• Matéria: Matemática
  • Autor: hta6almeida
  • Perguntado 4 anos atrás

02 - Determine a equação reduzida da reta que passa pelos pontos A(5, -2) e B(4, 2).

Respostas

respondido por: Math739
1

Primeira vamos encontrar o coeficiente angular da reta:

\Large\displaystyle\text{$\begin{gathered} \sf{ m=\frac{(y_2-y_1)}{(x_2-x_1)}} \end{gathered}$}

\Large\displaystyle\text{$\begin{gathered} \sf{m=\dfrac{(2-(-2))}{(4-5)} } \end{gathered}$}

\Large\displaystyle\text{$\begin{gathered} \sf{m=\dfrac{4}{-1} } \end{gathered}$}

\Large\displaystyle\text{$\begin{gathered} \sf{m=-4 } \end{gathered}$}

\Large\displaystyle\text{$\begin{gathered} \sf{ (y-y_1)=-4(x-x_1)} \end{gathered}$}

\Large\displaystyle\text{$\begin{gathered} \sf{y-2=-4(x-4) } \end{gathered}$}

\Large\displaystyle\text{$\begin{gathered} \sf{y-2=-4x+16 } \end{gathered}$}

\Large\displaystyle\text{$\begin{gathered} \sf{y=-4x+16+2 } \end{gathered}$}

\Large\displaystyle\text{$\begin{gathered} \sf{y=-4x+18 } \end{gathered}$}

Portanto, a equação reduzida da reta é:

\Large\displaystyle\text{$\begin{gathered} \boxed{\boxed{\bf{y=-4x+18 }}} \end{gathered}$}

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