Question

1) Calcule em cada caso a distância entre pontos:
a) A(-2,-1) e B(4,2)
b) C(-4,0) e D(-2,0)

​

Answer 1

Resposta:

istância entre os pontos \left(1,\,3 \right )(1,3) e \left(9,\,9 \right )(9,9) é

\begin{lgathered}d_{\left(1,\,3 \right ) \to \left(9,\,9 \right )}=\sqrt{\left(9-1 \right )^{2}+\left(9-3 \right )^{2}}\\ \\ d_{\left(1,\,3 \right ) \to \left(9,\,9 \right )}=\sqrt{\left(8 \right )^{2}+\left(6 \right )^{2}}\\ \\ d_{\left(1,\,3 \right ) \to \left(9,\,9 \right )}=\sqrt{64+36}\\ \\ d_{\left(1,\,3 \right ) \to \left(9,\,9 \right )}=\sqrt{100}\\ \\ \boxed{d_{\left(1,\,3 \right ) \to \left(9,\,9 \right )}=10\text{ u.c.}}\end{lgathered}

d

(1,3)→(9,9)

=

(9−1)

2

+(9−3)

2

d

(1,3)→(9,9)

=

(8)

2

+(6)

2

d

(1,3)→(9,9)

=

64+36

d

(1,3)→(9,9)

=

100

d

(1,3)→(9,9)

=10 u.c.

b) A distância entre os pontos \left(-3,\,1 \right )(−3,1) e \left(5,\,-14 \right )(5,−14) é

\begin{lgathered}d_{\left(-3,\,1 \right ) \to \left(5,\,-14 \right )}=\sqrt{\left(5-\left(-3 \right ) \right )^{2}+\left(-14-1 \right )^{2}}\\ \\ d_{\left(-3,\,1 \right ) \to \left(5,\,-14 \right )}=\sqrt{\left(5+3 \right )^{2}+\left(-14-1 \right )^{2}}\\ \\ d_{\left(-3,\,1 \right ) \to \left(5,\,-14 \right )}=\sqrt{\left(8 \right )^{2}+\left(-15 \right )^{2}}\\ \\ d_{\left(-3,\,1 \right ) \to \left(5,\,-14 \right )}=\sqrt{64+225}\\ \\ d_{\left(-3,\,1 \right ) \to \left(5,\,-14 \right )}=\sqrt{289}\\ \\ \boxed{d_{\left(-3,\,1 \right ) \to \left(5,\,-14 \right )}=17\text{ u.c.}}\end{lgathered}

d

(−3,1)→(5,−14)

=

(5−(−3))

2

+(−14−1)

2

d

(−3,1)→(5,−14)

=

(5+3)

2

+(−14−1)

2

d

(−3,1)→(5,−14)

=

(8)

2

+(−15)

2

d

(−3,1)→(5,−14)

=

64+225

d

(−3,1)→(5,−14)

=

289

d

(−3,1)→(5,−14)

=17 u.c.

Answer 2

Resposta:

a) 3$\sqrt{5}$

b) 2

Explicação passo-a-passo:

Distância entre pontos = $\sqrt{(Bx-Ax)^{2}+ (By-Ay)^{2} }$

a)

 $\sqrt{(4-(-2))^2+(2-(-1))^2 }$

$\sqrt{(6)^2+(3)^2}\\\sqrt{36+9}\\\sqrt{45} = 3\sqrt{5}$

b)

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