nOOCHBirj to pn-tix tin- muni-* si^'ji to mil. -i ..in- ,.f tin* two 



iiuinlMjn* wij or i/i, in formulas [6]. These formulas then 

 become 



_ mfy - nyi "'I'':- '":.'' i r?! 



*- fin-ia, '*- m,-*, 



COR. If P t IK* the middle point then ro, = ro r 



ami formulas [6] become 



- - [8] 



abscissa of tin- nulllr IM-IM: -m^ two 



given points is half the HIIIH of the abscissas of those points, 



In- ordinate is half th-- *\nn ..f tli.-ir ordinates. 

 The remarks in Art. 28 are well illustrated by formulas 

 [4] to [8]. 



EXERCISES 



1. By means of an appropriate figure, derive formulas [7] independ- 

 nUyof[6). 



2. The point P a = (2, 3) is one third of the distance from the point 

 P,B (-1,4) to the point /', = (r r y,): to find the coordinates 



1 l.-re P, and /*, are giv.-n. \\ itli ^, = - 1, jfj = 4, x, = 2, f, = 3, also 

 m, = 1, and m, = 2; therefore, from [0], 



which give x 9 = 8 and y t = 1 ; therefore the required point P, is (8, 1) 

 3. Find the points of triseetion of the line joining (1, -2) to (3, 4). 



4 1 ,1 the point which divides the line from (1, 3) to (-J 

 externally into segments whose numerical ratio is 3 : I. 



II 1. y, = 3, x, = -2, y t =* 4, m l = 3, and m, = 4, but the 



of division being an external one, the two segments are oppo- 

 sitely directed; therefore one of the numbers 3 or 4, say 4, must have 

 the minus sign prefixed to it Substituting these values in [], 



the required point is, therefore, J,s(l" 



