Explanation:
First, let's draw the quadrilateral. So:
Then, the distance d and slope m between two points with coordinates (x1, y1) and (x2, y2) can be calculated as:
[tex]\begin{gathered} d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ m=\frac{y_2-y_1}{x_2-x_1} \end{gathered}[/tex]
So, the distance and slope of AB where A is (-5,3) and B is (0, 6) are:
[tex]\begin{gathered} d=\sqrt[]{(0-(-5))^2+(6-3)^2} \\ d=\sqrt[]{(0+5)^2+3^2} \\ d=\sqrt[]{34} \\ m=\frac{6-3}{0-(-5)}=\frac{3}{0+5_{}}=\frac{3}{5} \end{gathered}[/tex]
The distance and slope of BC where B is (0,6) and C is (5, 3) are:
[tex]\begin{gathered} d=\sqrt[]{(5-0)^2+(3-6)^2}=\sqrt[]{34} \\ m=\frac{3-6}{5-0}=-\frac{3}{5} \end{gathered}[/tex]
The distance and slope of CD where C is (5,3) and D is (0, 0) is:
[tex]\begin{gathered} d=\sqrt[]{(0-5)^2+(0-3)^2}=\sqrt[]{34} \\ m=\frac{0-3}{0-5}=\frac{-3}{-5}=\frac{3}{5} \end{gathered}[/tex]
The distance and slope of AD where A is (-5,3) and D is (0, 0) are:
[tex]\begin{gathered} d=\sqrt[]{(0-(-5))^2+(0-3)^2} \\ d=\sqrt[]{(0+5)^2+(3)^2}=\sqrt[]{34} \\ m=\frac{0-3}{0-(-5)}=-\frac{3}{5} \end{gathered}[/tex]
Therefore, the correct answers are:
Option 1 : Opposite sides (AB, CD, and BC, AD) have equal slopes making them parallel to each other, making ABCD a parallelogram.
Option 4: The distance of the sides AB, BC, CD, and AD are all congruent
