Warren Turner
04/29/2023 · Middle School
77) On se place dans un repère orthonormé, dans chaque cas, calculer la distance \( A B \) \( \begin{array}{ll}\text { a. } A(1 ; 5) \text { et } B(4 ; 3) & \text { b. } A(-2 ; 3) \text { et } B(-1 ; 1) \\ \text { c. } A(2 ;-3) \text { et } B(6 ; 0) & \text { d. } A\left(\frac{1}{3} ; 2\right) \text { et } B\left(-1 ; \frac{4}{7}\right)\end{array} \)
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Step-by-step Solution
Pour calculer la distance \( AB \) entre deux points \( A(x_1, y_1) \) et \( B(x_2, y_2) \) dans un repère orthonormé, on utilise la formule de la distance :
\[
AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Calculons la distance pour chaque cas :
### a. \( A(1 ; 5) \) et \( B(4 ; 3) \)
\[
AB = \sqrt{(4 - 1)^2 + (3 - 5)^2} = \sqrt{(3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}
\]
### b. \( A(-2 ; 3) \) et \( B(-1 ; 1) \)
\[
AB = \sqrt{(-1 - (-2))^2 + (1 - 3)^2} = \sqrt{(1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}
\]
### c. \( A(2 ; -3) \) et \( B(6 ; 0) \)
\[
AB = \sqrt{(6 - 2)^2 + (0 - (-3))^2} = \sqrt{(4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5
\]
### d. \( A\left(\frac{1}{3} ; 2\right) \) et \( B\left(-1 ; \frac{4}{7}\right) \)
\[
AB = \sqrt{\left(-1 - \frac{1}{3}\right)^2 + \left(\frac{4}{7} - 2\right)^2}
\]
Calculons chaque terme :
1. \( -1 - \frac{1}{3} = -\frac{3}{3} - \frac{1}{3} = -\frac{4}{3} \)
2. \( \frac{4}{7} - 2 = \frac{4}{7} - \frac{14}{7} = -\frac{10}{7} \)
Maintenant, calculons la distance :
\[
AB = \sqrt{\left(-\frac{4}{3}\right)^2 + \left(-\frac{10}{7}\right)^2} = \sqrt{\frac{16}{9} + \frac{100}{49}}
\]
Pour additionner ces fractions, trouvons un dénominateur commun, qui est \( 441 \) (le produit de \( 9 \) et \( 49 \)) :
\[
\frac{16}{9} = \frac{16 \times 49}{441} = \frac{784}{441}
\]
\[
\frac{100}{49} = \frac{100 \times 9}{441} = \frac{900}{441}
\]
Donc,
\[
AB = \sqrt{\frac{784 + 900}{441}} = \sqrt{\frac{1684}{441}} = \frac{\sqrt{1684}}{21}
\]
Calculons \( \sqrt{1684} \) :
\[
1684 = 4 \times 421 \quad \Rightarrow \quad \sqrt{1684} = 2\sqrt{421}
\]
Finalement,
\[
AB = \frac{2\sqrt{421}}{21}
\]
### Résumé des distances :
- a. \( AB = \sqrt{13} \)
- b. \( AB = \sqrt{5} \)
- c. \( AB = 5 \)
- d. \( AB = \frac{2\sqrt{421}}{21} \)
Quick Answer
a. \( AB = \sqrt{13} \)
b. \( AB = \sqrt{5} \)
c. \( AB = 5 \)
d. \( AB = \frac{2\sqrt{421}}{21} \)
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