MOORE. — MINIMUM GEOMETRY. 211 



The distance between two points (xi, yi), (x 2 , y 2 ) is 



d = Xi y 2 — Xz 2/1. 

 For in the triangle XPiY, fig. 2, from the triangle relations, 



XPx + P:X + YX x x - yi - 1 



1 XPx + YX - Xl 



In the triangle XP 2 Y, 



XP 2 + P 2 X + YX x 2 -y 2 -l 



2 XP 2 • YX - x 2 

 Then, 



e = e - e- = yiX2 ~ y2Xl + X2 ~ Xl • 



XlX 2 



In the triangle XPiP 2 we have, 



XPi + P : P 2 + P 2 X 



e = 



XPi • P 2 X 



xi + d — x 2 yix 2 — xiy 2 + x 2 — xi 



— X\X 2 XiX 2 



This solved for d gives the value above. 



The equation of a straight line in this system of coordinates is 

 linear as can easily be seen by finding the locus of points equidistant 

 from a given point, this we saw was a straight line. 



For measuring angles we shall take a third point of reference 

 Z (1,-1). We will then write the equation of the straight line in the 

 form, 



ux + vy = 1 . 



The equations of ZY, ZX and XY are respectively, 



x = 1, y = -1, x - y = 1. 



The equation of a line passing through F is 



y = kx. 

 From the triangle PQZ, fig. 3, 



1 + v — u u — v — 1 v — u -\- 1 

 PQ + QZ + ZP = uv + v + u 



PQ • ZP v- u+ 1 _ v-u+ 1 = U ' 



u uv 



