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| Question 1064651:  What is the area of a triangle whose vertices are R(−4,2) , S(1,2) , and T(−5,−4)?
 
 
 Found 2 solutions by  KMST, Alan3354:
 Answer by KMST(5328)
      (Show Source): 
You can put this solution on YOUR website! That triangle has a horizontal base with y=2. The length of that base is RS=1-(-4)=1+4=5.
 The height of the triangle is the vertical distance
 from line y=2 to T(-5,-4).
 That height is
 2-(-4)=2+4=6.
 The area of the triangle is
 
  .Answer by Alan3354(69443)
      (Show Source): 
You can put this solution on YOUR website! What is the area of a triangle whose vertices are R(-4,2), S(1,2), and T(-5,-4)? -------------
 You can find the 3 side lengths, then use Heron's Law.
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 Or,
 
 
 R   S   T   R
-4   1  -5  -4
 2   2  -4   2
 Add the diagonal products starting at the upper left:
 = -8 - 4 - 10 = -22
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 Add the diagonal products starting at the lower left:
 = 2 - 10 + 16 = 8
 |-22 - 8| = 30
 The area is 1/2 of that, = 15 sq units.
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 This method works for any polygon, any # of sides.
 The points have to be in order around the perimeter.
 
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