Express the area of a triangle given by vertices, $A (x_{1}, y_{1}), B (x_{2}, y_{2}), C (x_{3}, y_{3})$ $A(x_1,y_1), B(x_2,y_2), C(x_3,y_3)$ . Show that it can be expressed as determinant of: $det(Delta) = [(1, 1, 1 ),(x_1, x_2, x_3),(y_1, y_2, y_3) ]$ .Calculate the area of A(3,6), B(7,8), & C(5,2)?

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Answer 1 · 2016-09-24T20:35:19

See below.

The triangle area can be computed as a closed circuit integral. So

$S = 1/2(y_1+y_3)(x_3-x_1) + 1/2(y_3+y_2)(x_2-x_3)+1/2(y_2+y_1)(x_1-x_2)$ .

Expanding and simplifying

$S = 1/2(x_2 y_1 - x_3 y_1 - x_1 y_2 + x_3 y_2 + x_1 y_3 - x_2 y_3)$

$S =1/2|(x_2,x_3),(y_2,y_3)|-1/2 |(x_1,x_3),(y_1,y_3)|+1/2|(x_1,x_2),(y_1,y_2)|$ which is equivalent to

$S =1/2 |(1,1,1),(x_1,x_2,x_3),(y_1,y_2,y_3)|$

For the example given we have

$S =1/2 |(1,1,1),(3,7,5),(6,8,2)| = -10$

Of course this must be considered in absolute value. So

$S = abs(-10)=10$