Select the correct answer from each drop-down menu. A quadrilateral has vertices A{11, -7), 89,-4), (111, -1), and D13,-4).
Quadrilateral ABCD is a
If the vertex c (11, -1) were shifted to the point c(11,1) quadrilateral Abcd would be a

A non-equilateral rhombus.

Step-by-step explanation:

We can solve this graphically.

We start with square:

ABCD

with:

A = (11, - 7)

B = (9, - 4)

C = (11, - 1)

D = (13, - 4)

Only with the vertices, we can see that ABCD is equilateral, as the length of each side is:

AB = √( (11 - 9)^2 + (-7 -(-4))^2) = √( (2)^2 + (3)^2) = √(4 + 9) = √13

BC =  √( (11 - 9)^2 + (-1 -(-4))^2) = √13

CD =  √( (11 - 13)^2 + (-1 -(-4))^2)  = √13

DA =  √( (11 - 13)^2 + (-7 -(-4))^2) = √13

And we change C by C' = (11, 1)

In the image you can see the 5 points and the figure that they make:

The figure ABCD is a rhombus, and ABC'D is also a rhombus, the only difference between the figures is that ABCD is equilateral while ABC'D is not equilateral.

two? xd

step-by-step explanation: