Sec. 10.1 Complex coordinates 147
from which we havey 4 −y 3 =t(y 2 −y 1 ).
Whenx 2 −x 1 =0, by (10.1.1) we must havex 4 −x 3 =0. We now let
t=y 4 −y 3
y 2 −y 1,
so thaty 4 −y 3 =t(y 2 −y 1 ). For thistwe also have, trivially,x 4 −x 3 =t(x 2 −x 1 ).
Thus in both casesx 4 −x 3 =t(x 2 −x 1 ),y 4 −y 3 =t(y 2 −y 1 ), and so on combining
these
x 4 −x 3 +ı(y 4 −y 3 )=t(x 2 −x 1 )+ıt(y 2 −y 1 ).
Thusz 4 −z 3 =t(z 2 −z 1 ).
(vi) By 6.5.1, Corollary (i), these lines are perpendicular if and only if
(y 2 −y 1 )(y 4 −y 3 )+(x 2 −x 1 )(x 4 −x 3 )= 0. (10.1.2)Suppose first thatz 4 −z 3 =tı(z 2 −z 1 )for somet∈R.Then
x 4 −x 3 +ı(y 4 −y 3 )=ıt(x 2 −x 1 )−t(y 2 −y 1 ),and so astis real,x 4 −x 3 =−t(y 2 −y 1 ),y 4 −y 3 =t(x 2 −x 1 ).Then
(y 2 −y 1 )(y 4 −y 3 )+(x 2 −x 1 )(x 4 −x 3 )
=(y 2 −y 1 )t(x 2 −x 1 )−t(x 2 −x 1 )(y 2 −y 1 )= 0 ,so that (10.1.2) holds, and hence the lines are perpendicular.
Conversely suppose that the lines are perpendicular so that (10.1.2) holds. When
x 1 =x 2 ,welet
t=y 4 −y 3
x 2 −x 1so thaty 4 −y 3 =t(x 2 −x 1 ). On inserting this in (10.1.2), we have
(y 2 −y 1 )t(x 2 −x 1 )+(x 2 −x 1 )(x 2 −x 1 )= 0 ,from which we havex 4 −x 3 =−t(y 2 −y 1 ).
Whenx 2 −x 1 =0, by (10.1.2) we must havey 4 −y 3 =0. We now let
t=−x 4 −x 3
y 2 −y 1,
so thatx 4 −x 3 =−t(y 2 −y 1 ). For thistwe also have, trivially,y 4 −y 3 =t(x 2 −x 1 ).
Thus in both casesx 4 −x 3 =−t(y 2 −y 1 ),y 4 −y 3 =t(x 2 −x 1 ), and so on combining
these
x 4 −x 3 +ı(y 4 −y 3 )=−t(y 2 −y 1 )+ıt(x 2 −x 1 )=ıt[x 2 −x 1 +ı(y 2 −y 1 )].Thusz 4 −z 3 =tı(z 2 −z 1 ).