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.