance d from A.

The required points are PI and P2, the intersection of E, and OA

3. § 11.9 2.

ST A TEMENTS

1 bisector

)oints

of BC, is eg uidistant [r

2

REASONS

1. Theorem 11.4.

Theorem ILl

2

The locus of points equidistant

from A and B is line l3 the

.1 bisector ofAB.

The loci of points equidistant from CD and EF are lb which bisects LCOE, and l2, which bisects

/f--'OD.

The solution set is PI and P2, the intersections of line l3 with lines II and l2'

of only one point.

set will consist of an infinite 3.

2. §

11.9. 11 7 Corollary.

GEOMETRIC LOCI [3/

t/

%pz

/1

Solution:

Given: Points A, B, and C.

Find: All points a distance d from A equidistant from Band C.

and

Illustrative Example 1

B ~//B

1< "'- //~

I d

)<

^PI ^C

"'. j'

, A / I

\ / I

/ /

"

_'''''''-I~ ^/ ^// ^/

/ Pz

I:

STATEMENTS Solution:

Find:

Points A and B; lines CD and EF intersecting at O.

All points equidistant from A and B and also equidistant from inter- secting lines CD and EF.

1.

REASONS

c -/

/

A~/

I~B

Theorem

illustrative ExamPle 2.

11 4 11.10. Illustrative Example

1. Find the set of all the points that are a

given distance d from a fixed point A and which are also equidistant from two points Band C.

326

Given:

327

328

FUNDAMENTALS OF COLLEGE GEOMETRY

2. Find all points within a given angle equidistant from the sides and a give distance from the vertex.

3. Find all the points equidistant from two 'parallel lines and equidistant from the sides of an angle.

4. Find all the points equidistant from the three vertices of L.ABC.

5. Find all the points equidistant from the three sides of L.ABC.

6. Find all the points that are equidistant from two parallel lines and lie on a third line.

7. Find all the points a distance d1 from a given line and dz from a given circle.

8. Find all points equidistant from two parallel lines and a given distance from a third line.

9. Find all points equidistant from two parallel lines and equidistant from two points.

10. Find all points equidistant from two intersecting lines and also at a given distance from a: given point.

Additional Loci (Optional)

11.12. Loci other than straight lines and circles. Euclidean geometry con- fines itself to figures formed by straight lines and circles. Our loci have thus far all resolved into straight lines and circles or combinations of them. The

locus of points equidistant from two straight lines, for example, is another straight line. The locus of points equidistant from two points is a straight line. The locus of points a given distance from a fixed point is a circle.

Let us briefly consider three other loci configurations. The early Greeks were familiar with these curves but were not able to relate them to our physical world. By the seventeenth century the advances of science and technology had produced a need for a clearer understanding of the properties of these curves. By that time mathematicians had developed powerful techniques of algebra and analysis which aided them in the study of these curves. We shall leave the algebraic analysis of these curves to the student when he studies analytic geometry in his future mathematical pursuits and limit ourselves in this text to a general discussion of these curves.

8.13. The parabola. As mathematicians studied loci of points equidistant from two points and loci of points equidistant from two straight lines, it was only natural that they should consider the locus of points equidistant from a point and a line. Such a locus is the parabola.

Definition: A parabola is the locus of the points whose distances from a fixed line and a fixed point are equal.

I::"_::"^."

;

fii:

, .

;

:'- .

it'

, '

"--

I ^~

I

i,.

",)i:

Thus, in Fig. 11.3, if P1Dl = P1F, 2Dz = P2F, . . ., the curve is a parabola. Conversely, if the curve is a parabola, P3D3

= P:JF,. ... With a

little study the student should dis- cover that the shape of the parabola varies as the distance from the fixed point to the fixed line varies.

We know that many moving objects travel in parabolic paths. A ball thrown in the air, the projectile fired from a cannon (see Fig. 11.4), the bomb released from an airplane, a stream of water from a garden hose

would all follow a parabolic path if the resistance of air could be neglected.

Fig. 11.4.

Thus, in firing an artillery shell, if the angle of elevation of the gun and its muzzle velocity are known, it is possible to calculate the equation of the path of flight. It is then possible to calculate in advance how far the projectile will go and how long it will take to go that distance. By varying the angle of elevation of the gun, the path of flight can be varied.

In like manner the equation of the path of the bomb released from an airplane can be determined (Fig. 11.5).

From the equation, the speed of the airplane, the height of plane and the position of the target, it can be deter-

GEOMETRIC LOCI 329

Fig. 113.

~IA I II II II II I II I I I II I I II

E'atb-oi plane

)

('~--'"'-.I.)'"⁰ 1'-)...1

L/I II II II II II I II II II II I,

Target X Fig. 11.5.

mined when to drop the bomb. Today the whole procedure is so mechanized ';

that the bombardier does not need to consider the equation, nor even be

i

aware of its existence. However, the persons responsible for the mechanized procedure had to use extensively these equations.

The parabolic curve is also useful in construction of many physical objects.

A parabolic arch is often used in constructing bridges since it is stronger than any other. (See Fig. 11.6.)

Fig, 11,6,

:

1

'\\;' t.';

),

:1: ^\

: I

I r

^:,',^,

^! I

;--

^~'i^I~,'

_~ ,

^.^.^'^'.

I

,

,~' J::$;;~1'

~,~,\'~

I.,^"

i' ~

;(, .:M:;.

-

331

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