Question 563408
Distance between two points	(1,2)(4,6)												
x1	y1	x2	y2										
d= 	{{{sqrt((y2-y1)^2+(x2-x1)^2)}}}												
1	2	4	6										
d= {{{sqrt((6-2	)^2+(4-1)^2	)}}}
d= 	{{{sqrt((4)^2+(	3)^2	)}}}				
d= 	{{{sqrt((25))}}}						
d=5

Distance between two points	(10,14)(4,6)

		
x1	y1	x2	y2										
d= 	{{{sqrt((y2-y1)^2+(x2-x1)^2)}}}												
10	14	4	6										
d= {{{sqrt((6-14)^2+(4-10)^2)}}}
d= 	{{{sqrt((-8)^2	+(-6)^2	)}}}				
d= 	{{{sqrt((100))}}}								
d= 	10.00	

										

Distance between two points	(10,14) (1,2)												
x1	y1	x2	y2										
d= 	{{{sqrt((y2-y1)^2+(x2-x1)^2)}}}												
10	14	1	2										
d= 	{{{sqrt((2-14)^2+(1-10)^2)}}}
d= 	{{{sqrt((-12)^2	+(-9)^2	)}}}				
d= 	{{{sqrt((225))}}}								
d= 	15.00		

5+10 = 15 
A=(4,6),    B=(10,14)     C=(1,2)

d(BC) = 15

d(AB)=10

d(AC)=5

d(AB) +d(AC)= d(BC) 
10+5=15
This is possible only if the points are colinear, else the sum of any two distances will not be equal to the third distance.


so the points are colinear.

BY slopes  (1,2)(4,6)

x1		y1	x2	y2				
1      		2    	4    	6    				
								
slope m =(y2-y1)/(x2-x1)						
(6-2)/(	4-1) 
(	4	/	3    	)  				
m=		4/ 3						 
										
(4,6)(10,14)

x1		y1	x2	y2				
4      		6    	10    	14    				
								
slope m =		(y2-y1)/(x2-x1)						
(14-6)/(10-4) 
(8/6)  				
m=		4/ 3		

the slopes are same