Given the points
W ( -2 , 1 ) , X ( -1 , 3 ) , Y ( 3 , 1 ) and Z ( 2 , -1 )
The distance formula is ;
[tex]\sqrt[]{(x2-x1)^2+(y2-y1)^2}[/tex]So, we will find the distances WX , XY , YZ and ZX
[tex]\begin{gathered} WX=\sqrt[]{(-2-(-1)^2+(1-3)^2}=\sqrt[]{1+4}=\sqrt[]{5} \\ XY=\sqrt[]{(-1-3)^2+(3-1)^2}=\sqrt[]{16+4}=\sqrt[]{20} \\ YZ=\sqrt[]{(3-2)^2+(1-(-1))^2}=\sqrt[]{1+4}=\sqrt[]{5} \\ ZX=\sqrt[]{(2-(-2))^2+(-1-1)^2}=\sqrt[]{16+4}=\sqrt[]{20} \end{gathered}[/tex]so, WX = YZ = √5
And XY = ZX = √20
so, WXYZ is rectangle provided that WX is perpendicular to XY
See the following image
The answer is yes
