# Python 2, <s>48</s> <s>47</s> 50 bytes

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Very simple; follows the standard equation:

    lambda a,b,c,d,e,f:abs(a*(d-f)+c*(f-b)+e*(b-d))/2.

The other, similarly simple approaches are longer:

    def a(a,b,c,d,e,f):return abs(a*(d-f)+c*(f-b)+e*(b-d))/2. # 57
    lambda t:abs(t[0]*(t[3]-t[5])+t[2]*(t[5]-t[1])+t[4]*(t[1]-t[3]))/2. # 67
    def a(t):return abs(t[0]*(t[3]-t[5])+t[2]*(t[5]-t[1])+t[4]*(t[1]-t[3]))/2. # 74

Python's access to a determinate function is through [numpy](http://stackoverflow.com/a/462514/2676531).

Thanks to [muddyfish](https://codegolf.stackexchange.com/users/32686/muddyfish) for 1 byte and [xnor](https://codegolf.stackexchange.com/users/20260/xnor) for catching an error.