202 Vector and complex-number methods Ch. 11
This must be the pointW 3 then. Similarly, on eliminatingZ 0 from the equations for
Z 3 andZ 6 we have that
( 1 −v)Z 3 −( 1 −q)Z 6 =q( 1 −v)Z 1 −( 1 −q)vZ 4 ,
and so
1 −v
1 −qv
Z 3 +
( 1 −q)v
1 −qv
Z 4 =
q( 1 −v)
1 −qv
Z 1 +
1 −q
1 −qv
Z 6.
This must be the pointW 2 then.
Now from the equations forZ 2 andZ 3 we have thatpZ 3 −qZ 2 =(p−q)Z 0 ,and
from the equations forZ 5 andZ 6 we have thatuZ 6 −vZ 5 =(u−v)Z 0. On combining
these, we have that
(v−u)(pZ 3 −qZ 2 )=(q−p)(uZ 6 −vZ 5 ).
From this we have that
u(q−p)
qv−pu
Z 6 +
q(v−u)
qv−pu
Z 2 =
p(v−u)
qv−pu
Z 3 +
(q−p)v
qv−pu
Z 5.
This must then be the pointW 1.
However, the left-hand sides of the representations forW 1 ,W 2 andW 3 contain
four pointsZ 1 ,Z 2 ,Z 5 ,Z 6 and we wish to reduce this to three non-collinear points.
For this purpose we eliminateZ 5. From the equations forZ 5 andZ 6 we have that
uZ 6 −vZ 5 =(u−v)Z 0 , while from the equation forZ 2 we haveZ 2 −pZ 1 =( 1 −p)Z 0.
Combining these gives
Z 5 =
u
v
Z 6 −
(u−v)
v( 1 −p)
(Z 2 −pZ 1 ).
On substitution, this gives that
W 3 =
pu( 1 −v)
v( 1 −pu)
Z 1 +
v−u
v( 1 −pu)
Z 2 +
u( 1 −p)
v( 1 −pu)
Z 6.
We note that the sum of the coefficients for each ofW 1 ,W 2 ,W 3 in terms ofZ 1 ,Z 2
andZ 6 is equal to 1, and so by repeated use of 10.5.3 and 11.4.5 we have that