Area of a Triangle
We know that the area of a triangle whose vertices are (x1 , y1 ),(x2 , y 2 ) and (x3 , y3 ) is equal to the absolute value of
1/2 (x1 y 2 − x2 y1 + x2 y3 − x3 y 2 + x3 y1 − x1 y3 ) .
This expression can be written in the form of a determinant as the absolute value of
If the area of the triangle with vertices (- 3, 0), (3, 0) and (0, k) is 9 square units, find the values of k .
The area of the triangle formed by three points is zero if and only if the three points are collinear. Also, we remind the reader that the determinant could be negative whereas area is always non-negative.
Find the area of the triangle whose vertices are (- 2, - 3), (3, 2), and (- 1, - 8).
Show that the points (a, b + c), (b, c + a), and (c, a + b) are collinear.
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