Answer:
3.5 square units
Explanation:
You want the area of ∆ABC with vertices A(2, 1), B(3, 3), C(1, 6) using segment AC as the base of the triangle.
Slope
The slope of line AC is found using the slope formula:
m = (y2 -y1)/(x2 -x1)
m = (6 -1)/(1 -2) = -5
Equation
The point-slope equation for the line AC is ...
y -k = m(x -h) . . . . . . line with slope m through point (h, k)
y -1 = -5(x -2) . . . . . . line with slope -5 through point A(2, 1)
Writing this in general form, we have ...
5x +y -11 = 0
Height
The distance from point B to this line is given by ...
h = |5x +y -11|/√(5² +1²) . . . . . . distance from point (x, y)
h = |5(3) +(3) -11|/√26 = 7/√26 . . . . distance from point B(3, 3)
Base
The length of line segment AC is given by the distance formula:
b = √((x2 -x1)² +(y2 -y1)²)
b = √((1 -2)² +(6 -1)²) = √26
Area
The area of the triangle is given by the formula ...
A = 1/2bh
A = 1/2(√26)(7/√26) = 7/2 = 3.5
The area of ∆ABC is 3.5 square units.
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Additional comment
There are several other ways the coordinates of the vertices could be used to find the area.
In this instance, one of the simplest methods is to count the grid points on the boundary of the triangle (there are b=3) and the grid points inside the triangle (there are i=3). Using Pick's theorem, the area is ...
A = i +b/2 -1
A = 3 +3/2 -1 = 3.5 . . . . square units
Pick's theorem applies when polygon vertices are on grid points.