Question 863143
Distance between two points	(AB)												
x1	y1	x2	y2										
													
5	5	-7	1										
d=	{{{sqrt((y2-y1)^2+(x2-x1)^2)}}}												
d=	{{{sqrt((		1	-	5	)^2	+	(	-7	-	5	)^2	)}}}
d=	{{{sqrt((		-4	)^2	+	(	-12	)^2	)}}}				
d=	{{{sqrt((		160	)  	)}}}								
d=	12.65		

Distance between two points (BC)												
x1	y1	x2	y2										
													
1	-7	-7	1										
d=	{{{sqrt((y2-y1)^2+(x2-x1)^2)}}}												
d=	{{{sqrt((		1	-	-7	)^2	+	(	-7	-	1	)^2	)}}}
d=	{{{sqrt((		8	)^2	+	(	-8	)^2	)}}}				
d=	{{{sqrt((		128	)  	)}}}								
d=	11.31												


Distance between two points	(AC)												
x1	y1	x2	y2										
													
-7	1	5	5										
d=	{{{sqrt((y2-y1)^2+(x2-x1)^2)}}}												
d=	{{{sqrt((		5	-	1	)^2	+	(	5	-	-7	)^2	)}}}
d=	{{{sqrt((		4	)^2	+	(	12	)^2	)}}}				
d=	{{{sqrt((		160	)  	)}}}								
d=	12.65												
										
Apply Herons formula for area of triangle

Area of triangle = {{{sqrt(s(s-a)(s-b)(s-c))}}}

where s = 1/2 perimeter
Perimeter36.61
s=18.31

Area ={{{ sqrt(18.31(18.31-11.31)^2+(18.31-12.65)^2+(18.31-12.65)^2)}}}
=31.004

area of parallelogram = 2 * area of triangle

=2*31.004
=62.008 sq. units
=