Question

Dados os pontos A (4;2) B (-1;1) e C (0;2) , ache a área do triangulo ABC. ME AJUDEM NESSA QUESTÃO !

Answer 1

Utilizando o método do determinante, temos:

$Area~=~\frac{1}{2}~.~\left|\left|\begin{array}{ccc}x_A&y_A&1\\x_B&y_B&1\\x_C&y_C&1\end{array}\right|\right|\\\\\\Area~=~\frac{1}{2}~.~\left|\left|\begin{array}{ccc}4&2&1\\-1&1&1\\0&2&1\end{array}\right|\right|\\\\\\Area~=~\frac{1}{2}~.~\left|~(4.1.1+(-1).2.1+0.2.1)~-~(1.1.0+1.2.4+1.2.(-1))~\right|\\\\\\Area~=~\frac{1}{2}~.~\left|~(4-2+0)~-~(0+8-2)~\right|\\\\\\Area~=~\frac{1}{2}~.~\left|~2~-~6~\right|\\\\\\Area~=~\frac{1}{2}~.~\left|~-4~\right|\\\\\\Area~=~\frac{1}{2}~.~4$

$\boxed{Area~=~2~unidades~de~area}$