The problem in the image asks to find the perimeter of a triangle with the following vertices:

• $A(-2, -1)$
• $B(1, -1)$
• $C(1, 3)$

Steps to solve:

1. Calculate the length of each side using the distance formula between two points: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

2. Find the lengths of the three sides:

   • Side AB (between points $A(-2, -1)$ and $B(1, -1)$): $AB = \sqrt{(1 - (-2))^2 + (-1 - (-1))^2} = \sqrt{(1 + 2)^2 + 0^2} = \sqrt{9} = 3$

   • Side BC (between points $B(1, -1)$ and $C(1, 3)$): $BC = \sqrt{(1 - 1)^2 + (3 - (-1))^2} = \sqrt{0^2 + (3 + 1)^2} = \sqrt{16} = 4$

   • Side AC (between points $A(-2, -1)$ and $C(1, 3)$): $AC = \sqrt{(1 - (-2))^2 + (3 - (-1))^2} = \sqrt{(1 + 2)^2 + (3 + 1)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

3. Perimeter of the triangle: The perimeter $P$ is the sum of the lengths of all sides: $P = AB + BC + AC = 3 + 4 + 5 = 12$

Thus, the perimeter of the triangle is 12 units.

Would you like any further clarification or details?

Here are 5 related questions to expand on this topic:

1. What is the area of the triangle formed by the same vertices?
2. How can you verify if the triangle is a right triangle?
3. What would the perimeter be if one of the coordinates changed to $(2, 3)$?
4. Can you find the centroid of this triangle using its vertices?
5. How would the perimeter calculation differ for a non-triangular polygon?

Tip: Always double-check the units in geometry problems to ensure consistency, especially in real-world applications!