ANSWER :
square
EXPLANATION :
From the problem, we have the coordinates of the quadrilateral :
[tex]\begin{gathered} W(-8,-3) \\ X(-7,7) \\ Y(3,6) \\ Z(2,-4) \end{gathered}[/tex]Plot these points to the rectangular coordinate system.
It looks like a square or a rhombus, but we need to make sure of it.
We need to check the distance between two points.
Note that the side lengths of a square and a rhombus are congruent or equal.
Using the distance formula :
[tex]\begin{gathered} d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ d_{WX}=\sqrt{(-7+8)^2+(7+3)^2}=\sqrt{101} \\ d_{XY}=\sqrt{(3+7)^2+(6-7)^2}=\sqrt{101} \\ d_{YZ}=\sqrt{(2-3)^2+(-4-6)^2}=\sqrt{101} \\ d_{ZW}=\sqrt{(-8-2)^2+(-3+4)}=\sqrt{101} \end{gathered}[/tex]The side lengths are all equal to √101
So it is either square or rhombus
A square has an interior angle of 90 degrees.
For the sides to have a 90 degrees angle, the slope must be negative reciprocal of each other.
The slope formula is :
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ \\ m_{WX}=\frac{7+3}{-7+8}=10 \\ \\ m_{XY}=\frac{6-7}{3+7}=-\frac{1}{10} \end{gathered}[/tex]Side WX has a slope of 10
and side XY has a slope of -1/10
The slopes are negative reciprocal of each other. Therefore, the quadrilateral is a square
