Question

M is the midpoint of `\bar (CF)` for the points C(3, 4) and F(9, 8). What is `\bar (MF)`?

Answer 1

Distance `bar (MF) ~~ color(red)(3.6056)`

Explanation:

#C (3,4), F(9,

Mid point M of CF will have coordinates,

`x_M = (x_C + x_F)/2 = (3 + 9) / 2 = 6`

`y_M = (y_C + y_F)/2 = (4 + 8) / 2 = 6`

Using distance formula,

Distance `bar (MF) = sqrt((x_M - x_F)^2 + (y_M - y_F)^2`

`= sqrt((6-9)^2 + (6-8)^2) = sqrt13 ~~ color(red)(3.6056)`

Answer 2

`vec(MF)=((3),(2))`

Explanation:

`"since M is the midpoint of CF then"`

`vec(MF)=color(red)(1/2)vec(CF)`

`color(white)(vec(MF))=1/2(ulf-ulc)`

`color(white)(vec(MF))=1/2[((9),(8))-((3),(4))]=1/2((6),(4))=((3),(2))`

`|vec(MF)|=sqrt(3^2+2^2)=sqrt13`