Exercise 3.3.9

Let points A,B and C in the plane ℝ2 have coordinates (x1,y1),(x2,y2) and (x3,y3) respectively. Show that the area of triangle ABC is the absolute value of

1 2 |1x1y1 1x2y2 1x3y3 | .

Hint: use row operation and geometric interpretation of 2 × 2 determinants (area).

Answers

Proof. The area of triangle ABC is half of the parallelogram defined by neighbouring sides AB,AC, which also can be computed by

S△ABC = 1 2abs( |x2 − x1y2 − y1 x3 − x1y3 − y1 | ) = 1 2|(x2 − x1)(y3 − y1) − (x3 − x1)(y2 − y1)|.

In the same time, if we use row reduction to check the determinant

1 2 |1x1y1 1x2y2 1x3y3 | = 1 2 |1 x1 y1 0x2 − x1y2 − y1 0x3 − x1y3 − y1 | = 1 2 |1 x1 y1 0x2 − x1 y2 − y1 0 0 y3 − y1 − (y2 − y1)x3−x1 x2−x1 | = 1 2((x2 − x1)(y3 − y1) − (x3 − x1)(y2 − y1)).

We assume that x2 − x1≠0 and it can be verified if x2 − x1 = 0, the result still holds. With the absolute value, we can see the conclusion holds. □

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2018-11-29 00:00
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