Answer: Sqaure
Step-by-step explanation:
We obtain the figure by either plotting or finding the distances between the points:
[tex]\begin{aligned}&\text { Distance } b / w \quad A B =\sqrt{(1-7)^{2}+(3-(-3))^{2}}\\&=\sqrt{6^{2}+6^{2}}\\&A B=\sqrt{72} \text { units }\end{aligned}[/tex]
[tex]\text { Distance } b / w \quad BC =\sqrt{(7-1)^{2}+(-3-(-9))^{2}} \\BC=\sqrt{72} \text { units }[/tex]
[tex]\begin{aligned}\text { Distance } b / w \ C D &=\sqrt{(1-(-5))^{2}+(-9-(-3))^{2}} \\C D=\sqrt{72} \text { units }\end{aligned}[/tex]
[tex]$ Distance $b/w \ A D=\sqrt{(1-(-5))^{2}+(3(-3))^{2}}$A D=\sqrt{72} \text { units }[/tex]
Since all the distances of the four sides are equal, we have to check if the shape is either a rhombus or a square by the measuring the length of the diagonals
[tex]\begin{aligned}A C &=\sqrt{(1-1)^{2}+(3-(-9))^{2}} \\&=\sqrt{12^{2}} \\A C &=12 \text { units } \\B D &=\sqrt{(7-(-5))^{2}+(-3-(-3))^{2}} \\&=\sqrt{12^{2}} \\A C &=12 \text { units }\end{aligned}[/tex]
The diagonals are also equal; therefore, the figure is a square
