Geometry with Trigonometry

98 Cartesian coordinates; applications Ch. 6

then

x 1 =

1

1 +μ

x 3 +

μ

1 +μ

x 4 ,y 1 =

1

1 +μ

y 3 +

μ

1 +μ

y 4.

If we defineμ′by
1
1 +μ′

=

1 +λ

2 λ

,

so that

μ′=

λ− 1

1 +λ

,

μ′

1 +μ′

=

λ− 1

2 λ

,

then

x 2 =

1

1 +μ′

x 3 +

μ′

1 +μ′

x 4 ,y 2 =

1

1 +μ′

y 3 +

μ′

1 +μ′

y 4.

Asμ′=−μ, this shows thatZ 1 andZ 2 divide{Z 3 ,Z 4 }internally and externally in
the same ratio.

6.7.3 Distancesfrommid-point

Let Z 0 be the mid-point of distinct points Z 1 and Z 2. Then points Z 3 ,Z 4 ∈Z 1 Z 2 divide
{Z 1 ,Z 2 }internally and externally in the same ratio if and only if Z 3 and Z 4 are on the
one side of Z 0 on the line Z 1 Z 2 and

|Z 0 ,Z 3 ||Z 0 ,Z 4 |=^14 |Z 1 ,Z 2 |^2.

Proof.WehaveZ 0 ≡(x 0 ,y 0 )wherex 0 =^12 (x 1 +x 2 ),y 0 =^12 (y 1 +y 2 ).Then

x 3 −x 0 =(t−^12 )(x 2 −x 1 ), y 3 −y 0 =(t−^12 )(y 2 −y 1 ),

x 4 −x 0 =(s−^12 )(x 2 −x 1 ), y 4 −y 0 =(s−^12 )(y 2 −y 1 ),

and so
|Z 0 ,Z 3 ||Z 0 ,Z 4 |=|

(

t−^12

)(

s−^12

)

||Z 1 ,Z 2 |^2.

By (6.7.1)Z 3 ,Z 4 divide{Z 1 ,Z 2 }internally and externally in the same ratio if and
only if

(

s−^12

)(

t−^12

)

=^14. This is equivalent to having|

(

s−^12

)(

t−^12

)

( |=^14 and
s−^12

)(

t−^12

)

0. The latter is equivalent to having eithers−^12 >0andt−^12 >0,
   ors−^12 <0andt−^12 <0, so thatZ 3 andZ 4 are on the one side ofZ 0 on the line
   Z 1 Z 2.

6.7.4 Distancesfromend-point

Let{Z 3 ,Z 4 }divide{Z 1 ,Z 2 }internally and externally in the same ratio with Z 2 ∈
[Z 1 ,Z 4 ].Then
1
2

(

1

|Z 1 ,Z 3 |

+

1

|Z 1 ,Z 4 |

)

=

1

|Z 1 ,Z 2 |

.

Proof.Wehaveasbefore

x 3 =x 1 +

λ

1 +λ

(x 2 −x 1 ),y 3 =y 1 +

λ

1 +λ

(y 2 −y 1 ),