Find the lengths of the medians of triangle whose vertices are
A (-1,1) , B (5,-3) and C (3,5).

Explanation

A(-1,1) , B (5,-3)and C(3,5)
Let `D(x_(1) ,y_(1) , E(x_(2),y_(2)) ` and
`F(x_(1) ,y_(2)) ` be the midpoints of sides
BC , AC and AB respectively.
By midpoint formula ,
` (##NVT_21_MAT_P2_X_C06_E01_038_S01.png" width="80%">
`x_(1) = (5+3)/(2) `
`therefore x_(1) = (8)/(2)`
` therefore x_(1) = 4 `
` y_(1) = (-3+5)/(2)`
` therefore y_(1) =2/2`
` therefore y_(1) = 1`
` therefore ` the coordinates of the point D are (4,1).
By midpoint formula ,
`x_(2) = (-1+3)/(2)`
` therefore x_(2) = (2)/(2) therefore x_(2) =1`
`y_(2) = (1 +5)/(2)`
` therefore y_(2) = (6)/(2) therefore y_(2) = 3`
` therefore ` the coordinates of the point E are (1,3),
By midpoint formula,
`x_(1) = (5 + (-1))/(2)`
` therefore x_(3)= (4)/(2) therefore x_(3) = 2`
` y_(1) = (-3+1)/(2)`
`therefore y_(3) = (-2)/(2) therefore y_(3) = -1`
` therefore ` the coordinates of the point R are (2,-1).
By distance formula ,
` AD = sqrt([4-(-1)]^(2) + (1-1)^(2))`
` therefore AD = sqrt(5^(2) + 0^(2)) `
` therefore AD = sqrt(25) therefore AD = 5`
Length of medium AD is 5.
By distance formula ,
`BE = sqrt((5-1)^(2) + (-3-3)^(2))`
` = sqrt(4^(2) + (-6)^(2))`
` = sqrt( 16+36) = sqrt(52)`
` sqrt(2xx2xx13) = 2sqrt(13)`
Length of medium BE is `2sqrt(13)`.
By distance formula ,
` CF = sqrt((3-2)^(2) + [5-(-1)]^(2))`
`= sqrt((1)^(2) + (6)^(2))`
` = sqrt(1 + 36 ) = sqrt(37)`
Length of medium CF is ` sqrt(37)`
Lengths of the medians of triangle whose vertices are
A (-1,1), B(5,-3) and C(3,5) are `5,2 sqrt(13) "and " sqrt(37)` .