Geometry with Trigonometry

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 ,