Question 1188225: Please help me solve this equation Find the area of a triangle whose vertices are (3,4), (0,4), (2,1). Thank you so much
Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! 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):
Find the 3 lengths, then use Heron's Law:
s = perimeter/2
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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.
Answer by ikleyn(52900)  (Show Source):
You can put this solution on YOUR website! .
The side of the triangle, connecting vertices (3,4) and (0,4), is horizontal y= 4.
The length of this side is 3-0 = 3 units.
The height of this triangle is the distance from the third vertex (2,1) to this horizontal line y= 4;
so, the height is 4-1 = 3 units.
Thus the area of the triangle is  = 4.5 square units. ANSWER
Solved.
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