How do you find the area of a triangle whose vertices are (2,-2), (8,5), (6,-10)?

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How do you find the area of a triangle whose vertices are (2,-2), (8,5), (6,-10)?

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Answer 1 · 2015-10-08T14:13:32

I fount $38$ $38$ ua

Explanation:

I would use a matrix method involving the Determinant of a square matrix.
The area of the triangle is $\pm \frac{1}{2}$ $+-1/2$ the determinant of the matrix formed by the coordinates of the vertecies of the triangle and a column of $1$ $1$ , i.e.:
Area $= \pm \frac{1}{2} \cdot det$ $=+-1/2*det$ $((2,-2,1),(8,5,1),(6,-10,1))=$
$=+-1/2(-72)=$ changing sign through $+-$ you get:
Area $=76/2=38$ units of area

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How do you find the area of a triangle whose vertices are (2,-2), (8,5), (6,-10)?

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