Plugging A = ( 3 , 2 ) {\displaystyle A=(3,2)} , B = ( 1 , 3 ) {\displaystyle B=(1,3)} and C = ( 2 , 5 ) {\displaystyle C=(2,5)} into the formula in the Hint, we get
1 2 | det ( 3 2 1 1 3 1 2 5 1 ) | = 1 2 | 3 ( 3 × 1 − 5 × 1 ) − 2 ( 1 × 1 − 2 × 1 ) + 1 ( 5 × 1 − 2 × 3 ) | = 1 2 | − 6 + 2 − 1 | = 5 2 . {\displaystyle {\frac {1}{2}}\left|\det {\begin{pmatrix}3&2&1\\1&3&1\\2&5&1\end{pmatrix}}\right|={\frac {1}{2}}{\Big |}3(3\times 1-5\times 1)-2(1\times 1-2\times 1)+1(5\times 1-2\times 3){\Big |}={\frac {1}{2}}|-6+2-1|={\frac {5}{2}}.}
Thus, the area of the triangle A B C {\displaystyle ABC} is 5 2 {\displaystyle \color {blue}{\frac {5}{2}}} .
