For this case we have to find the area through Heron's formula:
![A = \sqrt{s (s-a) (s-b) (s-c)}](/tpl/images/0174/2645/ef04d.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:
![S: (- 2, -2)\\Q: (- 1,2)\\R: (1, -4)](/tpl/images/0174/2645/ddde0.png)
![SQ=\sqrt {(x2-x1) ^ 2+(y2-y1) ^ 2}\\SQ=\sqrt {(- 1 + 2) ^ 2+ (2 + 2) ^ 2}\\SQ=\sqrt {1 ^ 2 +(4) ^2}\\SQ=\sqrt {17}\\SQ=4.12](/tpl/images/0174/2645/0fb6b.png)
We found QR:
![QR = \sqrt {(x2-x1) ^ 2 + (y2-y1) ^ 2}\\QR = \sqrt {(1 + 1) ^ 2 + (- 4-2) ^ 2}\\QR = \sqrt {(2) ^ 2 + (- 6) ^ 2}\\QR = \sqrt {40}\\QR = 6.33](/tpl/images/0174/2645/60802.png)
We found RS:
![RS=\sqrt{(x2-x1)^2+(y2-y1)^2}\\RS=\sqrt{(1 + 2)^2+(- 4 + 2)^2}\\RS=\sqrt{(3)^2+(- 2)^2}\\RS=\sqrt{9+4}\\RS=\sqrt {13}\\RS=3.60](/tpl/images/0174/2645/a8566.png)
So, half of the perimeter is:
![s = \frac {4.12 + 6.33 + 3.61} {2} = 7.03](/tpl/images/0174/2645/a1926.png)
Thus, the area is:
![A = \sqrt{7.03 (7.03-4.12) (7.03-6.33) (7.03-3.61)}\\A = \sqrt{7.03 (2.91) (0.7) (3.42)}\\A = \sqrt {48.97}\\A = 6,997](/tpl/images/0174/2645/4bc7d.png)
Rounding we have that the area is 7 square units
Option A