Question

AJUDA AQUI PFV 01. DETERMINE A DISTANCIA ENTRE OS PONTOS A = (3 , 5) E B = (1 , 2).

02.DETERMINE A DISTANCIA ENTRE OS PONTOS A = (-1, 2) E B = (1 , -2).


03. DETERMINE A DISTANCIA ENTRE OS PONTOS A = (0 , 5) E B = (1 , 0).


04. DETERMINE AS COORDENADAS DO PONTO MÉDIO ENTRE OS PONTOS A = (1 , 2) E B = ( 4 , -4)


05. DETERMINE AS COORDENADAS DO PONTO MÉDIO ENTRE OS PONTOS A = (0 , 2) E B = ( 4 , 0)

Answer 1

Temos questões de geometria analítica cujo objetivo em umas é determinar a distância entre dois pontos, e em outras é determinar as coordenadas do ponto médio entre dois pontos

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Para determinar a distancia entre dois pontos, use:

$\boxed{\boxed{\begin{array}{l}\\\quad\sf d=\sqrt{(x_b-x_a)^2+(y_b-y_a)^2}\quad\\\\\end{array}}}$

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Para determinar as coordenadas do ponto médio use:

$\boxed{\boxed{\begin{array}{l}\\\quad \sf M\bigg(\dfrac{x_a+x_b}{2}~~,~~\dfrac{y_a+y_b}{2}\bigg)\quad\\\\\end{array}}}$

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Obs.: lembre-se que:

•  $\sf A(x_a~,~y_a)\quad e\quad B(x_b~,~y_b)$

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$\large\begin{array}{l}\underbrace{\sf Resoluc_{\!\!\!,}\tilde{a}o}\end{array}$

01 )

Aqui devemos determinar a distância entre A(3 , 5) e B(1 , 2)

Assim temos: xa = 3, xb = 1, ya = 5, yb = 2

$\begin{array}{l}\\\sf d=\sqrt{(x_b-x_a)^2+(y_b-y_a)^2}\\\\ \sf d=\sqrt{(1-3)^2+(2-5)^2}\\\\ \sf d=\sqrt{(-2)^2+(-3)^2}\\\\ \sf d=\sqrt{4+9}\\\\ \!\boxed{\sf d=\sqrt{13}}\end{array}$

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02 )

Aqui devemos determinar a distância entre A(–1 , 2) e B(1 , –2)

Assim temos: xa = –1, xb = 1, ya = 2, yb = –2

$\begin{array}{l}\\\sf d=\sqrt{(x_b-x_a)^2+(y_b-y_a)^2}\\\\ \sf d=\sqrt{(1-(-1))^2+(-2-2)^2}\\\\ \sf d=\sqrt{(1+1)^2+(-4)^2}\\\\ \sf d=\sqrt{(2)^2+16}\\\\ \sf d=\sqrt{4+16}\\\\ \sf d=\sqrt{20}\\\\ \sf\Rightarrow~~simplificando:\\\\ \sf d=\sqrt{4\cdot5}\\\\\sf d=\sqrt{2^2\cdot5}\\\\ \!\boxed{\sf d=2\sqrt{5}}\end{array}$

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03 )

Aqui devemos determinar a distância entre A(0 , 5) e B(1 , 0)

Assim temos: xa = 0, xb = 1, ya = 5, yb = 0

$\begin{array}{l}\\\sf d=\sqrt{(x_b-x_a)^2+(y_b-y_a)^2}\\\\ \sf d=\sqrt{(1-0)^2+0-5)^2}\\\\ \sf d=\sqrt{(1)^2+(-5)^2}\\\\ \sf d=\sqrt{1+25}\\\\ \!\boxed{\sf d=\sqrt{26}}\end{array}$

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04 )

Aqui devemos determinar as coordenadas do ponto médio entre:

A(1 , 2) e B(4 , –4)

Assim temos: xa = 1, xb = 4, ya = 2, yb = –4

$\begin{array}{l}\sf M\bigg(\dfrac{x_a+x_b}{2}~~,~~\dfrac{y_a+y_b}{2}\bigg)\\\\\sf M\bigg(\dfrac{1+4}{2}~~,~~\dfrac{2+(-4)}{2}\bigg)\\\\\sf M\bigg(\dfrac{1+4}{2}~~,~~\dfrac{2-4}{2}\bigg)\\\\\sf M\bigg(\dfrac{5}{2}~~,~-\dfrac{2}{2}\bigg)\\\\ \!\boxed{\sf M\bigg(\dfrac{5}{2}~~,~-1\bigg)}\end{array}$

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05 )

Aqui devemos determinar as coordenadas do ponto médio entre:

A(0 , 2) e B(4 , 0)

Assim temos: xa = 0, xb = 4, ya = 2, yb = 0

$\begin{array}{l}\sf M\bigg(\dfrac{x_a+x_b}{2}~~,~~\dfrac{y_a+y_b}{2}\bigg)\\\\\sf M\bigg(\dfrac{0+4}{2}~~,~~\dfrac{2+0}{2}\bigg)\\\\\sf M\bigg(\dfrac{4}{2}~~,~~\dfrac{2}{2}\bigg)\\\\ \!\boxed{\sf M\bigg(2~~,~~1\bigg)}\end{array}$

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Att. Nasgovaskov

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