We have to find the perimeter of a polygon.
We have the coordinates of the vertices: (-2,5), (3,1), (2,-6). As we have three points, we know that we the polygon is a triangle.
We will graph the points in order to see the distance we need to calculate for the perimeter.
Now, we will calculate the perimeter as the sum of the length of the segments AB, BC and CA.
For each segment, the length is the distance between the endpoints:
Segment AB:
[tex]\begin{gathered} \bar{AB}=\sqrt[]{(y_b-y_a)^2+(x_b-x_a)^2} \\ \bar{AB}=\sqrt[]{(1-5)^2+(3-(-2))^2} \\ \bar{AB}=\sqrt[]{(-4)^2+(5)^2} \\ \bar{AB}=\sqrt[]{16+25} \\ \bar{AB}=\sqrt[]{41} \\ \bar{AB}\approx6.4 \end{gathered}[/tex]
Segment BC:
[tex]\begin{gathered} \bar{BC}=\sqrt[]{(y_b-y_c)^2+(x_b-x_c)^2} \\ \bar{BC}=\sqrt[]{(1-(-6))^2+(3-2)^2} \\ \bar{BC}=\sqrt[]{(7)^2+(1)^2} \\ \bar{BC}=\sqrt[]{49+1} \\ \bar{BC}=\sqrt[]{50} \\ \bar{BC}\approx7.1 \end{gathered}[/tex]
Segment CA:
[tex]\begin{gathered} \bar{CA}=\sqrt[]{(y_a-y_c)^2+(x_a-x_c)^2} \\ \bar{CA}=\sqrt[]{(5_{}-(-6))^2+(-2_{}-2)^2} \\ \bar{CA}=\sqrt[]{(11)^2+(-4)^2^{}} \\ \bar{CA}=\sqrt[]{121+16} \\ \bar{CA}=\sqrt[]{137} \\ \bar{CA}\approx11.7 \end{gathered}[/tex]
Then, the perimeter as the sum of the three segment's length is:
[tex]P=\bar{AB}+\bar{BC}+\bar{CA=6.4+7.1+11.7=25.2[/tex]
Answer: the perimeter of the polygon is 25.2 units.
