Explanation
We are given the following points:
[tex](0,0),(1,3),(5,0),(6,3)[/tex]
We are required to determine which quadrilateral it is with the given points as vertices.
This is achieved thus:
- The graph of the points is:
- Next, we determine if AB = CD and BC = AD as follows:
[tex]\begin{gathered} A(0,0)\to(x_1,y_1) \\ B(1,3)\to(x_2,y_2) \\ Distance(d)=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\ AB=\sqrt{(1-0)^2+(3-0)^2} \\ AB=\sqrt{1+9} \\ AB=\sqrt{10} \\ \\ C(6,3)\to(x_1,y_1) \\ D(5,0)\to(x_2,y_2) \\ \begin{equation*} Distance(d)=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \end{equation*} \\ CD=\sqrt{(5-6)^2+(0-3)^2} \\ CD=\sqrt{1+9} \\ CD=\sqrt{10} \end{gathered}[/tex][tex]\begin{gathered} B(1,3)\to(x_1,y_1) \\ C(6,3)\to(x_2,y_2) \\ \begin{equation*} Distance(d)=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \end{equation*} \\ BC=\sqrt{(6-1)^2+(3-3)^2} \\ BC=\sqrt{25+0}=\sqrt{25} \\ BC=5 \\ \\ A(0,0)\to(x_1,y_1) \\ D(5,0)\to(x_2,y_2) \\ \begin{equation*} Distance(d)=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \end{equation*} \\ AD=\sqrt{(5-0)^2+(0-0)^2} \\ AD=\sqrt{25+0}=\sqrt{25} \\ AD=5 \end{gathered}[/tex]
- Using the graph and the distances gotten above, the quadrilateral is a Parallelogram.
Option D is correct.
