Geometry with Trigonometry

(Marvins-Underground-K-12) #1

Sec. 11.6 Mobile coordinates 225


z( 16 iii)=z 2 +p 1 (z 3 −z 2 )+^12 ( 1 −

q 1
p 1

ı)[z 2 +^12 p 1 ( 1 −

p 1 − 1
q 1

ı)−z 2 −p 1 ](z 3 −z 2 )

=z 2 +p 1 (z 3 −z 2 )+^12 ( 1 −

q 1
p 1

ı)[^12 p 1 ( 1 −

p 1 − 1
q 1

ı)−p 1 ](z 3 −z 2 )

=z 2 +

[


p 1 +^12 ( 1 −

q 1
p 1

ı)[

1


2


p 1 −^12

p 1 (p 1 − 1 )
q 1

ı−p 1 ]

]


(z 3 −z 2 )

=z 2 +p 1

[


1 +^12 ( 1 −


q 1
p 1

ı)[−^12 −^12

p 1 − 1
q 1

ı]

]


(z 3 −z 2 )

=z 2 +p 1

[


1 −^14 ( 1 −


q 1
p 1

ı)( 1 +

p 1 − 1
q 1

ı)

]


(z 3 −z 2

=z 2 +p 1

[


1 −^14 [ 1 +


q 1
p 1

p 1 − 1
q 1

+(


p 1 − 1
q 1


q 1
p 1

)ı]

]


(z 3 −z 2 )

=z 2 +p 1

[


1 −^14 [ 1 +


p 1 − 1
p 1

+


p^21 −p 1 −q^21
p 1 q 1

ı]

]


(z 3 −z 2 )

=z 2 +p 1

[


1 −^14


2 p 1 − 1
p 1

−^14 [


p^21 +q^21 −p 1
p 1 q 1

ı− 2

q^21
p 1 q 1

ı]

]


(z 3 −z 2 )

=z 2 +

[


1
4 +

1
2 p^1 −

1
4 [

p^21 +q^21
q 1


p 1
q 1

− 2 q 1 ]ı

]


(z 3 −z 2 )

=z 2 +^14

[


1 + 2 p 1 −

p^21 +q^21
q 1

ı+

p 1
q 1

ı+ 2 q 1 ı

]


(z 3 −z 2 ).

By (11.6.16) this is the same as that forZ′ 16. The same argument applies toZ 17
andZ 19.
Thus we have identified six extra points on the circle which passes through the
mid-points of the sides of the original triangle,which is named from this property.
There is an interesting article on this circle in Wikipedia. By a coincidence we have
Z 10 =Z 18 in our diagram.


11.6.12 Parametric equations of lines. ....................


For general distinct pointsZ 4 andZ 5 and generalZ∈Z 4 Z 5 we have that


z 4 =z 2 +(p 4 +q 4 ı)(z 3 −z 2 ),z 5 =z 2 +(p 5 +q 5 ı)(z 3 −z 2 ),z=z 2 +(p+qı)(z 3 −z 2 ),


and then


z=z 4 +s(z 5 −z 4 )=( 1 −s)z 4 +sz 5 , (s∈R)
=( 1 −s)[z 2 +(p 4 +q 4 ı)(z 3 −z 2 )]+s[z 2 +(p 5 +q 5 ı)(z 3 −z 2 )]
=( 1 −s+s)z 2 +[( 1 −s)(p 4 +q 4 ı)+s(p 5 +q 5 ı)](z 3 −z 2 )
=z 2 +{( 1 −s)p 4 +sp 5 +[( 1 −s)q 4 +sq 5 ]ı}(z 3 −z 2 ).
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