Question

o coeficientes angular da reta formada pelos pontos A (1 - 2) e B( 5 , 4)é

Answer 1

✅ Tendo terminado os cálculos, concluímos que o coeficiente angular da referida reta é:

   $\Large\displaystyle\text{ \begin{gathered}\boxed{\boxed{\:\:\:\bf m = \frac{3}{2}\:\:\:}}\end{gathered} }$  

Sejam os pontos:

      $\Large\begin{cases} A(1, -2)\\B(5, 4)\end{cases}$

Pra calcular o coeficiente angular "m" da reta que passa pelos referidos pontos devemos utilizar a seguinte estratégia:

     $\Large\displaystyle\begin{gathered} m = \tan \theta\end{gathered}$

           $\Large\displaystyle\begin{gathered} = \frac{\sin \theta}{\cos \theta}\end{gathered}$

           $\Large\displaystyle\text{ \begin{gathered} = \frac{Y_{B} - Y_{A}}{X_{B} - X_{A}}\end{gathered} }$

           $\Large\displaystyle\begin{gathered} = \frac{4 - (-2)}{5 - 1}\end{gathered}$

           $\Large\displaystyle\begin{gathered} = \frac{4 + 2}{5 - 1}\end{gathered}$

           $\Large\displaystyle\begin{gathered} = \frac{6}{4}\end{gathered}$

           $\Large\displaystyle\begin{gathered} = \frac{3}{2}\end{gathered}$

✅ Portanto, o coeficiente nagular da reta é:

       $\Large\displaystyle\begin{gathered} m = \frac{3}{2}\end{gathered}$

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Veja asolução gráfica representada na figura:

Answer 2

$\sf m=\dfrac{y_2-y_1}{x_2-x_1}$

$\sf m=\dfrac{4-(-2)}{5-1}$

$\sf m=\dfrac{6\div2}{4\div2}$

$\red{\sf m=\dfrac{3}{2}}$