The vertices form s rectangle.
Step-by-step explanation:
Given,
Let, four vertices are A(-5,5), B(4,2), C(3,1), D(-6,2)
To find out ABCD will be A type of quadrilateral
Formula
The length of two points ([tex]x_{1}[/tex],[tex]y_{1}[/tex]) and ([tex]x_{2} ,y_{2}[/tex]) is [tex]\sqrt{(x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2} }[/tex]
Here,
AB ⇒ [tex]\sqrt{(4+5)^{2} +(2-5)^{2} }[/tex] = [tex]\sqrt{90}[/tex] [ [tex]x_{1} =-5,y_{1} =5,x_{2}=4 ,y_{2}=2[/tex]]
BC ⇒[tex]\sqrt{(3-4)^{2} +(-1-2)^{2} }[/tex] =[tex]\sqrt{10}[/tex] [[tex]x_{1}=4, y_{1}=2, x_{2}=3, y_{2}=-1[/tex]]
CD ⇒[tex]\sqrt{(-6-3)^{2}+(2+1)^{2} }[/tex] = [tex]\sqrt{90}[/tex] [[tex]x_{1}=3, y_{1}=-1, x_{2}=-6, y_{2}=2][/tex]
DA ⇒[tex]\sqrt{(-5+6)^{2}+(5-2)^{2} }[/tex] = [tex]\sqrt{10}[/tex] [ [tex]x_{1}=-6, y_{1}=2, x_{2}=-5, y_{2}= 5[/tex]
So,
AB= CD and BC = DA
Hence, it is a rectangle.
