We can consider the case where (x1, y1) and (x2, y2) are in the first quadrant. We will assume that, together with (0, 0), they are in order clockwise around the triangle.
Consider the lines x=x1, x=x2, the x-axis, and the segment (x1, y1) to (x2, y2). These will form a trapezoid whose area is
... (area of trapezoid) = (1/2)(y1 + y2)(x2 - x1)
(If x2 < x1, the area will be negative.)
LIkewise, the triangle formed by the segment (0, 0) to (x1, y1), the line x=x1, and the x-axis will have area
... (area of first triangle) = (1/2)(y1)(x1)
The triangle formed in similar fashion by the segment (0, 0) to (x2, y2), the line x=x2, and the x-axis will have area
... (area of second triange) = (1/2)(y2)(x2)
You can convince yourself that the total area of the triangle of interest is equal to the sum of the areas computed here:
... (area of triangle in the diagram) = (area of first triangle) + (area of trapezoid) - (area of second triangle)
Note that we have subtracted the area of the second triangle. Then, in terms of coordinates, we have
... (area of triangle in the diagram) = (1/2)(x1·y1 + (y1+y2)(x2-x1) - x2·y2)
... ... = (1/2)(x2·y1 - x1·y2)
If the points are in counterclockwise order, the area will be the negative of this. Thus, we can say ...
... triangle area = (1/2)·|x1·y2 - x2·y1|
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This formula will apply whether the points are all in the first quadrant or not, as long as one of them is at the origin.
Answer:
B. [tex]\frac{1}{2}y_{1}(x_{2}-x_{1} )[/tex]
Step-by-step explanation:
This was correct on plato :D
