Geometry with Trigonometry

Sec. 6.1 Frame of reference, Cartesian coordinates 83

Proof.

(i) This is clear from the definition of coordinates.

(ii) For ifZ 1 ,Z 2 ∈H 3 we have

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

and so asU 1 ,U 2 ∈[O,I,
|U 1 ,U 2 |=±(x 2 −x 1 )

according asU 1 ∈[O,U 2 ]orU 2 ∈[O,U 1 ]. Similarly ifZ 1 ,Z 2 ∈H 4 ,wehave

|O,U 1 |=−x 1 ,|O,U 2 |=−x 2

and
|U 1 ,U 2 |=±[−x 2 −(−x 1 )]

according asU 1 ∈[O,U 2 ]orU 2 ∈[O,U 1 ]. Finally ifZ 1 ∈H 3 ,Z 2 ∈H 4 then

|O,U 1 |=x 1 ,|O,U 2 |=−x 2

andO∈[U 1 ,U 2 ]so that
|U 1 ,U 2 |=x 1 +(−x 2 );

similarly ifZ 1 ∈H 4 ,Z 2 ∈H 3.
That|V 1 ,V 2 |=±(y 2 −y 1 )can be shown in the same way.
(iii) Now the lines throughZ 1 parallel toOIand throughZ 2 parallel toOJare
perpendicular to each other, and so meet in a unique pointZ 4. ClearlyπOI(Z 4 )=
πOI(Z 2 )=U 2 soZ 2 andZ 4 have the same first coordinate,x 2 ;πOJ(Z 4 )=πOJ(Z 1 )=V 1
soZ 1 andZ 4 have the same second coordinate,y 1. ThusZ 4 has coordinates(x 2 ,y 1 ).
If the pointsZ 1 ,Z 2 ,Z 4 are not collinear, then by Pythagoras’ theorem

|Z 1 ,Z 2 |^2 =|Z 1 ,Z 4 |^2 +|Z 4 ,Z 2 |^2 ;

if they are collinear we must haveZ 1 =Z 4 orZ 2 =Z 4 and this identity is trivially true.
But|Z 1 ,Z 4 |=|U 1 ,U 2 |as[Z 1 ,Z 4 ,U 2 ,U 1 ]is a rectangle, or elseZ 1 =Z 4 andU 1 =U 2 ,
orZ 1 =U 1 ,Z 4 =U 2. Similarly|Z 2 ,Z 4 |=|V 1 ,V 2 |. Thus we have the distance formula

|Z 1 ,Z 2 |^2 =|U 1 ,U 2 |^2 +|V 1 ,V 2 |^2

=(x 2 −x 1 )^2 +(y 2 −y 1 )^2 ,

which expresses the distance|Z 1 ,Z 2 |in terms of the coordinates ofZ 1 andZ 2.
(iv) IfZ 1 =Z 2 ,thenx 2 =x 1 ,y 2 =y 1 so thatx 3 =x 1 ,y 3 =y 1. ThusZ 3 =Z 1 =
mp(Z 1 ,Z 1 ),as required.
Suppose then thatZ 1 =Z 2. Note that

|Z 1 ,Z 3 |^2 =

[

x 1 +x 2

2

−x 1

] 2

+

[

y 1 +y 2

2

−y 1

] 2

=

[

x 2 −x 1

2

] 2

+

[

y 2 −y 1

2

] 2