To find the area and the perimeter for a right triangle, first, we need to calculate the values for the two legs and the hypotenuse. To do so, we will use the following definition for a distance of two points:
![d_(a-b)=\sqrt[]{(x_a-x_b)^2+(y_a-y_b)^2}](https://img.qammunity.org/2023/formulas/mathematics/college/9anh5sqtbujvh5f62gkhgi4ugdewcuzsjp.png)
Applying this for the present question, we have:
![\begin{gathered} d_(AB)=\sqrt[]{(-2-3)^2+(-4-8)^2}=\sqrt[]{(-5)^2+(-12)^2}=\sqrt[]{25+144} \\ d_(AB)=\sqrt[]{169}=13 \\ \\ d_(BC)=\sqrt[]{(3-3)^2+(-4-8)^2}=\sqrt[]{(0)^2+(-12)^2}=\sqrt[]{144} \\ d_(BC)=12_{} \\ \\ d_(AC)=\sqrt[]{(-2-3)^2+(-4-(-4))^2}=\sqrt[]{(-5)^2+(0)^2}=\sqrt[]{25} \\ d_(AC)=5 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/hh0v16n0s4l05aremou5jf4ilidnj8klnh.png)
Now, we know that the perimeter of a triangle is just the sum of each side, as follows:

And the area of a right triangle is given by half of the multiplication of the legs, as follows:

Where u stands for an arbitrary unit.