Question

Calcule a área do ∆ABC de vértices:
a) A (2, 3), B (5, 4) e C (6, –3)
b) A (0, 0), B (3, 4) e C (2,11)

Answer 1

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

a)

$\sf M=\begin{bmatrix}\sf2&\sf3&\sf1\\\sf5&\sf4&\sf1\\\sf6&\sf-3&\sf1\end{bmatrix}\\\sf det~M=2\cdot(4+3)-3\cdot(5-6)+1\cdot(-15-24)\\\sf det~M=14+3-39=-22\\\sf A=\dfrac{|det~M|}{2}\\\sf A=\dfrac{|-22|}{2}\\\sf A=\dfrac{22}{2}\\\huge\boxed{\boxed{\boxed{\boxed{\sf A=11~u\cdot a}}}}\blue{\checkmark}$

b)

$\sf M=\begin{bmatrix}\sf0&\sf0&\sf1\\\sf3&\sf4&\sf1\\\sf2&\sf11&\sf1\end{bmatrix}\\\sf M=0\cdot(4-11)-0\cdot(3-2)+1(33-8)\\\sf det~M=0+0+25=25\\\sf A=\dfrac{|det~M|}{2}\\\sf A=\dfrac{|25|}{2}\\\huge\boxed{\boxed{\boxed{\boxed{\sf A=\dfrac{25}{2}~u\cdot a}}}}\blue{\checkmark}$