MHB The area of a triangle and determinants

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The discussion centers on proving that the area of a triangle can be represented by a determinant formed by its three vertices. The original poster encounters issues with sign discrepancies in their proof, specifically regarding the orientation of the triangle. It is clarified that the determinant yields double the area for a counter-clockwise orientation, while a clockwise orientation results in a negative value. This indicates that the signs in the formulas must be adjusted based on the triangle's orientation. Understanding these orientation effects is crucial for aligning the two area representations.
Yankel
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Dear all,

I was trying to prove that the area of a triangle is equal to the determinant consisting of the three points of the triangle. I got to the end, and something ain't working out. The signs are all wrong.

In the attached pictures I include my proof. Can you please tell me how can the two formulas be identical ? The first is the area coming from trapezoid subtraction , while the second is the determinant.

Thank you !

Clarification: when I say signs are opposite, I mean (y2-y3) vs. (y3-y2) , etc...

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Hi Yankel,

The determinant gives double the area of a triangle that is oriented counter-clockwise.
In your case the triangle is oriented clockwise, meaning that we'll find the opposite.
 

Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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