Question

Calcule a àrea do triângulo ABC, cujos vértices são A (-3,-1); B (3,7) e C (4,-1):

Answer 1

Usando a fórmula:

$\boxed{A=|D|*\frac{1}{2}}$

$\left[\begin{array}{ccc}-3&-1&1\\3&7&1\\4&-1&1\end{array}\right] \\\\\ D=-21-4-3-28+3-3\\\\\ D=-56\\\\ A=|D|*\frac{1}{2}\\\\ A=|-56|*\frac{1}{2}\\\\ \boxed{A=28}$

Answer 2

A área de um triângulo de vértices A, B e C , é dada por

$\boxed{\boxed{A=\dfrac{|det~M|}{2}}}$

Onde M é a matriz formada pelos vértices do triângulo:

$\boxed{\boxed{M= \left[\begin{array}{ccc}x_{A}&y_{A}&1\\x_{B}&y_{B}&1\\x_{C}&y_{C}&1\end{array}\right] }}$
_____________________________

Sendo os vértices A(-3,-1), B(3,7) e C(4,-1):

$M=\left[\begin{array}{ccc}-3&-1&1\\3&7&1\\4&-1&1\end{array}\right]$

Achando o determinante de M:

$det~M=-3\left|\begin{array}{ccc}7&1\\-1&1\end{array}\right|-3\left|\begin{array}{ccc}-1&1\\-1&1\end{array}\right|+4\left|\begin{array}{ccc}-1&1\\7&1\end{array}\right|\\\\\\det~M=-3(7+1)-3(-1+1)+4(-1-7)\\\\\\det~M=-3(8)-3(0)+4(-8)\\\\\\det~M=-24-32\\\\\\\boxed{\boxed{det~M=-56}}$

Achando a área do triângulo ABC:

$A=\dfrac{|det~M|}{2}=\dfrac{|-56|}{2}=\dfrac{56}{2}\\\\\\\boxed{\boxed{A=28}}$