✅ Após resolver os cálculos, concluímos que a distância entre os referidos pontos é:
[tex]\Large\displaystyle\text{$\begin{gathered}\boxed{\boxed{\:\:\:\bf D_{AB} = 10\:u\:\:\:}}\end{gathered}$}[/tex]
Sejam os pontos:
[tex]\Large\begin{cases} A(1, 6)\\B(-5, -2)\end{cases}[/tex]
Para calcular a distância entre os respectivos pontos, devemos:
[tex]\Large\displaystyle\text{$\begin{gathered} D_{AB} = \sqrt{(X_{B} - X_{A})^{2} + (Y_{B} - Y_{A})^{2}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \sqrt{(-5 - 1)^{2} + (-2 - 6)^{2}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \sqrt{(-6)^{2} + (-8)^{2}}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \sqrt{36 + 64}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = \sqrt{100}\end{gathered}$}[/tex]
[tex]\Large\displaystyle\text{$\begin{gathered} = 10\end{gathered}$}[/tex]
✅ Portanto, a distância procurada é:
[tex]\Large\displaystyle\text{$\begin{gathered} D_{AB} = 10\:u\end{gathered}$}[/tex]
Saiba mais:
Veja a solução gráfica da questão representada na figura:
[tex]\large\boxed{\begin{array}{l}\sf d_{AB}=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}\\\sf d_{AB}=\sqrt{(-5-1)^2+(-2-6)^2}\\\sf d_{AB}=\sqrt{(-6)^2+(-8)^2}\\\sf d_{AB}=\sqrt{36+64}\\\sf d_{AB}=\sqrt{100}\\\sf d_{AB}=10\end{array}}[/tex]
