Question

Geometria analítica distância entre os pontos: A(1/2, -1/3) e B(5/2, 1/3) A resposta é 2√10 sobre 3. mas, de todas as formas que tentei sempre deu 2√10 sobre 9 me ajudem por favor.

Answer 1

$<var>d = \sqrt{(x_{f} - x_{i})^{2} + (y_{f}-y_{i})^{2}} \\\\ d = \sqrt{(\frac{5}{2} - \frac{1}{2})^{2} + (\frac{1}{3}-(-\frac{1}{3}))^{2}} \\\\ d = \sqrt{(\frac{5}{2} - \frac{1}{2})^{2} + (\frac{1}{3}+\frac{1}{3})^{2}} \\\\ d = \sqrt{(\frac{4}{2})^{2} + (\frac{2}{3})^{2}} \\\\ d = \sqrt{(2)^{2} + (\frac{2}{3})^{2}} \\\\ d = \sqrt{4 + \frac{4}{9}} \\\\ MMC = 9 \\\\ d = \sqrt{\frac{36}{9} + \frac{4}{9}} \\\\ d = \sqrt{\frac{40}{9}} \\\\ d = \frac{\pm \sqrt{40}}{3}</var>$

 

Racionalizando o 40

 

40 | 2

20 | 2

10 | 2

5   | 5

1

 

$<var>d = \frac{\sqrt{2^{2} \cdot 2 \cdot 5}}{3} \\\\ d = \frac{2\sqrt{2 \cdot 5}}{3} \\\\ \boxed{\boxed{d = \frac{2\sqrt{10}}{3}}}</var>$

Answer 2

$<var>d_{AB}=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}</var>$

$<var>d_{AB}=\sqrt{(\frac{5}{2}-\frac{1}{2})^2+(\frac{1}{3}-(-\frac{1}{3}))^2 }</var>$

 

$<var>d_{AB}=\sqrt{(\frac{4}{2})^2+(\frac{2}{3})^2}=\sqrt{4+\frac{4}{9}}=\sqrt\frac{40}{9}}=\frac{2\sqrt{10}}{3}</var>$