Geometry with Trigonometry

(Marvins-Underground-K-12) #1

186 Vector and complex-number methods Ch. 11


(x)If Z 1 =Z 2 and Z 3 ∈Z 1 Z 2 ,then(Z 1 ,Z 2 )↑(Z 3 ,Z 4 )if and only if
[Z 1 ,Z 2 ,Z 4 ,Z 3 ]is a parallelogram.
Proof.
(i) By the mid-point formula,

mp(Z 1 ,Z 4 )≡

(


x 1 +x 4
2

,


y 1 +y 4
2

)


,mp(Z 2 ,Z 3 )≡

(


x 2 +x 3
2

,


y 2 +y 3
2

)


,


and the result follows immediately from this.
(ii) By part (i) it is necessary and sufficient that we chooseZ 4 so thatx 4 =x 2 +
x 3 −x 1 ,y 4 =y 2 +y 3 −y 1.
(iii) This is immediate asx 1 +x 2 =x 1 +x 2 ,y 1 +y 2 =y 1 +y 2.
(iv) This is immediate asx 2 +x 3 =x 1 +x 4 ,y 2 +y 3 =y 1 +y 4. (v) We are given that
x 1 +x 4 =x 2 +x 3 ,y 1 +y 4 =y 2 +y 3 ,
x 3 +x 6 =x 4 +x 5 ,y 3 +y 6 =y 4 +y 5.


By addition(x 1 +x 6 )+(x 3 +x 4 )=(x 2 +x 5 )+(x 3 +x 4 ), so by cancellation ofx 3 +x 4
we havex 1 +x 6 =x 2 +x 5. Similarlyy 1 +y 6 =y 2 +y 5 and so the result follows.
(vi) For by (i) above we havex 1 +x 4 =x 3 +x 2 ,y 1 +y 4 =y 3 +y 2.
(vii) For by (i) above(x 2 −x 1 )^2 +(y 2 −y 1 )^2 =(x 4 −x 3 )^2 +(y 4 −y 3 )^2 , and now
we apply the distance formula.
(viii) For ifx 1 =x 2 ,y 1 =y 2 , then (i) above is satisfied if and only ifx 3 =x 4 ,y 3 =y 4.






Z 1


Z 2 O I


Z 3 Z^4


Figure 11.1.





Z 1


Z 3 Z^4


Z 2


T


(ix) For suppose first that(Z 1 ,Z 2 )↑(Z 3 ,Z 4 ).ThenZ 3 ∈land
mp(Z 3 ,Z 4 )∈l,soZ 4 ∈l. By (vii) above we have|Z 1 ,Z 2 |=|Z 3 ,Z 4 |. Suppose first that
lis not perpendicular toOIand that, as in 6.4.1(ii), the correspondence between≤l
and the natural order≤OI, under whichO≤OII, is direct. ThenπOI(Z 1 )≤OIπOI(Z 2 ),
sox 1 ≤x 2. By (i) abovex 3 ≤x 4 , and so by this argument traced in reverse we have
Z 3 ≤lZ 4. If the correspondence is indirect, we havex 2 ≤x 1 ,x 4 ≤x 3 instead. Whenl
is perpendicular toOI, we project toOJinstead and make use of they-coordinates.
Conversely suppose thatZ 4 ∈l,|Z 1 ,Z 2 |=|Z 3 ,Z 4 |andZ 3 ≤lZ 4 .Nowlhas para-
metric equationsx=x 1 +t(x 2 −x 1 ), y=y 1 +t(y 2 −y 1 )(t∈R). Suppose thatZ 3
andZ 4 have parameterst 3 ,t 4 , respectively, so that


x 3 =x 1 +t 3 (x 2 −x 1 ), y 3 =y 1 +t 3 (y 2 −y 1 ),
x 4 =x 1 +t 4 (x 2 −x 1 ), y 4 =y 1 +t 4 (y 2 −y 1 ).
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