Question

quadrilateral WXYZ has verticals of W ( -2 , 1 ) , X ( -1 , 3 ) , Y ( 3 , 1 ) and Z ( 2 , -1 ) determine whether WXYZ is a rectangle by using the distance formula

Answer 1

Given the points

W ( -2 , 1 ) , X ( -1 , 3 ) , Y ( 3 , 1 ) and Z ( 2 , -1 )

The distance formula is ;

`\sqrt[]{(x2-x1)^2+(y2-y1)^2}`

So, we will find the distances WX , XY , YZ and ZX

`\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}`

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