Question

A área do triângulo de vértices A (1,4), B (3,2) e C (– 2, 5) é: *
a) 17
b) 8,5
c) 2
d) - 4
e) - 2

Answer 1

$\Large\boxed{\begin{array}{l}\sf Sejam\,A(x_A,y_A),B(x_B,y_C)\,e\,C(x_C,y_C)\\\sf v\acute ertices\,de\,um\,tri\hat angulo,\\\sf a\,\acute area\,deste\,tri\hat angulo\,\acute e\,dada\,por\\\sf A=\dfrac{|det\,M|}{2}\\\sf onde\\\sf M=\begin{vmatrix}\sf x_A&\sf y_A&\sf1\\\sf x_B&\sf y_B&\sf1\\\sf x_C&\sf y_C&\sf1\end{vmatrix}\end{array}}$

$\Large\boxed{\begin{array}{l}\sf M=\begin{vmatrix}\sf1&\sf4&\sf1\\\sf3&\sf2&\sf1\\\sf-2&\sf5&\sf1\end{vmatrix}\\\sf \\\\\sf det\,M=1\cdot(2-5)-4\cdot(3+2)+1\cdot(15+4)\\\sf det\,M=-3-20+19\\\sf det\,M=-4\\\sf A=\dfrac{|det\,M|}{2}\\\\\sf A=\dfrac{|-4|}{2}\\\\\sf A=\dfrac{4}{2}\\\\\sf A=2\\\huge\boxed{\boxed{\boxed{\boxed{\sf\red{\maltese}~\blue{alternativa~c}}}}} \end{array}}$