• Matéria: Matemática
  • Autor: leonardogerrard
  • Perguntado 9 anos atrás

(3,4) e (2,1) a distância

Respostas

respondido por: LuanaSC8
0
a~(3~;~4)~~~e~~~b~(2~;~1)\\\\\\ d_{ab}= \sqrt{(x_b-x_a)^2+(y_B-y_a)^2}\\\\\\d_{ab}= \sqrt{(2-3)^2+(1-4)^2}\to\\\\  d_{ab}= \sqrt{(-1)^2+(-3)^2}\to\\\\  d_{ab}= \sqrt{1+9}\to\\\\   \boxed{d_{ab}= \sqrt{10~}}
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