Question

Qual a área do triângulo cujos vértices são os pontos

(−4,3), (2, −1) (3,2)?​

Answer 1

$\large\boxed{\begin{array}{l}\sf Seja~A(x_A,y_A),B(x_B,y_B)~e~C(x_C,y_C)\\\sf os~v\acute ertices~de~um~tri\hat angulo~no~plano~cartesiano.\\\sf A~\acute area~deste~tri\hat angulo~\acute e~dada~por\\\sf A=\dfrac{|det~M|}{2}\\\sf onde\\\sf M=\begin{vmatrix}\sf x_A&\sf y_A&\sf1\\\sf x_B&\sf y_B&\sf1\\\sf x_C&\sf y_C&\sf1\end{vmatrix}\end{array}}$

$\large\boxed{\begin{array}{l}\rm M=\begin{vmatrix}\rm -4&\rm3&\rm1\\\rm 2&\rm -1&\rm1\\\rm 3&\rm2&\rm1\end{vmatrix}\\\rm det\,M=-4\cdot(-1-2)-3\cdot(2-3)+1\cdot (4+3)\\\rm det\,M=12+3+7\\\rm det\,M=22\\\rm A=\dfrac{|det\,M|}{2}\\\\\rm A=\dfrac{|22|}{2}\\\\\rm A=\dfrac{22}{2}\\\\\rm A=11~u\bullet a\end{array}}$