Question 672501
A(-3,1)B(1,4)C(4,0) D(0,-3)

Find the slopes of AB,BC,CD,DA

Slope of AB
x1	y1	x2	y2					
-3	1	1	4					
								
slope m =		(y2-y1)/(x2-x1)						
(	4	-	1	)/(	1	-	-3	)
(	3	/	4	)  				
m(AB)=		 3/4	

M(BC)
x1	y1	x2	y2					
4	0	1	4					
								
slope m =		(y2-y1)/(x2-x1)						
(	4	-	0	)/(	1	-	4	)
(	4	/	-3	)  				
m=		-4/3	

m(CA)
x1	y1	x2	y2					
4	0	0	-3					
								
slope m =		(y2-y1)/(x2-x1)						
(	-3	-	0	)/(	0	-	4	)
(	-3	/	-4	)  				
m=		 3/4	

m(AD)
x1	y1	x2	y2					
-3	1	0	-3					
								
slope m =		(y2-y1)/(x2-x1)						
(	-3	-	1	)/(	0	-	-3	)
(	-4	/	3	)  				
m=		-4/3						 
					 
Two lines are having same slopes and two are having slopes negative reciprocal

So the angles formed are 90 degrees.

If the distances aer equal then the co ordinates form a square.

l(AB)
x1	y1	x2	y2										
													
-3	1	1	4										
d=	{{{sqrt((y2-y1)^2+(x2-x1)^2)}}}												
d=	{{{sqrt((		4	-	1	)^2	+	(	1	-	-3	)^2	)}}}
d=	{{{sqrt((		3	)^2	+	(	4	)^2	)}}}				
d=	{{{sqrt((		25	)  	)}}}								
d=	5	

l(BC)
x1	y1	x2	y2										
													
4	0	1	4										
d=	{{{sqrt((y2-y1)^2+(x2-x1)^2)}}}												
d=	{{{sqrt((4-0)^2	+(1-4)^2)}}}
d=	{{{sqrt((4)^2	+(-3)^2	)}}}				
d=	{{{sqrt((		25	)  	)}}}								
d=	5	


l(CD)
x1	y1	x2	y2										
													
4	0	0	-3										
d=	{{{sqrt((y2-y1)^2+(x2-x1)^2)}}}												
d=	{{{sqrt((-3-0)^2+(0-4)^2)}}}
d=	{{{sqrt((-3)^2	+(-4)^2	)}}}				
d=	{{{sqrt((		25	)  	)}}}								
d=	5

l(DA)

x1	y1	x2	y2										
													
-3	1	0	-3										
d=	{{{sqrt((y2-y1)^2+(x2-x1)^2)}}}												
d=	{{{sqrt((		-3	-	1	)^2	+	(	0	-	-3	)^2	)}}}
d=	{{{sqrt((		-4	)^2	+	(	3	)^2	)}}}				
d=	{{{sqrt((		25	)  	)}}}								
d=	5												
The lengths are all 5 units. and they are at right angles. so it is a square.