Question

Why area of triangle in determinant form last column contains 1

Answer 1

Answer:

One way to see this is to start with the fact that the area of the triangle with one vertex at the origin and the other two at ( x₁, y₁ ) and ( x₂, y₂ ) is given by the 2×2 determinant

$A = \dfrac12\left|\begin{array}{cc}x_1&y_1\\x_2&y_2\end{array}\right|$

To get the area of a triangle with vertices at ( x₁, y₁ ) ,  ( x₂, y₂ )  and  ( x₃, y₃ ), we just translate this so that one of the vertices is at the origin then use the formula above.

So the area of our triangle is the same as the area of the congruent triangle with vertices at ( 0, 0 ),  ( x₂ - x₁ , y₂ - y₁ ) and ( x₃ - x₁ , y₃ - y₁ ).

To get this area, use the formula above.  Then be tricky to move to 3×3 determinants.

$A = \frac12\left|\begin{array}{cc}x_2-x_1&y_2-y_1\\x_3-x_1&y_3-y_1\end{array}\right|\\= \frac12\left|\begin{array}{ccc}x_1&x_2&1\\x_2-x_1&y_2-y_1&0\\x_3-x_1&y_3-y_1&0\end{array}\right|\\= \frac12\left|\begin{array}{ccc}x_1&x_2&1\\x_2&y_2&1\\x_3&y_3&1\end{array}\right|$

In the last step, we added the first row to the second row, and added the first row to the third row, and these are operations that don't change the value of a determinant.