SOLUTION: Use the graphs of y=2x+6, 3x+2y=19 and y=2, which contain the sides of a triangle. 1. Find the coordinates of the vertices of the triangle. 2. Find the area of the triangle

Question 386488: Use the graphs of y=2x+6, 3x+2y=19 and y=2, which contain the sides of a triangle.
1. Find the coordinates of the vertices of the triangle.
2. Find the area of the triangle
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Use the graphs of y=2x+6, 3x+2y=19 and y=2, which contain the sides of a triangle.
1. Find the coordinates of the vertices of the triangle.
2. Find the area of the triangle
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I don't know of a "slick" or elegant way to do this, just grind it out.
Sub y=2 into the other eqns to find the vertices.
y=2x+6 and y=2
2 = 2x+6
x = -2 --> (-2,2) vertex A
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3x+2y=19 and y=2
3x+4 = 19
x = 5 --> (5,2) vertex B
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y=2x+6
3x+2y=19
Sub for y
3x + 2(2x+6) = 19
7x + 12 = 19
x = 1 --> (1,8) vertex C
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To find the area, list the points in columns and repeat the 1st point
The + signs are for alignment
.A...B...C...A
-2 +5 +1 -2
+2 +2 +8 +2
Starting at the left, add the products of the upward diagonals, the the downward
Upward:
2*5 + 2*1 + 8*-2 = 10 + 2 - 16 = -4
Downward:
-2*2 + 5*8 + 1*2 = -4 + 40 + 2 = 38
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The difference is 38 -(-4) = 42
The area is 1/2 that, = 21 sq units
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PS The method for calculating area works for polygons of any shape, and any number of sides. It doesn't seem to be well known.

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