For this case we have to find the area through Heron's formula:
![A=\sqrt{s(s-a)(s-b)(s-c)}](/tpl/images/0174/2639/558f1.png)
Where:
s: It's half the perimeter of the triangle
a, b, c are the sides.
We can find the sides by equating the distance between two points:
![A (1,1)\\B: (4,0)\\C: (3,5)\\](/tpl/images/0174/2639/4f353.png)
![AB=\sqrt {(x2-x1) ^ 2+(y2-y1) ^ 2}\\AB =\sqrt {(4-1) ^ 2+(0-1) ^ 2}\\AB=\sqrt {3 ^ 2+(- 1) ^ 2}\\AB =\sqrt {10}\\AB = 3.16](/tpl/images/0174/2639/daf72.png)
We found BC:
![BC =\sqrt {(x2-x1) ^ 2 + (y2-y1) ^ 2}\\BC =\sqrt {(3-4) ^ 2 + (5-0) ^ 2}\\BC =\sqrt {(- 1) ^ 2 + 5 ^ 2}\\BC = \sqrt {26}\\BC = 5.10](/tpl/images/0174/2639/f80ef.png)
We found CA:
![CA=\sqrt {(x2-x1) ^ 2 + (y2-y1) ^ 2}\\CA=\sqrt {(1-3) ^ 2 + (1-5) ^ 2}\\CA=\sqrt {(- 2) ^ 2 + (- 4) ^ 2}\\CA=\sqrt {4 + 16}\\CA = \sqrt {20}\\CA = 4.47](/tpl/images/0174/2639/c9988.png)
So, half of the perimeter is:
![s = \frac{3.16 + 5.10 + 4.47} {2} = 6.365](/tpl/images/0174/2639/2f38d.png)
Thus, the area is:
![A=\sqrt {6.365 (6.365-3.16) (6.365-5.10) (6.365-4.47)}\\A=\sqrt {6,365 (3,205) (1,265) (1,895)}\\A = \sqrt {48.90}\\A = 6.99](/tpl/images/0174/2639/f96b3.png)
Rounding we have that the area is 7 square units
Option B