Sec. 11.4 Components of a vector 193
11.4.2Arealcoordinates ...........................
Given non-collinear pointsZ 1 ,Z 2 ,Z 3 , the position vector of any pointZof the plane
can be expressed in the form
−→
OZ=p
−−→
OZ 1 +q
−−→
OZ 2 +r
−−→
OZ 3 , withp+q+r=1. This is
equivalent to havingq,rsuch that
q(x 2 −x 1 )+r(x 3 −x 1 )=x−x 1 ,
q(y 2 −y 1 )+r(y 3 −y 1 )=y−y 1.
These equations have the unique solution
q=
δF(Z,Z 3 ,Z 1 )
δF(Z 1 ,Z 2 ,Z 3 )
,r=
δF(Z,Z 1 ,Z 2 )
δF(Z 1 ,Z 2 ,Z 3 )
,
and now we takep= 1 −q−rso that by 10.5.4
p=
δF(Z,Z 2 ,Z 3 )
δF(Z 1 ,Z 2 ,Z 3 )
For non-collinear pointsZ 1 ,Z 2 ,Z 3 ,foranyZwe write
α=δF(Z,Z 2 ,Z 3 ),β=δF(Z,Z 3 ,Z 1 ),γ=δF(Z,Z 1 ,Z 2 ),
and call(α,β,γ)areal point coordinatesofZwith respect to(Z 1 ,Z 2 ,Z 3 ). Note that
we have
p=
α
δF(Z 1 ,Z 2 ,Z 3 )
,q=
β
δF(Z 1 ,Z 2 ,Z 3 )
,r=
γ
δF(Z 1 ,Z 2 ,Z 3 )
,
andα+β+γ=δF(Z 1 ,Z 2 ,Z 3 ). These were first used by Möbius in 1827.
11.4.3Cartesiancoordinatesfromarealcoordinates .............
With the notation in 11.4.2, we have
(y 2 −y 3 )x−(x 2 −x 3 )y= 2 α−x 2 y 3 +x 3 y 2 ,
(y 3 −y 1 )x−(x 3 −x 1 )y= 2 β−x 3 y 1 +x 1 y 3 ,
and if we solve these we obtain
x=
x 1 α+x 2 β+x 3 γ
δF(Z 1 ,Z 2 ,Z 3 )
,y=
y 1 α+y 2 β+y 3 γ
δF(Z 1 ,Z 2 ,Z 3 )
11.4.4 .....................................
The representation in 11.4.2 is in fact independent of the originO. For we have
x=px 1 +qx 2 +rx 3 ,y=py 1 +qy 2 +ry 3 ,