# Express the area of a triangle given by vertices, 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)?

Sep 24, 2016

See below.

#### Explanation:

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

$S = \frac{1}{2} \left({y}_{1} + {y}_{3}\right) \left({x}_{3} - {x}_{1}\right) + \frac{1}{2} \left({y}_{3} + {y}_{2}\right) \left({x}_{2} - {x}_{3}\right) + \frac{1}{2} \left({y}_{2} + {y}_{1}\right) \left({x}_{1} - {x}_{2}\right)$.

Expanding and simplifying

$S = \frac{1}{2} \left({x}_{2} {y}_{1} - {x}_{3} {y}_{1} - {x}_{1} {y}_{2} + {x}_{3} {y}_{2} + {x}_{1} {y}_{3} - {x}_{2} {y}_{3}\right)$

$S = \frac{1}{2} | \left({x}_{2} , {x}_{3}\right) , \left({y}_{2} , {y}_{3}\right) | - \frac{1}{2} | \left({x}_{1} , {x}_{3}\right) , \left({y}_{1} , {y}_{3}\right) | + \frac{1}{2} | \left({x}_{1} , {x}_{2}\right) , \left({y}_{1} , {y}_{2}\right) |$ which is equivalent to

$S = \frac{1}{2} | \left(1 , 1 , 1\right) , \left({x}_{1} , {x}_{2} , {x}_{3}\right) , \left({y}_{1} , {y}_{2} , {y}_{3}\right) |$

For the example given we have

$S = \frac{1}{2} | \left(1 , 1 , 1\right) , \left(3 , 7 , 5\right) , \left(6 , 8 , 2\right) | = - 10$

Of course this must be considered in absolute value. So

$S = \left\mid - 10 \right\mid = 10$