Sec. 11.2 Sum of couples, multiplication of a couple by a scalar 187
Recall thatZ 1 ,Z 2 have parameters 0 and 1 and 0<1. As in the last paragraph above,
iflis not perpendicular toOIand the correspondence between≤land≤OIis direct,
thenx 1 <x 2 ,x 3 <x 4 ; hencet 3 <t 4 and we obtain this same conclusion when the
correspondence is inverse. Whenlis perpendicular toOIwe project toOJinstead,
and use they-coordinates. Moreover
|Z 3 ,Z 4 |^2 =[(t 4 −t 3 )(x 2 −x 1 )]^2 +[(t 4 −t 3 )(y 2 −y 1 )]^2 =(t 4 −t 3 )^2 |Z 1 ,Z 2 |^2.Hence|t 4 −t 3 |=1,andsoast 3 <t 4 we havet 4 = 1 +t 3 .Then
x 1 +x 4 = 2 x 1 +( 1 +t 3 )(x 2 −x 1 ),x 2 +x 3 =x 2 +x 1 +t 3 (x 2 −x 1 )and these are equal. Similarly
y 1 +y 4 = 2 y 1 +( 1 +t 3 )(y 2 −y 1 ),y 2 +y 3 =y 2 +y 1 +t 3 (y 2 −y 1 )and these are equal. By (i) above we now have(Z 1 ,Z 2 )↑(Z 3 ,Z 4 ).
(x) If[Z 1 ,Z 2 ,Z 4 ,Z 3 ]is a parallelogram, then mp(Z 1 ,Z 4 )=mp(Z 2 ,Z 3 ). Conversely
suppose thatZ 1 =Z 2 ,Z 3 ∈Z 1 Z 2 and mp(Z 1 ,Z 4 )=mp(Z 2 ,Z 3 ).ThenZ 1 Z 2 ,Z 3 Z 4 have
equations
−(y 2 −y 1 )(x−x 1 )+(x 2 −x 1 )(y−y 1 )= 0 ,
−(y 4 −y 3 )(x−x 3 )+(x 4 −x 3 )(y−y 3 )= 0.By (i) above,x 2 −x 1 =x 4 −x 3 ,y 2 −y 1 =y 4 −y 3 , so these lines are parallel. Similarly
Z 1 Z 3 ,Z 2 Z 4 have equations
−(y 3 −y 1 )(x−x 1 )+(x 3 −x 1 )(y−y 1 )= 0 ,
−(y 4 −y 2 )(x−x 2 )+(x 4 −x 2 )(y−y 2 )= 0 ,and by (i) abovex 3 −x 1 =x 4 −x 2 ,y 3 −y 1 =y 4 −y 2 , so that these lines are parallel.
Thus[Z 1 ,Z 2 ,Z 4 ,Z 3 ]is a parallelogram.
11.2 Sum of couples, multiplication of a couple by a scalar .....
11.2.1 .....................................
Z 2
O
Z 1
Z 3
T
Z 1 Z ^4
OI
(t> 0 )Figure 11.2.
Z 1
Z 4 O I
(t< 0 )