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Show up to sign the area of a triangle in $\mathbb{R^2}$ with vertices $(x_{0}, y_{0}), (x_{1},y_{1}), (x_{2}, y_{2})$ is

$$\frac{1}{2} \det \begin{bmatrix} x_{1} - x_{0} & x_{2} - x_{0} \\ y_{1} - y_{0} & y_{2} - y_{0} \end{bmatrix} $$

I'm having trouble conjuring up the right idea to show this. We don't know if the triangle is an equilateral triangle. Since all I could do is compute the lengths of the sides I would still need to figure out the height of the triangle. I was thinking of trying to use Pythagoras, but in order for me to do that I would need to find out which side provides the greatest height. Some guidance on how to proceed would be appreciated. I don't need the solution, just the process to do it and perhaps the reasoning. Figuring out the determinant is straightforward, it is using just the points that is an issue.

nmasanta
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D.C. the III
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    You can consider the issue as half the computation of the area of a parallelogram. see the graphics and explanations in https://math.stackexchange.com/q/29128 and in https://www.toppr.com/guides/maths/determinants/area-triangle-using-determinants/ – Jean Marie Sep 15 '19 at 20:38
  • Put $A=(x_0,y_0), B=(x_1,y_1), C=(x_2,y_2) $ and take vectorial product of $AB$ and $AC$ – Piquito Sep 16 '19 at 11:00

1 Answers1

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You can think of triangle area as sum of 2 trapezoids, then remove the base.

$$Area = \left|{y_1 + y_0 \over 2}(x_1 - x_0) + {y_2 + y_1 \over 2}(x_2 - x_1) - {y_2 + y_0 \over 2}(x_2 - x_0) \right|$$

The middle term can be rewrite as:

$${(y_2+y_1)(x_2-x_0) - (y_2+y_1)(x_1-x_0) \over 2}$$

$$Area = {1\over2}\left|(y_0-y_2)(x_1 - x_0) - (y_0 - y_1)(x_2 - x_0) \right|$$

Flip the sign inside absolute function, we get your ½ Det expression.

albert chan
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