Caucule o valor de t sabendo os pontos A (8, t) B (3,0) e C (-1,6)
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Olá !
Eis aqui a resolução :
![\left[\begin{array}{ccc}8&t&1\\3&0&1\\-1&6&1\end{array}\right\left]\begin{array}{ccc}8&t&\\3&0&\\-1&6&\end{array}\right \left[\begin{array}{ccc}8&t&1\\3&0&1\\-1&6&1\end{array}\right\left]\begin{array}{ccc}8&t&\\3&0&\\-1&6&\end{array}\right](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D8%26amp%3Bt%26amp%3B1%5C%5C3%26amp%3B0%26amp%3B1%5C%5C-1%26amp%3B6%26amp%3B1%5Cend%7Barray%7D%5Cright%5Cleft%5D%5Cbegin%7Barray%7D%7Bccc%7D8%26amp%3Bt%26amp%3B%5C%5C3%26amp%3B0%26amp%3B%5C%5C-1%26amp%3B6%26amp%3B%5Cend%7Barray%7D%5Cright+)

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Eis aqui a resolução :
Resposta :
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