Sec. 6.3 Cartesian equation of a line 87
Note thatb=0, asb=0wouldimplya=0 here. On inserting these values foraand
cabove we see thatZ∈lif and only if
−b(y 1 −y 0 )
x 1 −x 0x+by+−b(x 1 y 0 −x 0 y 1 )
x 1 −x 0= 0 ,
and so asb/(x 1 −x 0 )=0, if and only if
−(y 1 −y 0 )x+(x 1 −x 0 )y−x 1 y 0 +x 0 y 1 = 0.This is equivalent to the stated equation.
CASE 2. Lety 0 =y 1. We rewrite our equations as
by 1 +c=−ax 1 ,by 0 +c=−ax 0 ,and note that, asy 1 −y 0 =0, by 6.2.1 we must have
b=−a(x 1 −x 0 )
y 1 −y 0,c=−a(y 1 x 0 −y 0 x 1 )
y 1 −y 0.
Note thata=0, asa=0wouldimplyb=0 here. On inserting these values forband
cabove we see thatZ∈lif and only if
ax+−a(x 1 −x 0 )
y 1 −y 0y+−a(y 1 x 0 −y 0 x 1 )
y 1 −y 0= 0 ,
and so as−a/(y 1 −y 0 )=0, if and only if
−(y 1 −y 0 )x+(x 1 −x 0 )y−x 1 y 0 +x 0 y 1 = 0.This is equivalent to the stated equation.
Now either CASE 1 or CASE 2 (or both) must hold, as otherwise we havex 0 =
x 1 ,y 0 =y 1 and soZ 0 =Z 1 , contrary to what is given.
Definition.Ifl∈Λandl={Z≡(x,y):ax+by+c= 0 }, we callax+by+c= 0
aCartesian equationoflrelative toF, and we writel≡Fax+by+c=0. When
Fcan be understood we relax this tol≡ax+by+c=0.
Let l∈Λbe a line, with Cartesian equation
(i)
ax+by+c= 0.Then l also has
(ii)
a 1 x+b 1 y+c 1 = 0 ,
as an equation if and only if
(iii)
a 1 =ja,b 1 =jb,c 1 =jc,