Question 810650
A(–2, 3), B(2, 0), and C(–1, –4)

find the distance between AB, BC, & CA

AB
Distance between two points													
x1	y1	x2	y2										
													
-2	3	2	0										
d=	{{{sqrt((y2-y1)^2+(x2-x1)^2)}}}												
d=	{{{sqrt((		0	-	3	)^2	+	(	2	-	-2	)^2	)}}}
d=	{{{sqrt((		-3	)^2	+	(	4	)^2	)}}}				
d=	{{{sqrt((		25	)  	)}}}								
d=	5.00												
distance BC

Distance between two points													
x1	y1	x2	y2										
													
2	0	-1	-4										
d=	{{{sqrt((y2-y1)^2+(x2-x1)^2)}}}												
d=	{{{sqrt((		-4	-	0	)^2	+	(	-1	-	2	)^2	)}}}
d=	{{{sqrt((		-4	)^2	+	(	-3	)^2	)}}}				
d=	{{{sqrt((		25	)  	)}}}								
d=	5.00												
Distance CA
x1	y1	x2	y2										
													
-2	3	-1	-4										
d=	{{{sqrt((y2-y1)^2+(x2-x1)^2)}}}												
d=	{{{sqrt((		-4	-	3	)^2	+	(	-1	-	-2	)^2	)}}}
d=	{{{sqrt((		-7	)^2	+	(	1	)^2	)}}}				
d=	{{{sqrt((		50	)  	)}}}								
d=	7.07												
AB=5,BC=5 & AC=7.07

ab^2+BC^2= is not equal to AC^2

therefore it is not a right triangle