Answer:
34 square units
Explanation:
You want the area of the triangle with vertices X(6, 8), Y(3, 3), and Z(13, -3).
Area from vertices
There are several ways the area of a triangle can be calculated from the coordinates of the vertices. One relatively easy method is to find half the absolute value of the sum of the determinants of adjacent pairs of points.
This calculation is shown in the second attachment.
The area of ∆XYZ is 34 square units.
Right triangle
A slope calculation will tell you side XY is perpendicular to side YZ. This means you can find the area from the lengths of these two sides. The distance formula can tell you the lengths.
XY = √((3 -6)² +(3 -8)²) = √(9 +25) = √34
YZ = √((13 -3)² +(-3 -3)²) = √(100 +36) = 2√34
Area = 1/2bh = (1/2)(√34)(2√34) = 34 . . . . square units
Pick's theorem
Pick's theorem tells you the area of a polygon with vertices on grid points can be found by counting the grid points on the boundary and interior to the boundary.
A = i + (b/2) -1
This triangle has the three given vertices on grid points, along with point (8, 0), which is the midpoint of YZ. There are 33 interior points, so the area is ...
A = 33 +(4/2) -1 = 34 . . . . square units
Heron's formula
The length of XZ is ...
XZ = √((13 -6)² +(-3 -8)²) = √(49 +121) = √170
Heron's formula tells you the area is given by ...
A = √(s(s -a)(s -b)(s -c)) . . . . . . where s = (a+b+c)/2, the semiperimeter
The third attachment shows this calculation gives an area of 34 square units.
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Additional comments
The slope calculation tells you the slope of XY is ...
m = (y2 -y1)/(x2 -x1) = -5/-3 = 5/3
and the slope of YZ is ...
m = -6/10 = -3/5 . . . . . . . . the opposite reciprocal of the slope of XY
Hence these sides are perpendicular, as we noted above.
We like Pick's theorem for finding the areas of small figures with irrational side lengths (such as this one). It just involves counting, which is often easier than using a bunch of different formulas for slope or length.
The determinant formula is easy to compute, but tedious to enter on a calculator. It is much easier to program a spreadsheet for this. As with Pick's theorem, the method applies to a polygon with any number of vertices.
In these 2×2 matrices, the coordinates can be listed in rows or in columns. It makes no difference to the calculation. The key is to work in one direction around the figure. (Here, we went CCW.)
The triangle is drawn in the first attachment. The geometry program that drew it also calculated its area to be 34 square units.
You may have noticed that the "determinant" method and Pick's theorem guarantee that the area of any polygon with integer coordinates will be an integer multiple of 1/2 square unit. That is, it will never be irrational. (You might not see that using Heron's formula.)
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