Question

um triângulo tem vértices p = (2, 1), q = (2, 5) e r = (x0, 4), com x0 > 0. sabendo-se que a área do triângulo é 20, a abscissa x0 do ponto r é:

Answer 1

✅ Após resolver os cálculos, concluímos que a abscissa do ponto "R" é:

            $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf x_{o} = 12\:\:\:}}\end{gathered} }$

Sejam os dados:

                  $\Large\begin{cases} P(2, 1)\\Q(2, 5)\\R(x_{o},4)\\S = 20\,u.\,a.\end{cases}$

Sabemos, analiticamente, que a área de um triângulo a partir de seus vértices é igual à metade do módulo do determinante da matriz "M" formado pelos vértices do referido triângulo.

Desta forma temos:

    $\Large\displaystyle\begin{gathered} S = \frac{|\det M|}{2}\end{gathered}$

Então, temos:

    $\Large\displaystyle\text{ \begin{gathered} 20 = \frac{\left|\begin{vmatrix} 2 & 1 & 1\\2 & 5 & 1\\x_{o} & 4 & 1\end{vmatrix}\right|}{2}\end{gathered} }$

     $\Large\displaystyle\text{ \begin{gathered} 20 = \frac{\left|\begin{vmatrix} 5 & 1\\4 & 1\end{vmatrix}\cdot2 - \begin{vmatrix} 2 & 1\\x_{o} & 1\end{vmatrix}\cdot1 + \begin{vmatrix} 2 & 5\\x_{o} & 4\end{vmatrix}\cdot1\right|}{2}\end{gathered} }$

    $\Large\displaystyle\begin{gathered} 40 = |(5 - 4)\cdot2 - (2 - x_{o})\cdot1 + (8 - 5x_{o})\cdot1|\end{gathered}$

     $\Large\displaystyle\begin{gathered} 40 = |1\cdot2 - (2 - x_{o}) + (8 - 5x_{o})|\end{gathered}$

     $\Large\displaystyle\begin{gathered} 40 = |2 - 2 + x_{o} + 8 - 5x_{o}|\end{gathered}$

     $\Large\displaystyle\begin{gathered} 40 = |- 4x_{o} + 8|\end{gathered}$

Invertendo os membros sem perda alguma de generalidades, temos:

    $\Large\displaystyle\begin{gathered} |-4x_{o} + 8| = 40\end{gathered}$

    $\Large\displaystyle\begin{gathered} (-4x_{o} + 8)^{2} = 40^{2}\end{gathered}$

 $\Large\displaystyle\begin{gathered} 16x_{o}^{2} - 64x_{o} + 64 = 1600\end{gathered}$

$\Large\displaystyle\begin{gathered} 16x_{o}^{2} - 64x_{o} + 64 - 1600 = 0\end{gathered}$

$\Large\displaystyle\begin{gathered} 16x_{o}^{2} - 64x_{o} - 1536 = 0\end{gathered}$

$\Large\displaystyle\text{ \begin{gathered} \frac{16x_{o}^{2}}{16} - \frac{64x_{o}}{16} - \frac{1536}{16} = \frac{0}{16}\end{gathered} }$

$\Large\displaystyle\begin{gathered} x_{o}^{2} - 4x_{o} - 96 = 0\end{gathered}$

Aplicando a fórmula resolutiva da equação do segundo grau, temos:

$\Large\displaystyle\text{ \begin{gathered} x_{o} = \frac{-(-4) \pm\sqrt{(-4)^{2} - 4\cdot1\cdot(-96)}}{2\cdot1}\end{gathered} }$

      $\Large\displaystyle\text{ \begin{gathered} = \frac{4 \pm\sqrt{16 + 384}}{2}\end{gathered} }$

      $\Large\displaystyle\text{ \begin{gathered} = \frac{4\pm\sqrt{400}}{2}\end{gathered} }$

      $\Large\displaystyle\begin{gathered} = \frac{4\pm20}{2}\end{gathered}$

Encontrando as raízes, temos:

     $\LARGE\begin{cases} x_{o}' = \frac{4 - 20}{2} = \frac{-16}{2} = -8\\x_{o}'' = \frac{4 + 20}{2} = \frac{24}{2} = 12\end{cases}$

Como a questão  nos diz que:

                     $\Large\displaystyle\begin{gathered} x_{o} > 0\end{gathered}$

✅ Então, o resultado do nosso x0 é:

                    $\Large\displaystyle\begin{gathered} x_{o} = 12\end{gathered}$

Desta forma, os vértices do triângulo são:

                    $\Large\begin{cases} P(2, 1)\\Q(2, 5)\\R(12, 4)\end{cases}$

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

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