Respuesta :
Given that
The vertices of a quadrilateral are A(-5, 7), B(6,-3), C(10, 2), and D(-1, 12).
So we have to find the type of quadrilateral ABCD.
Explanation -
First, we have to find the length of each side AB, BC, CD, and DA.
We will use the distance formula to find the length of each side
[tex]\begin{gathered} T\text{he distance formula for point \lparen x}_1,y_1)\text{ and \lparen x}_2,y_2)\text{ is } \\ d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \end{gathered}[/tex]Now we will find the sides by substituting the value,
[tex]\begin{gathered} For\text{ AB,} \\ AB=\sqrt{(6-(-5))^2+(-3-7)^2} \\ \\ AB=\sqrt{(6+5)^2+(-10)^2}=\sqrt{11^2+10^2}=\sqrt{121+100}=\sqrt{221} \\ \\ AB=\sqrt{221}\text{ units} \end{gathered}[/tex][tex]\begin{gathered} For\text{ BC,} \\ BC=\sqrt{(10-6)^2+(2-(-3))^2} \\ BC=\sqrt{4^2+(2+3)^2} \\ BC=\sqrt{16+25}=\sqrt{41} \\ BC=\sqrt{41}\text{ }units \end{gathered}[/tex][tex]\begin{gathered} For\text{ CD} \\ CD=\sqrt{(-1-10)\placeholder{⬚}^2+(12-2)\placeholder{⬚}^2} \\ CD=\sqrt{(-11)^2+(10)^2} \\ CD=\sqrt{121+100} \\ CD=\sqrt{221}\text{ units} \end{gathered}[/tex][tex]\begin{gathered} For\text{ DA,} \\ DA=\sqrt{(-1-(-5))^2+(12-7)^2} \\ DA=\sqrt{(-1+5)^2+5^2} \\ DA=\sqrt{4^2+5^2}=\sqrt{16+25} \\ DA=\sqrt{41}\text{ units} \end{gathered}[/tex]So we found that the opposite sides are equal. So it can be a parallelogram or a rectangle.
Now, we will check the triangular part ABC, by applying Pythagoras' theorem.
[tex]\begin{gathered} In\text{ trianlge ABC, } \\ AC=\sqrt{(10-(-5))^2+(2-7)^2}=\sqrt{(10+5)^2+(-5)^2} \\ AC=\sqrt{15^2+5^2} \\ AC=\sqrt{225+25}=\sqrt{250} \\ Now\text{ using pythagoras theorem we have,} \\ AC^2=AB^2+BC^2 \\ \sqrt{250}^2=\sqrt{221}^2+\sqrt{41}^2 \\ 250=221+41 \\ 250\ne262 \end{gathered}[/tex]So it is a parallelogram. Then option D is correct.
Final answer -
The final answer is a parallelogram.