Question 1188225
Find the area of a triangle whose vertices are A(3,4), B(0,4), C(2,1).


Find the 3 side lengths.
For AC(side b):
{{{b = sqrt(diffy^2 + diffx^2) = sqrt(1^2 + 3^2) = sqrt(10)}}}
Find the 3 lengths, then use Heron's Law:
s = perimeter/2
{{{Area = sqrt(s*(s-a)*(s-b)*(s-c))}}}
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Heron's Law is messy for this one.
Notice that A and B have the same y value, 4.
Point C is 3 units from the line AB, so the height is 3 and base is 3.
Area = b*h/2 = 3*3/2 = 4.5 sq units
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PS  That's not an equation.