Question

Calcule a área do triangulo cujo os vértices são A(a+1,a+2) B(a,a-1) e C(a+2,a)

Answer 1

Utilizando o método do determinante, temos:

$Area~=~\frac{1}{2}~.~\left|\left|\begin{array}{ccc}x_A&y_A&1\\x_B&y_B&1\\x_C&y_C&1\end{array}\right|\right|\\\\\\\\Area~=~\frac{1}{2}~.~\left|\left|\begin{array}{ccc}a+1&a+2&1\\a&a-1&1\\a+2&a&1\end{array}\right|\right|\\\\\\Area~=~\frac{1}{2}~.~\left|~(~(a+1).(a-1).1+a.a.1+(a+2.)(a+2).1~)~-~\right\\\\~~~~~~~~~~~~~~~~~~~~\,\left(~1.(a-1).(a+2)+1.a.(a+1)+1.(a+2).a~)\right|$

$Area~=~\frac{1}{2}~.~\left|~(~(a^2-1)+(a^2)+(a^2+4a+4)~)~-~\right\\\\~~~~~~~~~~~~~~~~~~~~\,\left(~(a^2+a-2)+(a^2+a)+(a^2+2a)~)~\right|\\\\\\Area~=~\frac{1}{2}~.~\left|~(~3a^2+4a+3~)~-~(~3a^2+4a-2~)~\right|\\\\\\Area~=~\frac{1}{2}~.~\left|~3a^2+4a+3-3a^2-4a+2~\right|\\\\\\Area~=~\frac{1}{2}~.~\left|~5~\right|\\\\\\Area~=~\frac{1}{2}~.~5\\\\\\\boxed{Area~=~2,5~unidades~de~area}$