Question

Graph the parallclogram with vertices A (4,-6), B(-1,-6), C(3,-3), and D(8,-3).

Answer 1

Explanation

We are given the following coordinates:

`\begin{gathered} A(4,-6) \\ B(-1,-6) \\ C(3,-3) \\ D(8,-3) \end{gathered}`

We are required to graph the coordinates given and find the distance of the following:

`\begin{gathered} AB \\ BC \\ CD \\ DA \end{gathered}`

The points can be graphed as follows:

We know that the distance between two points can be calculated as:

Therefore, we can determine the distance of AB as:

`\begin{gathered} A(4,-6)\to(x_1,y_1) \\ B(-1,-6)\to(x_2,y_2) \\ \therefore AB=√((-1-4)^2+(-6-(-6))^2) \\ AB=√((-1-4)^2+(-6+6)^2) \\ AB=√((-5)^2+(0)^2) \\ AB=√(25+0)=√(25) \\ AB=5\text{ }units\text{ } \end{gathered}`

We can determine the distance of BC as:

`\begin{gathered} B(-1,-6)\to(x_1,y_1) \\ C(3,-3)\to(x_2,y_2) \\ \therefore BC=√((3-(-1))^2+(-3-(-6))^2) \\ BC=√((3+1)^2+(-3+6)^2) \\ BC=√((4)^2+(3)^2) \\ BC=√(16+9)=√(25) \\ BC=5\text{ }units \end{gathered}`

The value of CD is:

`\begin{gathered} C(3,-3)\to(x_1,y_1) \\ D(8,-3)\to(x_2,y_2) \\ \therefore CD=√((8-3)^2+(-3-(-3))^2) \\ CD=√((8-3)^2+(-3+3)^2) \\ CD=√((5)^2+(0)^2) \\ CD=√(25+0)=√(25) \\ CD=5\text{ }units \end{gathered}`

Finally, the value of DA is:

`\begin{gathered} D(8,-3)\to(x_1,y_1) \\ A(4,-6)\to(x_2,y_2) \\ \therefore DA=√((4-8)^2+(-6-(-3))^2) \\ DA=√((4-8)^2+(-6+3)^2) \\ DA=√((-4)^2+(-3)^2) \\ DA=√(16+9)=√(25) \\ DA=5\text{ }units \end{gathered}`

Hence, the answers are:

`\begin{gathered} AB=5\text{ }units \\ BC=5\text{ }un\imaginaryI ts \\ CD=5\text{ }un\imaginaryI ts \\ DA=5\text{ }un\imaginaryI ts \end{gathered}`