Question

Determine a área do triangulo retangulo ABC, cujos vértices são A(-1,2), B(-1,-2) e C(6,-2).

Answer 1

✅Tendo terminado os cálculos, concluímos que a área do referido triângulo é:

          $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf S_{\triangle} = 14\,u^{2}\:\:\:}}\end{gathered} }$

Sejam os vértices do triângulo:

               $\Large\begin{cases} A(-1, 2)\\B(-1, -2)\\C(6, -2)\end{cases}$

Como temos apenas os vértices do triângulo então podemos aplicar conceitos de geometria analítica para calcular sua área. Para isso, basta calcular a metade do módulo do determinante da matriz "M".

Se a matriz "M" é:

     $\Large\displaystyle\begin{gathered}M = \begin{bmatrix} -1 & 2 & 1\\ -1 & -2 & 1\\6 & -2 & 1\end{bmatrix}\end{gathered}$

Então, temos:

     $\Large\displaystyle\begin{gathered} S_{\triangle} = |\det M|/ 2\end{gathered}$

              $\Large\displaystyle\begin{gathered} = \left|\begin{vmatrix}-1 & 2 & 1\\-1 & -2 & 1\\6 & -2 & 1 \end{vmatrix}\right| / 2\end{gathered}$

              $\Large\displaystyle\begin{gathered} = \left|\begin{vmatrix} -2 & 1\\-2 & 1\end{vmatrix}\cdot(-1) - \begin{vmatrix}-1 & 1\\6 & 1 \end{vmatrix}\cdot2 + \begin{vmatrix}-1 & -2\\6 & -2 \end{vmatrix}\cdot1\right| / 2\end{gathered}$

              $\large\displaystyle\begin{gathered} = \left|(-2 + 2)\cdot(-1) - (-1 - 6)\cdot2 + (2 + 12)\cdot1\right| / 2\end{gathered}$

              $\Large\displaystyle\begin{gathered} = |0\cdot(-1) - (-7)\cdot2 + 14\cdot1| / 2\end{gathered}$

              $\Large\displaystyle\begin{gathered} = |0 + 14 + 14| / 2\end{gathered}$

              $\Large\displaystyle\begin{gathered} = |28| / 2\end{gathered}$

              $\Large\displaystyle\begin{gathered} = 28 / 2\end{gathered}$

              $\Large\displaystyle\begin{gathered} = 14\end{gathered}$

Portanto, a área do triângulo é:

     $\Large\displaystyle\begin{gathered} S_{\triangle} = 14\,u^{2}\end{gathered}$

$\LARGE\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Bons \:estudos!!\:\:\:Boa\: sorte!!\:\:\:}}}\end{gathered} }$

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$\Large\displaystyle\text{ \begin{gathered} \underline{\boxed{\boldsymbol{\:\:\:Observe \:o\:Gr\acute{a}fico!!\:\:\:}}}\end{gathered} }$