Question

determine a distancia entre os pontos indicados: A) A (5, 6) e B (6, 7)
B) A (3, 6) e B (-3, -5)
C) A (-1, -2) e B (-4, -9)
D) A (0, 1) e B (0, 3)
E) A (-1, -2) e B (1/2, 2)
F) A (1/2, 3/4) e B (3/8, 1/6)

Answer 1

Consideramos os pontos A(Xa,Ya) e B(Xb,Yb), a distância entre dois pontos é a hipotenusa do triângulo retângulo, que pode ser calculada aplicando o Teorema de Pitágoras (“o quadrado da hipotenusa é igual à soma dos quadrados dos catetos”):

$D^2=(x_b-x_a)^2+(y_b-y_a)^2\\\\D=\sqrt{(x_b-x_a)^2+(y_b-y_a)^2}$

Substituímos então os valores:

A) A (5, 6) e B (6, 7)

$D=\sqrt{(6-5)^2+(7-6)^2}\\\\D=\sqrt{(1)^2+(1)^2}\\\\D=\sqrt{2} \cong 1,41$

B) A (3, 6) e B (-3, -5)

$D=\sqrt{(-3-3)^2+(-5-6)^2}\\\\D=\sqrt{(-6)^2+(-11)^2}\\\\D=\sqrt{36+121}\\\\D=\sqrt{157} \cong 12,53$

C) A (-1, -2) e B (-4, -9)

$D=\sqrt{(-4(-1))^2+(-9-(-2))^2}\\\\D=\sqrt{(-3)^2+(-7)^2}\\\\D=\sqrt{9+49}\\\\D=\sqrt{58} \cong 7,62$

D) A (0, 1) e B (0, 3)

$D=\sqrt{(0-0)^2+(3-1)^2}\\\\D=\sqrt{2^2}\\\\D=\sqrt{4}\\\\D=2$

E) A (-1, -2) e B (1/2, 2)

$D=\sqrt{(\frac{1}{2}-(-1))^2+(2-(-2))^2}\\\\D=\sqrt{(\frac{3}{2})^2+(4)^2}\\\\D=\sqrt{\frac{9}{4}+16}\\\\D=\sqrt{\frac{73}{4}}\\\\D=\frac{\sqrt{73}}{4} \cong 4,27$

F) A (1/2, 3/4) e B (3/8, 1/6)

$D=\sqrt{(\frac{3}{8}-\frac{1}{2})^2+(\frac{1}{6}-\frac{3}{4})^2}\\\\D=\sqrt{(-\frac{1}{8})^2+(-\frac{7}{12})^2}\\\\D=\sqrt{\frac{1}{64}+\frac{49}{144}}\\\\D=\sqrt{\frac{205}{576}}\\\\D=\frac{\sqrt{205}}{24} \cong 0,60$