determine o ponto medio do segmento de extremidades: a) A(1,-7) e B(3,-5) b)A(-1,5) eB (5,-2) c)A (-4,-2) eB (-2,-4) d)A(-1,0) e B(0,-2)

A)

A(1,-7) e B(3,-5)
xM=(1+3) / 2 = 4/2 = 2
yM=(-7-5) / 2 = -12/2= -6
M(2,-6)

B)

A(-1,5) eB (5,-2)
xM=(5-1)/2 = 4 / 2 = 2
yM=(5-2) /2 = 3/2
M(2, 3/2)

C)

A (-4,-2) eB (-2,-4)
xM=(-4-2) / 2 = -6 / 2 = -3
yM=(-4-2) / 2 = -6 / 2 = -3
M=(-3;-3)

D)

A(-1,0) e B(0,-2)
xM=(-1+0) / 2 = -1/2
yM=(-2+0) / 2 = -2/2 = -1
M=(-1/2: -1)

Nesse exercício vamos estudar ponto médio.

O ponto médio de um segmento de que conhecemos os extremos $A$ e $B$ é dado por:

$$M=\dfrac12(A+B)$$

a) $M=\dfrac12[(1,-7)+(3,-5)]\Rightarrow \boxed{M=(2,-6)}$

b) $M=\dfrac12[(-1,5)+(5,-2)]\Rightarrow \boxed{M=\left(2,\dfrac32\right)}$

c) $M=\dfrac12[(-4,-2)+(-2,-4)]\Rightarrow \boxed{M=\left(-3,-3\right)}$

d) $M=\dfrac12[(-1,0)+(0,-2)]\Rightarrow \boxed{M=\left(-\dfrac12,-1\right)}$