Explanation
We are given the following points:
![(0,0),(1,3),(5,0),(6,3)](https://img.qammunity.org/2023/formulas/mathematics/college/gf05pp4eycqf2tt4mviz8215fco1g0vsm0.png)
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:
![\begin{gathered} A(0,0)\to(x_1,y_1) \\ B(1,3)\to(x_2,y_2) \\ Distance(d)=√((x_2-x_1)^2+(y_2-y_1)^2) \\ AB=√((1-0)^2+(3-0)^2) \\ AB=√(1+9) \\ AB=√(10) \\ \\ C(6,3)\to(x_1,y_1) \\ D(5,0)\to(x_2,y_2) \\ \begin{equation*} Distance(d)=√((x_2-x_1)^2+(y_2-y_1)^2) \end{equation*} \\ CD=√((5-6)^2+(0-3)^2) \\ CD=√(1+9) \\ CD=√(10) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/rin86pic7p6wvp3jazjo3ceytcbs3wu50j.png)
![\begin{gathered} B(1,3)\to(x_1,y_1) \\ C(6,3)\to(x_2,y_2) \\ \begin{equation*} Distance(d)=√((x_2-x_1)^2+(y_2-y_1)^2) \end{equation*} \\ BC=√((6-1)^2+(3-3)^2) \\ BC=√(25+0)=√(25) \\ BC=5 \\ \\ A(0,0)\to(x_1,y_1) \\ D(5,0)\to(x_2,y_2) \\ \begin{equation*} Distance(d)=√((x_2-x_1)^2+(y_2-y_1)^2) \end{equation*} \\ AD=√((5-0)^2+(0-0)^2) \\ AD=√(25+0)=√(25) \\ AD=5 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/7k22642oa7sx0rih6qu1x1qykh50ad8t9p.png)
- Using the graph and the distances gotten above, the quadrilateral is a Parallelogram.
Option D is correct.