88 Cartesian coordinates; applications Ch. 6
for some j= 0.
Proof.
Necessity. Suppose first thatlcan be expressed in each of the forms (i) and (ii)
above. We subdivide into four cases as follows.
CASE 1. Suppose thata= 0 ,b=0andc=0. Then we note from (i) that the
pointsA≡(−c/a, 0 )andB≡( 0 ,−c/b)are inl, and are in fact the only points oflin
eitherOIorOJ,asAis the only point withy=0andBis the only point withx=0.
We now note that none ofa 1 ,b 1 ,c 1 can be equal to 0. For ifa 1 =0, by (ii) we
would havey=−c 1 /b 1 for all pointsZinl; this would makelparallel toOIand give
a contradiction. Similarlyb 1 =0wouldimplythatx=−c 1 /a 1 for all pointsZinl,
makinglparallel toOJand again giving a contradiction. Moreover ifc 1 =0, by (ii)
we would have thatO∈l, again a contradiction.
We note from (ii) that the pointsA 1 ≡(−c 1 /a 1 , 0 ),B 1 ≡( 0 ,−c 1 /b 1 )are inland
are in fact the only points oflin eitherOIorOJ. Thus we must haveA 1 =A,B 1 =B
and so
−c 1
a 1=−
c
a,−
c 1
b 1=−
c
b.
Thus
a 1
a
=
b 1
b=
c 1
c,
and if we denote the common value of these byj,wehavej=0 and (iii).
CASE 2. Suppose thata=0. Thenb=0 and by (i) for everyZ∈lwe have
y=−c/b,sothatlcontainsBand is parallel toOI;whenc=0,lhas no point in
common withOI,andwhenc= 0 ,lcoincides withOI.Nowwemusthavea 1 =0, as
otherwiselwould meetOIin the unique pointA 1 , and that would give a contradiction.
Thenb 1 =0 and for everyZ∈lwe havey=−c 1 /b 1 ,sothatlcontainsB 1 and is
parallel toOI. Thus we must have
−
c 1
b 1=−
c
b.
Whenc=0, this implies thatc 1 =0, so that if we takej=b 1 /b, we have satisfied
(iii). Whenc=0, we must have thatb 1 /b=c 1 /c, and if we takejto be the common
value of these we have (iii) again.
CASE 3. Suppose thatb=0. This is treated similarly to CASE 2.
CASE 4. Finally suppose thata= 0 ,b=0andc=0. Then by (i) we see that
O∈land then by (ii) we must havec 1 =0. We see from (i) thatC≡( 1 ,−a/b)is in
l, and on using this information in (ii) we find thata 1 +b 1 (−a/b)=0. This implies
thata 1 /a=b 1 /b, and if we takejto be the common value of these, we must have
(iii).
This establishes the necessity of (iii).
SufficiencySuppose now that (iii) holds. Thena 1 x+b 1 y+c 1 =j(ax+by+c)
and asj=0wehavea 1 x+b 1 y+c 1 =0 if and only ifax+by+c= 0.