We will have the following:
First, we have the graph of the problem:
Now, we determine the slope of the diagonals, and if those are perpendiullar we then have that it will be a square, that is:
[tex]\begin{cases}m_{AC}=\frac{0-(-2)}{2-(-6)}\Rightarrow m_{AC}=\frac{1}{4} \\ \\ m_{BD}=\frac{-5-3}{-1-(-3)})\Rightarrow m_{BD}=-4\end{cases}[/tex]
From this, we can see that the slopes are perpendicular. This is a condition for a square or a rhombus.
Now, we determine if the graph belongs to a square by determining if the slopes of AB & BC are perpendicular:
[tex]\begin{cases}m_{AB}=\frac{3-(-2)}{-3-(-6)}\Rightarrow m_{AB}=\frac{5}{3} \\ \\ m_{BC}=\frac{0-3}{2-(-3)}\Rightarrow m_{BC}=-\frac{3}{5}\end{cases}[/tex]
From this we can see that those segmens are also perpendicular, so in this particular case the graph is a square. [Which technically speaking is also a rhombus].
The reasoning is that the diagonals are perpendicular and the external segments are also perpendicular, a property that belong to squares.
Now, we find the intersection point of the diagonals, that is:
[tex]M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})_{}[/tex]
[tex]M(\frac{2-6}{2},\frac{0-2}{2})\rightarrow M(-2,-1)[/tex]
Now, we determine the distance of all 4 segments AM, BM, CM & DM:
[tex]\begin{cases}d_{AM}=\sqrt[]{(-2+6)^2+(-1+2)^2}\Rightarrow d_{AM}=\sqrt[]{17} \\ \\ d_{BM}=\sqrt[]{(-2+3)^2+(-1-3)^2}\Rightarrow d_{AM}=\sqrt[]{17} \\ \\ d_{CM}=\sqrt[]{(-2-2)^2+(-1-0)^2}\Rightarrow d_{CM}=\sqrt[]{17} \\ \\ d_{DM}=\sqrt[]{(-2+1)^2+(-1+5)^2}\Rightarrow d_{DM}=\sqrt[]{17}\end{cases}[/tex]
So, the distance of all segments that divide the diagonals are equal, thus the points describe a square.
