Geometry with Trigonometry

214 Vector and complex-number methods Ch. 11

Assumingq 1 =0 we have the pair of equations

1

√

p^21 +q^21

r−

1

√

( 1 −p 1 )^2 +q^21

s= 0 ,

⎛

⎝ 1 +√ p^1

p^21 +q^21

⎞

⎠r+

⎛

⎝ 1 +√^1 −p^1

( 1 −p 1 )^2 +q^21

⎞

⎠s= 2 |Z 2 ,Z 3 |. (11.6.8)

We denote byD 1 the value of the determinant of coefficients on the left-hand side

∣∣

∣∣

∣

∣

√^1

p^21 +q^21 −

√^1

( 1 −p 1 )^2 +q^21

1 +√pp 21

1 +q^21

1 +√( 11 −−pp^1

1 )^2 +q^21

∣∣

∣∣

∣

∣

and have

D 1 =

√

( 1 −p 1 )^2 +q^21 +

√

p^21 +q^21 + 1

√

p^21 +q^21

√

( 1 −p 1 )^2 +q^21

.

We denote byD 2 the value of the determinant

∣∣

∣∣

∣∣

0 −√( 1 −p^1

1 )^2 +q^21

2 |Z 2 ,Z 3 | 1 +√( 11 −−pp^1

1 )^2 +q^21

∣∣

∣∣

∣∣

which has the value

D 2 =

√^2 |Z^2 ,Z^3 |

( 1 −p 1 )^2 +q^21

.

Then we have the solution

r=

D 2

D 1

=

2 |Z 2 ,Z 3 |

√

p^21 +q^21

√

( 1 −p 1 )^2 +q^21

√

p^21 +q^21 + 1

.

Thus the point of intersection from this bisector is

z 2 +

√

p^21 +q^21

√

( 1 −p 1 )^2 +q^21 +

√

p^21 +q^21 + 1

⎡

⎣ 1 +√p^1 +ıq^1

p^21 +q^21

⎤

⎦(z 3 −z 2 ), (11.6.9)

which is thus the incentreZ 15.