Geometry with Trigonometry

100 Cartesian coordinates; applications Ch. 6

respectively, and soW 3 has coordinates

u 3 =

x 3 (u 1 −x 2 )−x 1 (u 1 −x 3 )

x 3 −x 2

, v 3 =v 1

x 3 −x 1

x 3 −x 2

.

On forming the equation ofW 2 W 3 and finding where it meetsZ 1 Z 2 we obtain

x 4 =

−x 3 (x 1 +x 2 )+ 2 x 1 x 2

x 1 +x 2 − 2 x 3

,

from which it follows that

x 4 −x 1 =

(x 1 −x 2 )(x 3 −x 1 )

x 1 +x 2 − 2 x 3

, x 2 −x 4 =

(x 1 −x 2 )(x 3 −x 2 )

x 1 +x 2 − 2 x 3

.

From these we see that
x 4 −x 1
x 2 −x 4

=−

x 3 −x 1

x 2 −x 3

.

Exercises

6.1 Suppose thatZ 1 ,Z 2 ,Z 3 are non-collinear points andZ 5 =mp{Z 3 ,Z 1 },

Z 6 =mp{Z 1 ,Z 2 }. Show that if|Z 2 ,Z 5 |=|Z 3 ,Z 6 |,then|Z 3 ,Z 1 |=|Z 1 ,Z 2 |.

(Hint. Select a frame of reference to simplify the calculations).

6.2 Letl 1 ,l 2 be distinct intersecting lines andZ 0 a point not on either of them.

Show that there are unique pointsZ 1 ∈l 1 ,Z 2 ∈l 2 such thatZ 0 is the mid-point

ofZ 1 andZ 2.

6.3 Suppose thatZ 1 ,Z 2 ,Z 3 are non-collinear points. Show that the pointsZ≡(x,y),

the perpendicular distances from which to the linesZ 1 Z 2 ,Z 1 Z 3 are equal, are

those the coordinates of which satisfy

−(y 2 −y 1 )(x−x 1 )+(x 2 −x 1 )(y−y 1 )

√

(x 2 −x 1 )^2 +(y 2 −y 1 )^2

±

−(y 3 −y 1 )(x−x 1 )+(x 3 −x 1 )(y−y 1 )

√

(x 3 −x 1 )^2 +(y 3 −y 1 )^2

= 0.

Show that ifx=( 1 −t)x 2 +tx 3 ,y=( 1 −t)y 2 +ty 3 thenZlies on the line with

equation

−(y 2 −y 1 )(x−x 1 )+(x 2 −x 1 )(y−y 1 )

√

(x 2 −x 1 )^2 +(y 2 −y 1 )^2

+

−(y 3 −y 1 )(x−x 1 )+(x 3 −x 1 )(y−y 1 )

√

(x 3 −x 1 )^2 +(y 3 −y 1 )^2

= 0 ,

if and only

t=

|Z 1 ,Z 2 |^2

|Z 1 ,Z 2 |^2 +|Z 2 ,Z 3 |^2

.

Deduce that this latter line is the mid-line of|Z 2 Z 1 Z 3.