Question

Dados os pontos:

A (1,3)
B (5,3)
C (1,7)
D (0,3)
E (-4,1)
F (-6,7)
G (-4, 1)
H (1, -5)
I (-2,-7)


Calcule a distância entre:

AB
BC
CA
DE
EF
FD
GH
HI
IG

Answer 1

Olá !

Distância do segmento $\mathsf{\overline{AB}}$

$\mathsf{D_{\overline{AB}}=\sqrt{\left(X_{B} - X_{A})^{2} + (Y_{B} - Y_{A}\right)^{2}}} \\\\\\ \mathsf{D_{\overline{AB}}=\sqrt{\left(5 - 1)^{2} + (3 - 3\right)^{2}}} \\\\\\ \mathsf{D_{\overline{AB}}=\sqrt{(4)^{2} + (0)^{2}}} \\\\\\ \mathsf{D_{\overline{AB}}=\sqrt{(16 + 0)}} \\\\\\ \mathsf{D_{\overline{AB}}=\sqrt{16}} \\\\\\ \mathsf{D_{\overline{AB}}=4~~U.C}$

Portanto , a distância de $\mathsf{\overline{AB}}$ será 4 U.C

Distância do segmento $\mathsf{\overline{BC}}$

$\mathsf{D_{\overline{BC}}=\sqrt{\left(X_{C} - X_{B})^{2} + (Y_{C} - Y_{B}\right)^{2}}} \\\\\\ \mathsf{D_{\overline{BC}}=\sqrt{\left(1 - 5)^{2} + (7 - 3\right)^{2}}} \\\\\\ \mathsf{D_{\overline{BC}}=\sqrt{(-4)^{2} + (4)^{2}}} \\\\\\ \mathsf{D_{\overline{BC}}=\sqrt{(16 + 16)}} \\\\\\ \mathsf{D_{\overline{BC}}=\sqrt{32}} \\\\\\ \mathsf{D_{\overline{BC}}=4\sqrt{2}~~U.C}$

Portanto , a distância do segmento $\mathsf{\overline{BC}}$ será$\mathsf{4\sqrt{2}~~U.C}$

Distância do segmento $\mathsf{\overline{CA}}$ ou $\mathsf{\overline{AC}}$

$\mathsf{D_{\overline{CA}}=\sqrt{\left(X_{C} - X_{A})^{2} + (Y_{C} - Y_{A}\right)^{2}}} \\\\\\ \mathsf{D_{\overline{CA}}=\sqrt{\left(1 - 1)^{2} + (7 - 3\right)^{2}}} \\\\\\ \mathsf{D_{\overline{CA}}=\sqrt{(0)^{2} + (4)^{2}}} \\\\\\ \mathsf{D_{\overline{CA}}=\sqrt{(0^{2} + 16)}} \\\\\\ \mathsf{D_{\overline{CA}}=\sqrt{16}} \\\\\\ \mathsf{D_{\overline{CA}}=4~~U.C}$

Portanto , a distância do segmento $\mathsf{\overline{CA}}$ será $\mathsf{4~~U.C}$

Distância do segmento $\mathsf{\overline{DE}}$

$\mathsf{D_{\overline{DE}}=\sqrt{\left(X_{E} - X_{D})^{2} + (Y_{E} - Y_{D}\right)^{2}}} \\\\\\ \mathsf{D_{\overline{DE}}=\sqrt{\left(-4 - 0)^{2} + (1 - 3\right)^{2}}} \\\\\\ \mathsf{D_{\overline{DE}}=\sqrt{(-4)^{2} + (-2)^{2}}} \\\\\\ \mathsf{D_{\overline{DE}}=\sqrt{(16 + 4)}} \\\\\\ \mathsf{D_{\overline{DE}}=\sqrt{20}} \\\\\\ \mathsf{D_{\overline{DE}}=2\sqrt{5}~~U.C}$

Portanto , a distância do segmento. $\mathsf{\overline{DE}}$ $\mathsf{2\sqrt{5}~~U.C}$

Distância do segmento $\mathsf{\overline{EF}}$

$\mathsf{D_{\overline{EF}}=\sqrt{\left(X_{B} - X_{A})^{2} + (Y_{B} - Y_{A}\right)^{2}}} \\\\\\ \mathsf{D_{\overline{EF}}=\sqrt{\left((-6) - (-4))^{2} + (7 - 1\right)^{2}}} \\\\\\ \mathsf{D_{\overline{EF}}=\sqrt{(-2)^{2} + (6)^{2}}} \\\\\\ \mathsf{D_{\overline{EF}}=\sqrt{(4 + 36)}} \\\\\\ \mathsf{D_{\overline{EF}}=\sqrt{40}} \\\\\\ \mathsf{D_{\overline{EF}}=2\sqrt{10}~~U.C}$

Portanto , a distância do segmento $\mathsf{\overline{EF}}$ será $\mathsf{2\sqrt{10}}$

Distância do segmento $\mathsf{\overline{FD}}$ ou $\mathsf{\overline{DF}}$

$\mathsf{D_{\overline{DF}}=\sqrt{\left(X_{F} - X_{D})^{2} + (Y_{F} - Y_{D}\right)^{2}}} \\\\\\ \mathsf{D_{\overline{DF}}=\sqrt{\left(-6 - 0)^{2} + (7 - 3\right)^{2}}} \\\\\\ \mathsf{D_{\overline{DF}}=\sqrt{(-6)^{2} + (4)^{2}}} \\\\\\ \mathsf{D_{\overline{DF}}=\sqrt{(36 + 16)}} \\\\\\ \mathsf{D_{\overline{DF}}=\sqrt{52}} \\\\\\ \mathsf{D_{\overline{DF}}=2\sqrt{13}}$

Portanto , a distância do segmento $\mathsf{\overline{DF}}$ $\mathsf{2\sqrt{13}}$

Distância do segmento $\mathsf{\overline{GH}}$

$\mathsf{D_{\overline{GH}}=\sqrt{\left(X_{H} - X_{G})^{2} + (Y_{H} - Y_{G}\right)^{2}}} \\\\\\ \mathsf{D_{\overline{GH}}=\sqrt{\left(1 - (-4)^{2} + (-5 - 1\right)^{2}}} \\\\\\ \mathsf{D_{\overline{GH}}=\sqrt{(5)^{2} + (-6)^{2}}} \\\\\\ \mathsf{D_{\overline{GH}}=\sqrt{(25 + 36)}} \\\\\\ \mathsf{D_{\overline{GH}}=\sqrt{61}~~U.C}$

Portanto , a distância do segmento $\mathsf{\overline{GH}}$ será $\mathsf{\sqrt{61}~~U.C}$

Distância do segmento $\mathsf{\overline{HI}}$

$\mathsf{D_{\overline{HI}}=\sqrt{\left(X_{I} - X_{H})^{2} + (Y_{I} - Y_{H}\right)^{2}}} \\\\\\ \mathsf{D_{\overline{HI}}=\sqrt{\left(-2 - 1)^{2} + (-7 - (-5)\right)^{2}}} \\\\\\ \mathsf{D_{\overline{HI}}=\sqrt{(-3)^{2} + (-2)^{2}}} \\\\\\ \mathsf{D_{\overline{HI}}=\sqrt{9 + 4}} \\\\\\ \mathsf{D_{\overline{HI}}=\sqrt{13}~~U.C}$

Portanto , a distância do segmento $\mathsf{\overline{HI}}$ será $\mathsf{\sqrt{13}~~U.C}$

Distância do segmento $\mathsf{\overline{IG}}$ ou $\mathsf{\overline{GI}}$

$\mathsf{D_{\overline{GI}}=\sqrt{\left(X_{I} - X_{G})^{2} + (Y_{G} - Y_{I}\right)^{2}}} \\\\\\ \mathsf{D_{\overline{GI}}=\sqrt{\left(-2 - (-4)^{2} + (-7 - (-5)\right)^{2}}} \\\\\\ \mathsf{D_{\overline{GI}}=\sqrt{(2)^{2} + (-2)^{2}}} \\\\\\ \mathsf{D_{\overline{GI}}=\sqrt{(4 + 4)}} \\\\\\ \mathsf{D_{\overline{GI}}=\sqrt{8}~U.C} \\\\\\ \mathsf{D_{\overline{GI}}=2\sqrt{2}}$

Portanto , a distância do segmento $\mathsf{\overline{GI}}$ será $\mathsf{2\sqrt{2}~U.C}$

Answer 2

A (1,3)
B (5,3)
C (1,7)
D (0,3)
E (-4,1)
F (-6,7)
G (-4, 1)
H (1, -5)
I (-2,-7)

Calcule a distância entre:

AB


Dist=√(∆x)^2+(∆y)^2=√(5-1)^2+(3-3)^2

Dist=√(4)^2+(0)^2

Dist=√16

Dist=4



BC

Dist=√(∆x)^2+(∆y)^2

Dist=√(5-1)^2+(7-3)^2

Dist=√(4)^2+(4)^2

dist=√16+16

Dist=√2.(16)

Dist=4√2


CA

Dist=√(∆x)^2+(∆y)^2

Dist=√(1-1)^2+(7-3)^2

Dist=√(0)^2+(4)^2

Dist=√0+16

dist=√16

Dist=4


DE

Dist=√(∆x)^2+(∆y)^2

Dist=√(-4-0)^2+(3-1)^2

Dist=√(-4)^2+(2)^2

Dist=√16+4

dia=√4.(5)

Dist=2√5


EF

E (-4,1)
F (-6,7)

Dist=√(∆x)^2+(∆y)^2

Dist=√(-6+4)^2+(7-1)^2

Dist=√(-2)^2+(6)^2

Dist=√4+36

Dist=2√10


FD

D (0,3)

F (-6,7)

Dist=√(∆x)^2+(∆y)^2

Dist=√(-6-0)^2+(7-3)^2

Dist=√(-6)^2+(4)^2

Dist=√36+16

Dist=√52



GH
G (-4, 1)
H (1, -5)


Dist=√(∆x)^2+(∆y)^2

Dist=√(-4-1)^2+(-5-1)^2

Dist=√(25)+(36)

Dist=√61


HI


H (1, -5)
I (-2,-7)


Dist=√(∆x)^2+(∆y)^2

Dist=√(-2-1)^2+(-7+5)^2

Dist=√(-3)^2+(-2)^2

Dist=√9+4

Dist=√13


IG

G (-4, 1)
I (-2,-7)

Dist=√(∆x)^2+(∆y)^2

Dist=√(-4+2)^2+(-7-1)^2

Dist=√(-2)^2+(-8)^2

Dist=√4+64

Dist=√70

espero ter ajudado!

boa tarde!