Drawing the points A, B, C and D in the coordinate plane, we have:
In order to find the most precise name of ABCD, first let's find the slope of each side, using the formula:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]So calculating the slope of each side, we have:
[tex]\begin{gathered} A\text{ and }B\colon \\ m=\frac{-7-0}{1-(-5)}=-\frac{7}{6} \\ B\text{ and }C\colon \\ m=\frac{-1-(-7)}{8-1}=\frac{6}{7} \\ C\text{ and }D\colon \\ m=\frac{6-(-1)}{2-8}=-\frac{7}{6} \\ D\text{ and }A\colon \\ m=\frac{0-6}{-5-2_{}}=\frac{6}{7} \end{gathered}[/tex]Since the slopes of opposite sides are the same, they are parallel. Also, since the slopes of adjacent sides follow the rule m1 = -1 / m2, they are perpendicular.
So we already know that the figure is a rectangle. In order to find if the figure is a square, let's find the length of all sides using the formula for the distance between two points:
[tex]\begin{gathered} d=\sqrt[]{(y_2-y_1)^2+(x_2-x_1)^2_{}} \\ d_{AB}=\sqrt[]{(-7-0)^2+(1-(-5))^2}=\sqrt[]{49+36}=\sqrt[]{85} \\ d_{BC}=\sqrt[]{(-1-(-7))^2+(8-1)^2}=\sqrt[]{36+49}=\sqrt[]{85} \\ d_{CD}=\sqrt[]{(6-(-1))^2+(2-8)^2_{}}=\sqrt[]{49+36}=\sqrt[]{85} \\ d_{DA}=\sqrt[]{(0-6)^2+(-5-2)^2}=\sqrt[]{36+49}=\sqrt[]{85} \end{gathered}[/tex]All sides have the same length, so the figure ABCD is a square.
