Find the lengths of the medians of a ΔABC whose vertices are A(0,-1) , B(2,1) and C (0.3).
Solution
The vertices of ABC A(0,-1) , B(2,1) and C (0.3)
Let AD, BE and CF be the medians of Δ ABC.
Let D be the midpoint of BC. So, the coordinates of D ar
`D ((2+0)/2 , (1+3)/2) i.e D (2/2 , 4/2) i.e D (1,2)`
Let E be the midpoint of AC. So the coordinate of E are
`E ((0+0)/2 , (-1+3)/2) i.e . E (0/2,0/2) i.e E (0,1) `
Let F be the midpoint of AB. So, the coordinates of F are
` F ((0+2)/2 , (-1+1)/2) i.e F (2/2 , 0/2) i.e F (1,0) `
`AD = sqrt((1-0)^2 +(2-(-1))^2) = sqrt((1)^2 +(3)^2) = sqrt(1+9) = sqrt(10) units`
` BE = sqrt((0-2)^2 +(1-1)^2) = sqrt((-2)^2 +(0)^2) = sqrt(4+0) = sqrt(4)= 2 units`
`CF = sqrt((1-0)^2 +(0-3)^2) = sqrt((1)^2 +(-3)^2) = sqrt(1+9) = sqrt(10) units`
`"Therefore, the lengths of the medians:" AD = sqrt(10) units . BE=2 units and CF = sqrt(10) units .`