✅ Após resolver os cálculos, concluímos que o valor da incógnita "x" é:
[tex]\Large\displaystyle\text{$\begin{gathered}\boxed{\boxed{\:\:\:\bf x = \frac{5}{4}\:\:\:}}\end{gathered}$}[/tex]
Sejam os pontos:
[tex]\Large\begin{cases} A = (2, -3)\\B = (x, 3)\\C = (1, 5)\end{cases}[/tex]
Para que os referidos pontos sejam colineares, isto é, alinhados em uma linha reta, devemos ter o determinante da matriz "M", formada pelas coordenadas dos pontos igual a "0".
Se a matriz "M" é:
[tex]\Large\displaystyle\text{$\begin{gathered}M = \begin{pmatrix} 2 & -3 & 1\\x & 3 & 1\\1 & 5 & 1\end{pmatrix}\end{gathered}$}[/tex]
Então, temos:
[tex]\Large\displaystyle\text{$\begin{gathered} \det M = 0\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered}\begin{vmatrix} 2 & -3 & 1 \\x & 3 & 1\\1 & 5 & 1\end{vmatrix}\begin{matrix}2 & -3\\x & 3\\ 1 & 5\end{matrix} = 0\end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered} 2\cdot3\cdot1 + (-3)\cdot1\cdot1 + 1\cdot x\cdot 5 - (-3)\cdot x\cdot 1 - 2\cdot1\cdot5 - 1\cdot3\cdot1 = 0\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered}6 - 3 + 5x + 3x - 10 - 3 = 0 \end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} 5x + 3x = -6 + 3 + 10 + 3\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} 8x = 10\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} x = \frac{10}{8}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} x = \frac{5}{4}\end{gathered}$}[/tex]
✅ Portanto, o valor de "x" é:
[tex]\Large\displaystyle\text{$\begin{gathered} x = \frac{5}{4}\end{gathered}$}[/tex]
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