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A quadrilateral has vertices [tex][tex]$A(11, -7)$[/tex][/tex], [tex][tex]$B(9, -4)$[/tex][/tex], [tex][tex]$C(11, -1)$[/tex][/tex], and [tex][tex]$D(13, -4)$[/tex][/tex].

Quadrilateral [tex][tex]$ABCD$[/tex][/tex] is a [tex][tex]$\square$[/tex][/tex].

If point [tex][tex]$C$[/tex][/tex] is moved to [tex][tex]$C^{\prime}(11, 1)$[/tex][/tex], quadrilateral [tex][tex]$ABC^{\prime}D$[/tex][/tex] would be a [tex][tex]$\square$[/tex][/tex].



Answer :

Given the vertices [tex]\(A(11,-7), B(9,-4), C(11,-1)\)[/tex], and [tex]\(D(13,-4)\)[/tex], and comparing them to the provided answers, we proceed as follows:

First, we determine the lengths of the sides of the quadrilateral ABCD.
For [tex]\(AB\)[/tex]:
[tex]\[ AB = \sqrt{(9 - 11)^2 + (-4 - (-7))^2} = \sqrt{(-2)^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.605551275463989 \][/tex]

For [tex]\(BC\)[/tex]:
[tex]\[ BC = \sqrt{(11 - 9)^2 + (-1 - (-4))^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.605551275463989 \][/tex]

For [tex]\(CD\)[/tex]:
[tex]\[ CD = \sqrt{(13 - 11)^2 + (-4 - (-1))^2} = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.605551275463989 \][/tex]

For [tex]\(DA\)[/tex]:
[tex]\[ DA = \sqrt{(13 - 11)^2 + (-4 - (-7))^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.605551275463989 \][/tex]

All sides of quadrilateral [tex]\(ABCD\)[/tex] are equal to [tex]\(\sqrt{13} \approx 3.605551275463989\)[/tex].

Next, we need to determine if this is a parallelogram. For it to be a parallelogram, opposite sides must be equal and parallel.

Additionally, we check the diagonals to see if they're equal:
[tex]\[ AC = \sqrt{(11 - 11)^2 + (-1 - (-7))^2} = \sqrt{0 + 36} = 6 \][/tex]
[tex]\[ BD = \sqrt{(13 - 9)^2 + (-4 - (-4))^2} = \sqrt{4^2 + 0} = 4 \][/tex]

Since AC and BD do not equal, ABCD is not a parallelogram; however, as all sides are equal and its diagonals do not bisect each other equally, it is a rhombus.

Next, we consider [tex]\(C^{\prime}(11, 1)\)[/tex] instead of [tex]\(C\)[/tex]. We compute the distances:
For [tex]\(B C'\)[/tex]:
[tex]\[ BC' = \sqrt{(11 - 9)^2 + (1 - (-4))^2} = \sqrt{2^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29} \approx 5.385164807134504 \][/tex]

For [tex]\(C' D\)[/tex]:
[tex]\[ C'D = \sqrt{(11 - 13)^2 + (1 - (-4))^2} = \sqrt{(-2)^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29} \approx 5.385164807134504 \][/tex]

For [tex]\(DA\)[/tex]:
[tex]\[ DA = \sqrt{(13 - 11)^2 + (-4 - (-7))^2} = \sqrt{2^2 + 3^2} = \sqrt{13} \approx 3.605551275463989 \][/tex]

Given the difference in side lengths [tex]\( \sqrt{13}\)[/tex] and [tex]\( \approx \sqrt{29}\)[/tex], [tex]\(AB' C' D\)[/tex] cannot be a parallelogram or a rhombus. Hence, it’s a general quadrilateral, not fitting any special categories of quadrilaterals listed here.

Therefore:

Quadrilateral [tex]\(ABCD\)[/tex] is a rhombus. Quadrilateral [tex]\(AB'C'D\)[/tex] would be a quadrilateral.
Final Answer:
Quadrilateral [tex]\(ABCD\)[/tex] is a [tex]\(\boxed{\text{Rhombus}}\)[/tex] and [tex]\(AB'C'D\)[/tex] would be a [tex]\(\boxed{\text{Quadrilateral}}\)[/tex].