Find the lengths of the medians of a ΔABC whose vertices are A(0, – 1), B(2, 1) and C(0, 3).
For coordinates of median AD segment BC will be taken
X = (m1x2 + m2x1)/ m1 + m2
= (1 × 0 + 1x 2)/1 + 1
= (0 + 2) /2
= 2/2 = 1
Y = (m1y2 + m2y1)/ m1 + m2
= (1x 3 + 1x 1)/2
= (3 + 1)/2
= 4 / 2 = 2
D(1,2)
By distance Formula
AD = √(1 – 0)2 + (2 + 1)2
= √1 + 9
= √10
For coordinates of BE, segment AC will be taken
X = (m1x2 + m2x1)/ m1 + m2
= (1 × 0 + 1x 0)/1 + 1
= (0 + 0) /2
= 0/2 = 0
Y = (m1y2 + m2y1)/ m1 + m2
= (1x 3 + 1x (– 1))/2
= (3 – 1)/2
= 2 / 2 = 1
∴ E(0,1)
By distance Formula
BE = √(0 – 2)2 + (1 – 1)2
= √4 + 0
= √4 = 2
For coordinates of median CF segment AB will be taken
X = (m1x2 + m2x1)/ m1 + m2
= (1 × 2 + 1x 0)/1 + 1
= (2 + 0) /2
= 2/2 = 1
Y = (m1y2 + m2y1)/ m1 + m2
= (1x(– 1) + 1x 1)/2
= (– 1 + 1)/2
= 0 / 2 = 0
F(1,0)
By distance Formula
CF = √(1 – 0)2 + (0 – 3)2
= √1 + 9
= √10
AD = √10 units, BE = 2 units, CF = √10 units