485] 
PROBLEMS AND SOLUTIONS. 
579 
which shows that the lines 14 and 23 intersect at right angles; similarly the lines 12 
and 34, and also the lines 13 and 24, intersect at right angles; or starting from the 
given points 1, 2, 3, the point 4 is the intersection of the perpendiculars let fall from 
the angles 1, 2, 3 of the triangle 123 on the opposite sides respectively. 
Again combining the equations as follows, 
(12)+ (13)-(23), 
we obtain 
n*=(+ - «2) («i - a») + (ft - ft) (ft - /3;»)- 
The entire system of equations will remain unaltered if we pass from the original axes 
to any other system of rectangular axes; hence taking the axes of x in the sense 
from 1 to 2 along the line 12, ft — ft 2 becomes = 0, and we have 
a 2 — aj = 12, a 3 - ai =l(12, 34); 
viz. a 2 — ol x is the distance 12 of the points 1 and 2, a 3 — cq is the distance 1(12, 34) 
of the point 1 from the point (12, 34) which is the intersection of the lines 12 and 
34 ; we have therefore 
^=12.1(12, 34). 
But similarly 
r 2 2 = 21.2 (12, 34), = 12.(12, 34)2, 
(since 21 = — 12 and 2(12, 34) = —(12, 34)2). And we have therefore 
Write 
r,* : r,> = 1(12, 34) : (12. 34)2, or ~ : -ft (12, 34)2 : 1(12, 34). 
/1 '9 
X = 
(12, 34)2 
12 
_ 1 (12, 34) 
12 
where 1 (12, 34) and (12, 34) 2 are as above the distances from 1 to (12, 34) and from 
(12, 34) to 2; and, in the denominators, 12 is the distance from 1 to 2; we have 
X +//. = 1; the coordinates of (12, 34) are Xo^+^ou, X/3 l +y!3 2 , and the values of X, yu 
are obtained by writing X+ + /¿a.,, Aft + /¿ft, X + y for x, y, 1 in the equations 
x, y , 1 
a $> A, l 
«4 , ßi, 1 
of the line 34. Making this substitution, we find 
where as above 
X (134) + y (234) = 0, 
(134)= 
«1, ft, 1 
«3, ft) 1 
ft, 1 
&C., 
«4, 
73—2