Question

calcule o ponto médio do segmento AV nos seguintes casos: A) A (2,0) e B (-4,3)
B) A (3,2) e B (1,-2)
C) A (1,2) e B (2,4)
D) A (-3,5) e B (3,-5)

me ajuda pfv​

Answer 1

Explicação passo-a-passo:

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Lembrando que:

$\boxed{x_{m}=\dfrac{x_{a}+x_{b} }{2}}$

$\boxed{y_{m}=\dfrac{y_{a}+y_{b} }{2}}$

a)Temo os seguintes pontos A(2, 0) e B(-4, 3).

Calculando $x_{m}$ :

$x_{m}=\dfrac{2-4}{2} \\\\x_{m}=\dfrac{-2}{2}\\\\\boxed{x_{m}=-1}$

Calculando $y_{m}$ :

$y_{m}=\dfrac{0+3}{2} \\\\\boxed{y_{m}=\dfrac{3}{2} }$

O ponto médio é:

$\boxed{\boxed{M\bigg(-1\ ,\dfrac{3}{2} \bigg)}}$

b)Temo os seguintes pontos A(3, 2) e B(1, -2).

Calculando $x_{m}$ :

$x_{m}=\dfrac{3+1}{2} \\\\x_{m}=\dfrac{4}{2}\\\\\boxed{x_{m}=2}$

Calculando $y_{m}$ :

$y_{m}=\dfrac{2-2}{2}\\\\y_{m}=\dfrac{0}{2} \\\\\boxed{y_{m}=0 }$

O ponto médio é:

$\boxed{\boxed{M(2,0)}}$

c)Temo os seguintes pontos A(1, 2) e B(2, 4).

Calculando $x_{m}$ :

$x_{m}=\dfrac{1+2}{2}\\\\\boxed{x_{m}=\dfrac{3}{2} }$

Calculando $y_{m}$ :

$y_{m}=\dfrac{2+4}{2}\\\\y_{m} =\dfrac{6}{2} \\\\\boxed{y_{m}=3 }$

O ponto médio é:

$\boxed{\boxed{M\bigg(\dfrac{3}{2}\ ,3 \bigg)}}$

d)Temo os seguintes pontos A(-3, 5) e B(3, -5).

Calculando $x_{m}$ :

$x_{m}=\dfrac{-3+3}{2} \\\\x_{m}=\dfrac{0}{2}\\\\\boxed{x_{m}=0}$

Calculando $y_{m}$ :

$y_{m}=\dfrac{5-5}{2} \\\\y_{m}=\dfrac{0}{2}\\\\\boxed{y_{m}=0}$

O ponto médio é:

$\boxed{\boxed{M(0,0)}}$