Sec. 6.6 Projection and axial symmetry 93
Asab 1 −a 1 b=0, by 6.2.1 these will have a unique solution, yielding a point which
we shall denote byZ 0 ≡(x 0 ,y 0 ).Thenby6.4.1
l={Z≡(x,y):x=x 0 +bt,y=y 0 −at,t∈R},
m={Z≡(x,y):x=x 0 +b 1 t,y=y 0 −a 1 t,t∈R}.We chooseZ 1 ∈l,Z 2 ∈mas above, and from (6.5.1) find that|Z 0 ,Z 1 |^2 +|Z 0 ,Z 2 |^2 =
|Z 1 ,Z 2 |^2. By 6.4.1 we can conclude thatl⊥m.
(ii) By 6.2.1 the equationsax+by+c= 0 ,a 1 x+b 1 y+c 1 =0 have either no
solution or more than one if and only if (6.5.2) holds.
Alternatively, by (i) above we havel‖mif and only if there is some(a 2 ,b 2 )=
( 0 , 0 )such that
aa 2 +bb 2 = 0 ,a 1 a 2 +b 1 b 2 = 0.
But the equations
au+bv= 0 ,a 1 u+b 1 v= 0 ,
have a solution(u,v)other than( 0 , 0 )if and only ifab 1 −a 1 b=0. Thus (6.5.2) is a
condition forlandmto be parallel.
COROLLARY.
(i)The lines Z 1 Z 2 and Z 3 Z 4 are perpendicular if and only if(y 2 −y 1 )(y 4 −y 3 )+(x 2 −x 1 )(x 4 −x 3 )= 0.(ii)These lines are parallel if and only if−(y 2 −y 1 )(x 4 −x 3 )+(y 4 −y 3 )(x 2 −x 1 )= 0.6.6 Projectionandaxialsymmetry
6.6.1
Let l≡ax+by+c= 0 and Z 0 ≡(x 0 ,y 0 ).Then
(i)
|Z 0 ,πl(Z 0 )|=|ax 0 +by 0 +c|
√
a^2 +b^2.
(ii)πl(Z 0 )≡(
x 0 −a
a^2 +b^2(ax 0 +by 0 +c),y 0 −b
a^2 +b^2(ax 0 +by 0 +c)