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For three points in 2D, $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$, show that the determinant of

\begin{bmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 2\\ x_3 & y_3 & 3\\ \end{bmatrix}

is proportional to the area of the triangle whose corners are the three points.

I know that a $3\times3$ matrix

\begin{bmatrix} a & b & c\\ d & e & f\\ g & h & i\\ \end{bmatrix}

has this determinant:

$aei+bfg+cdh-ceg-bdi-afh$

So the above matrix has a determinant of $x_1y_23+y_12x_{3}+1x_2y_3-1y_2x_3-y_1x_23-x_12y_3$

But then I don't know how to prove that this is "proportional to the area of the triangle whose corners are the three points".

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  • $\begingroup$ Have you any knowledge about the relationship between determinants and the notion of volume (of a poyhedron)? See this link: math.ucdavis.edu/~daddel/linear_algebra_appl/Applications/… $\endgroup$ – Marc Bogaerts Sep 2 '14 at 5:01
  • $\begingroup$ I did not have any knowledge about the relationship between determinants and the notion of volume. Thank you for the link! $\endgroup$ – user173243 Sep 2 '14 at 6:43

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