Question

1°) Calcule a distância entre os pontos dos dados a seguir :

a) A ( 3 , 7 ) e B ( 1 , 4 )

b) C ( 3 , -1 ) e D ( 3 , 5 )

c) D ( -2 , -5 ) e E ( 0 , 0 )

d) E ( 0 , -2 ) e F ( 3 , 4 )

e) F ( 3 , 3 ) e G ( -3 , 3 )

f) G ( - 4 , 0 ) e H ( 0 , 3 )

g) H ( -1 , 2 ) e I ( 2 , 6 )

h) I ( -3 , 5 ) e J ( 3 , 13 )​

Answer 1

Explicação passo-a-passo:

a) A ( 3 , 7 ) e B ( 1 , 4 )

D=√(3-1)²+(7-4)²

D=√(2)²+(3)²

D=√4+9

D=√13

b) C ( 3 , -1 ) e D ( 3 , 5 )

D=√(3-3)²+(-1-5)²

D=√(0)²+(-6)²

D=√0+36

D=√36

D=6

c) D ( -2 , -5 ) e E ( 0 , 0 )

D=√(-2-0)²+(-5-0)²

D=√(-2)²+(-5)²

D=√4+25

D=√29

d) E ( 0 , -2 ) e F ( 3 , 4 )

D=√(0-3)²+(-2-4)²

D=√(-3)²+(-6)²

D=√9+36

D=√45

D=√9.5

D=3√5

e) F ( 3 , 3 ) e G ( -3 , 3 )

D=√(3+3)²+(3-3)2

D=√(6)²+(0)²

D=√36+0

D=√36

D=6

f) G ( - 4 , 0 ) e H ( 0 , 3 )

D=√(-4-0)²+(0-3)²

D=√(-4)²+(-3)²

D=√16+9

D=√25

D=5

g) H ( -1 , 2 ) e I ( 2 , 6 )

D=√(-1-2)²+(2-6)²

D=√(-3)²+(-4)²

D=√9+16

D=√25

D=5

h) I ( -3 , 5 ) e J ( 3 , 13 )

D=√(-3-3)²+(5-13)²

D=√(-6)²+(-8)²

D=√36+64

D=√100

D=10

Answer 2

a) A(3,7) B(1,4)

$\Delta\:x=x_{A}-x_{B}=3-1=2$

$\Delta\:y=y_{7}-y_{4}=3$

$D_{A,B}=\sqrt{{(\Delta\:x)}^{2}+{(\Delta\:y)}^{2}}$

$D_{A,B}=\sqrt{{2}^{2}+{3}^{2}}$

$D_{A,B}=\sqrt{4+9}$

$D_{A,B}=\sqrt{13}$

b) C(3,-1) D(3,5)

$\Delta\:x=x_{C}-x_{D}=3-3=0$

$\Delta\:y=y_{C}-y_{D}=-1-5=-6$

$D_{C,D}=\sqrt{{0}^{2}+{(-6)}^{2}}$

$D_{C,D}=\sqrt{0+36}$

$D_{C,D}=\sqrt{36}$

$D_{C,D}=6$

c)D(-2,-5) E(0,0)

$\Delta\:x=x_{E}-x_{D}=0-(-2)=0+2=2$

$\Delta\:y=y_{E}-y_{D}=0-(-5)=0+5=5$

$D_{E,D}=\sqrt{{\Delta\:x}^{2}+{\Delta\:y}^{2}}$

$D_{E,D}=\sqrt{{2}^{2}+{5}^{2}}$

$D_{E,D}=\sqrt{4+25}$

$D_{E,D}=\sqrt{29}$

d) E(0,-2) F(3,4)

$\Delta\:x=x_{F}-x_{E}=3-0=3$

$\Delta\:y=y_{F}-y_{E}=4-(-2)=4+2=6$

$D_{F,E}=\sqrt{{\Delta\:x}^{2}+{\Delta\:y}^{2}}$

$D_{F,E}=\sqrt{{3}^{2}+{6}^{2}}$

$D_{F,E}=\sqrt{9+36}\\D_{F, E} =\sqrt{45}=3\sqrt{5}$

e) F(3,3) G(-3,3)

$\Delta\:x=x_{F}-x_{G}=3-(-3)=3+3=6$

$\Delta\:y=y_{F}-y_{G}=3-3=0$

$D_{F,G}=\sqrt{{\Delta\:x}^{2}+{\Delta\:y}^{2}}$

$D_{F,G}=\sqrt{{6}^{2}+{0}^{2}}=\sqrt{36}=6$

f) G(-4,0) H(0,3)

$\Delta\:x=x_{H}-x_{G}=0-(-4)=4$

$\Delta\:y=y_{H}-y_{G}=3-0=3$

$D_{H,G}=\sqrt{{\Delta\:x}^{2}+{\Delta\:y}^{2}}$

$D_{H,G}=\sqrt{{4}^{2}+{3}^{2}}$

$D_{H,G}=\sqrt{16+9}=\sqrt{25}=5$

g) H(-1,2) I(2,6)

$\Delta\:x=x_{I}-x_{H}=2-(-1)=3$

$\Delta\:y=y_{I}-y_{H}=6-2=4$

$D_{I, H}=\sqrt{{\Delta\:x}^{2}+{\Delta\:y}^{2}}$

$D_{I,H}=\sqrt{{3}^{2}+{4}^{2}}$

$D_{I,H}=\sqrt{9+16}=\sqrt{25}=5$

h) I(-3,5) J(3,13)

$\Delta\:x=x_{J}-x_{I}=3-(-3)=6$

$\Delta\:y=y_{J}-y_{I}=13-5=8$

$D_{J, I}=\sqrt{{6}^{2}+{8}^{2}}$

$D_{I, J}=\sqrt{36+64}=\sqrt{100}=10$

$\bf{\it{Bons\:Estudos!}}$