Question

A área do triângulo de vertical A(-1,2),B(2,0) e C(-1,-2)

Answer 1

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

$\Large\boxed{\begin{array}{l}\sf A(-1,2),B(2,0),C(-1,-2)\\\sf M=\begin{vmatrix}\sf-1&\sf2&\sf1\\\sf2&\sf0&\sf1\\\sf-1&\sf-2&\sf1\end{vmatrix}\\\sf det\,M=-1(0+2)-2\cdot(2+1)+1\cdot(-4)\\\sf det~M=-2-6-4\\\sf det~M=-12\\\sf A=\dfrac{|det~m|}{2}\\\\\sf A=\dfrac{|-12|}{2}\\\\\sf A=\dfrac{12}{2}\\\\\sf A=6~u\cdot a \end{array}}$

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