IDNLearn.com provides a reliable platform for finding accurate and timely answers. Get comprehensive and trustworthy answers to all your questions from our knowledgeable community members.
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