Question

Calcule a distância entre os pontos A e B nos seguintes casos:

a) A (3, 7) B (1, 4)
b) A (3, -1) B (3, 5)

Answer 1

a)

$\mathsf{{(x_{B}-x_{A})}^{2}={(1-3)}^{2}={(-2)}^{2}=4}\\\mathsf{{(y_{B}-y_{A})}^{2}={(4-7)}^{2}={(-3)}^{2}=9}$

$\mathsf{d_{A,B}=\sqrt{{(x_{B}-x_{A})}^{2}+{(y_{B}-y_{A})}^{2}}}\\\mathsf{d_{A,B}=\sqrt{4+9}}\\\huge\boxed{\boxed{\mathsf{d_{A,B}=\sqrt{13}}}}$

b)

$\mathsf{{(x_{B}-x_{A})}^{2}={(3-3)}^{2}={(0)}^{2}=0}\\\mathsf{{(y_{B}-y_{A})}^{2}={(5-[-1])}^{2}={(5+1)}^{2}={6}^{2}=36}$

$\mathsf{d_{A,B}=\sqrt{{(x_{B}-x_{A})}^{2}+{(y_{B}-y_{A})}^{2}}}\\\mathsf{d_{A,B}=\sqrt{0+36}}\\\huge\boxed{\boxed{\mathsf{d_{A,B}=\sqrt{36}=6}}}$

Answer 2

Explicação passo-a-passo:

A distância entre dois pontos $\sf A(x_A,y_A)~e~B(x_B,y_B)$ é dada por:

$\sf d_{AB}=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}$

a)

$\sf d_{AB}=\sqrt{(3-1)^2+(7-4)^2}$

$\sf d_{AB}=\sqrt{2^2+3^2}$

$\sf d_{AB}=\sqrt{4+9}$

$\sf \red{d_{AB}=\sqrt{13}}$

b)

$\sf d_{AB}=\sqrt{(3-3)^2+(-1-5)^2}$

$\sf d_{AB}=\sqrt{0^2+(-6)^2}$

$\sf d_{AB}=\sqrt{0+36}$

$\sf d_{AB}=\sqrt{36}$

$\sf \red{d_{AB}=6}$