Question

Os vértices de um triângulo são os pontos A(0,4),B(2,-6) e C(-4,2).Calcular os comprimentos das medianas do triângulo

Answer 1

A(0,4) e B (2 , -6)
..Xa,Ya......Xb,Yb

Mx = Xa + Xb
..........______
................2

Mx = 0 + 2
........._____
...............2

Mx = 2
.........__
...........2

Mx = 1

My = Ya + Yb
..........______
.................2

My = 4 + (-6)
.........._____
...............2

My = 4 - 6
..........____
..............2

My = - 2
..........___
.............2

My = - 1

M(AB) = ( 1 , - 1)


A(0,4) e C(- 4, 2)
..Xa,Ya........Xc ,Yc

Mx = Xa + Xc
..........______
.................2

Mx = 0 + (-4)
.........._____
...............2

Mx = 0 - 4
..........____
..............2

Mx = - 4
..........___
.............2

Mx = - 2

My = Ya + Yc
..........______
.................2

My = 4 + 2
........._____
..............2

My = 6
.........__
..........2

My = 3

M(AC) = ( -2 , 3 )


B( 2 , - 6) e C(- 4 , 2)
..Xb ,Yb...........Xc ,Yc

Mx = Xb + Xc
........._______
.................2

Mx = 2 + (-4)
.........______
..............2

Mx = 2 - 4
.........____
.............2

Mx = - 2
..........___
.............2

Mx = - 1

My = Yb + Yc
.........._______
.................2

My = - 6 + 2
..........______
.................2

My = - 4
..........___
.............2

My = - 2

M(BC) = (- 1 , - 2)


C (-4,2) e M(AB)= (1 ,-1)

=<> (-4 - 1)^2 + (2 + 1)^2
=<> (-5)^2 + (3)^2
=<> 25 + 9
=<> \|34


B(2,-6) e M(AC) = (-2,3)

=<> (2 - (-2))^2 + (- 6 - 3)^2
=<> (2+2)^2 + (-9)^2
=<> (4)^2 +81
=<> 16 + 81
=<> \|97

A(0,4) e M(-1,-2)

=<> (0 - (-1))^2 + (4 - (-2))^2
=<> (0+ 1)^2 +( 4+ 2)^2
=<> (1)^2 + (6)^2
=<> 1 + 36
=> \|37