86 Cartesian coordinates; applications Ch. 6
6.3 Cartesianequationofaline
6.3.1
Given any line l∈Λ, there are numbers a,b and c, with the case a=b= 0 excluded,
such that Z≡(x,y)∈l if and only if
ax+by+c= 0.Proof. Take any pointZ 2 ≡(x 2 ,y 2 )∈land letZ 3 ≡(x 3 ,y 3 )=sl(Z 2 ).ThenZ 2 =
Z 3 .Nowlis the perpendicular bisector of[Z 2 ,Z 3 ], so by 4.1.1Z∈lif and only
if|Z,Z 2 |=|Z,Z 3 |. As these are both non-negative, this is the case if and only if
|Z,Z 2 |^2 =|Z,Z 3 |^2. By 6.1.1 this happens if and only if
(x−x 2 )^2 +(y−y 2 )^2 =(x−x 3 )^2 +(y−y 3 )^2.This simplifies to
2 (x 3 −x 2 )x+ 2 (y 3 −y 2 )y+x^22 +y^22 −x^23 −y^23 = 0.On writing
a= 2 (x 3 −x 2 ),b= 2 (y 3 −y 2 ),c=x^22 +y^22 −x^23 −y^23 ,we see thatZ≡(x,y)∈lif and onlyax+by+c=0. Nowa=b=0 corresponds to
x 2 =x 3 ,y 2 =y 3 , which is ruled out asZ 2 =Z 3.
COROLLARY.Let Z 0 ≡(x 0 ,y 0 ),Z 1 ≡(x 1 ,y 1 )be distinct points and Z≡(x,y).
Then Z∈Z 0 Z 1 if and only if
−(y 1 −y 0 )(x−x 0 )+(x 1 −x 0 )(y−y 0 )= 0.Proof. By the theorem, there exist numbersa,b,c, with the casea=b=0ex-
cluded, such thatZ∈Z 0 Z 1 if and only ifax+by+c=0. AsZ 0 ,Z 1 ∈Z 0 Z 1 we then
have
ax 0 +by 0 +c= 0 ,
ax 1 +by 1 +c= 0.We subdivide into two cases as follows.
CASE 1. Letx 0 =x 1. We rewrite our equations as
ax 1 +c=−by 1 ,
ax 0 +c=−by 0 ,and regard these as equations in the unknownsaandc.Asx 1 −x 0 =0, we note that
by 6.2.1 we must have
a=−b(y 1 −y 0 )
x 1 −x 0,c=−b(x 1 y 0 −x 0 y 1 )
x 1 −x 0