Geometry with Trigonometry

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