Question

Calcule a distância entre os pontos dados:
a) A (-3,-5) e B (-2,-4)
b) C (0, - 2) e D ( 3/2, 2/3)
c) E (1/2, 1/3) e F (1/3, 12)

Answer 1

Resposta:

$a) \sqrt{2}$

$b) 3$

$c)\sqrt{\frac{4901}{36} }$

Explicação passo-a-passo:

$Dab = \sqrt{(x'' - x')^2 + (y'' - y')^2}$

$Dab = \sqrt{(-2 - (-3))^2 + ((-4) - (-5))^2}$

$Dab = \sqrt{(-2 + 3)^2 + (-4 + 5)^2}$

$Dab = \sqrt{1^2 + 1^2}$

$Dab = \sqrt{2}$

$Dab = \sqrt{(\frac{3}{2} - 0)^2 + (\frac{2}{3} - (-2))^2}$

$Dab = \sqrt{(\frac{3}{2})^2 + (\frac{2}{3} + 2)^2}$

$Dab = \sqrt{(\frac{3}{2})^2 + (\frac{8}{3})^2}$

$Dab = \sqrt{\frac{9}{4} + \frac{64}{9}}$

$Dab = \sqrt{\frac{81 + 256}{36}}$

$Dab = \sqrt{\frac{324}{36}}$

$Dab = 3$

$Dab = \sqrt{(\frac{1}{3} - \frac{1}{2})^2 + (12 - \frac{1}{3})^2}$

$Dab = \sqrt{(\frac{-1}{6})^2 + (\frac{35}{3})^2}$

$Dab = \sqrt{\frac{1}{36} + \frac{1225}{9}}$

$Dab = \sqrt{\frac{4901}{36}$