Geometry with Trigonometry

84 Cartesian coordinates; applications Ch. 6

and so|Z 1 ,Z 3 |=^12 |Z 1 ,Z 2 |. Similarly

|Z 3 ,Z 2 |^2 =

[

x 1 +x 2

2

−x 2

] 2

+

[

y 1 +y 2

2

−y 2

] 2

=

[

x 1 −x 2

2

] 2

+

[

y 1 −y 2

2

] 2

and so|Z 3 ,Z 2 |=^12 |Z 1 ,Z 2 |.Then

|Z 1 ,Z 3 |+|Z 3 ,Z 2 |=|Z 1 ,Z 2 |.

It follows by 3.1.2 and 4.3.1 thatZ 3 ∈[Z 1 ,Z 2 ]⊂Z 1 Z 2 .As|Z 1 ,Z 3 |=|Z 3 ,Z 2 |it then

follows thatZ 3 =mp(Z 1 ,Z 2 ).

O I

J H 1

H 2

H 4 H 3

Z 1

U 1

V 1

Z 4

U 2

Z 2 V 2

Figure 6.3. The distance formula.

U 1 O U 2

O U 1 U 2

U 1 U 2 O

Order of points on thex-axis.

(v) By 2.1.4 at least one of

(a)O∈[U 1 ,U 2 ],(b)U 1 ∈[O,U 2 ],(c)U 2 ∈[O,U 1 ],

holds.
In (a),U 1 andU 2 are in different half-lines with end-pointO. We cannot have
U 1 ∈[O,Ias then we would havex 1 ≥ 0 ,x 2 ≤0, a contradiction. ThusU 2 ∈[O,Iso
thatU 1 ≤lO,O≤lU 2 and thusU 1 ≤lU 2.
In (b) we cannot haveU 1 ≤lO. For then we would haveU 2 ≤lOand

|O,U 1 |=−x 1 ,|O,U 2 |=−x 2.

AsU 1 ∈[O,U 2 ]we have|O,U 1 |≤|O,U 2 |which yields−x 1 ≤−x 2 and sox 1 ≥x 2 ,a

contradiction. HenceO≤lU 1 and so asU 1 ∈[O,U 2 ],U 1 ≤lU 2.

In (c) we cannot haveO≤lU 1. For then we would haveO≤lU 2 and so

|O,U 1 |=x 1 ,|O,U 2 |=x 2.

AsU 2 ∈[O,U 1 ]we have|O,U 2 |≤|O,U 1 |,sothatx 2 ≤x 1 , a contradiction. Hence
U 1 ≤lOsoU 1 ≤lU 2.