94 Cartesian coordinates; applications Ch. 6
(iii)sl(Z 0 )≡(
x 0 −2 a
a^2 +b^2(ax 0 +by 0 +c),y 0 −2 b
a^2 +b^2(ax 0 +by 0 +c))
Proof.Letmbe the line such thatl⊥mandZ 0 ∈m.Thenasl⊥m, by 6.5.1 we
will havem≡−bx+ay+c 1 =0forsomec 1 ,andasZ 0 ∈mwe havec 1 =bx 0 −ay 0.
To find the coordinates(x,y)ofπl(Z 0 )we need to solve simultaneously the equations
ax+by=−c,−bx+ay=−bx 0 +ay 0.As for (i) we shall then go on to apply 6.1.1 it is(x−x 0 )^2 and(y−y 0 )^2 that we shall
actually use, and it is easier to work directly with these. We rewrite the equations as
a(x−x 0 )+b(y−y 0 )=−(ax 0 +by 0 +c),
−b(x−x 0 )+a(y−y 0 )= 0.Now on squaring each of these and adding, we find that
(a^2 +b^2 )[(x−x 0 )^2 +(y−y 0 )^2 ]=(ax 0 +by 0 +c)^2.The conclusion (i) now readily follows.
For (ii) we solve these equations, obtaining
x−x 0 =−a
a^2 +b^2(ax 0 +by 0 +c),y−y 0 =−b
a^2 +b^2(ax 0 +by 0 +c)).For (iii) we recall that if sl(Z 0 )≡(x 1 ,y 1 ) and πl(Z 0 )≡(x 2 ,y 2 ),thenas
mp(Z 0 ,sl(Z 0 )) =πl(Z 0 )we havex 1 +x 0 = 2 x 2 ,y 1 +y 0 = 2 y 2 .Nowx 2 andy 2 are
given by (ii) of the present theorem, and the result follows.
6.6.2 Formula for area of a triangle
Let Z 1 ≡F(x 1 ,y 1 ),Z 2 ≡F(x 2 ,y 2 )and Z 3 ≡F(x 3 ,y 3 )be non- collinear points. Then
the areaΔ[Z 1 ,Z 2 ,Z 3 ]is equal to|δF(Z 1 ,Z 2 ,Z 3 )|where
δF(Z 1 ,Z 2 ,Z 3 )=^12 [x 1 (y 2 −y 3 )−y 1 (x 2 −x 3 )+x 2 y 3 −x 3 y 2 ]=^12 det⎛
⎝
x 1 y 1 1
x 2 y 2 1
x 3 y 3 1⎞
⎠.
Proof. By 6.3.1 Z 2 Z 3 ≡
−(y 3 −y 2 )(x−x 2 )+(x 3 −
x 2 )(y−y 2 )= 0 , so by 6.6.1
|Z 1 ,πZ 2 Z 3 (Z 1 )| is equal to
|−(y 3 −√y 2 )(x 1 −x 2 )+(x 3 −x 2 )(y 1 −y 2 )|
(y 3 −y 2 )^2 +(x 3 −x 2 )^2.
But Δ[Z 1 ,Z 2 ,Z 3 ]=
1
2 |Z^2 ,Z^3 ||Z^1 ,πZ^2 Z^3 (Z^1 )|,and
the denominator above is equal
to |Z 2 ,Z 3 |. Hence the area is
equal to half the numerator.
Z 3
Z 1
Z 2
πZ 2 Z 3 (Z 1 )
Figure 6.6. Area of a triangle.