The vertices of the square are the coordinates of the endpoints
The third vertex of the square could be (0,7) or (-5,1)
Two vertices of the square are:
(0,-3) and (4,2)
Calculate the distance between these vertices using:
[tex]d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}[/tex]
Square both sides
[tex]d^2 = (x_2 - x_1)^2+(y_2 - y_1)^2[/tex]
The above expression represents the area of the square.
So, we have:
[tex](x_2 - x_1)^2+(y_2 - y_1)^2 = 41[/tex]
Next, we test the options with the given vertices (0,-3) and (4,2)
So, we have:
A. (0,7)
[tex](0 - 0)^2+(7 + 3)^2 = 41[/tex]
[tex]100 \ne 41[/tex]
[tex](0 - 4)^2+(7 - 2)^2 = 41[/tex]
[tex]41 = 41[/tex]
B. (-5,1)
[tex](-5 - 0)^2+(1 + 3)^2 = 41[/tex]
[tex]41 = 41[/tex]
[tex](-5 - 4)^2+(1 -2)^2 = 41[/tex]
[tex]82 \ne 41[/tex]
C.(5,13)
[tex](5 - 0)^2+(13 + 3)^2 = 41[/tex]
[tex]281 \ne 41[/tex]
[tex](5 - 4)^2+(13 -2)^2 = 41[/tex]
[tex]122 \ne 41[/tex]
D.(-2,-2)
[tex](-2 - 0)^2+(-2 + 3)^2 = 41[/tex]
[tex]5 \ne 41[/tex]
[tex](-2 - 4)^2+(-2 -2)^2 = 41[/tex]
[tex]52 \ne 41[/tex]
E.(4,-6)
[tex](4 - 0)^2+(-6 + 3)^2 = 41[/tex]
[tex]25 \ne 41[/tex]
[tex](4 - 4)^2+(-6 -2)^2 = 41[/tex]
[tex]64 \ne 41[/tex]
The vertices where at least one of the equations is true could be a third vertex of the square
Hence, a third vertex of the square could be (0,7) or (-5,1)
Read more about square vertices at:
https://brainly.com/question/1292795
