How to find the length of a median of a triangle with vertices

HOW TO FIND THE LENGTH OF A MEDIAN OF A TRIANGLE WITH VERTICES

Example 1 :

Find the length of the medians of the triangle whose vertices are (1 , -1) (0, 4) and (-5, 3)

Solution :

Let A (1, -1) B (0, 4) and C (-5, 3) are the points vertices of the triangle

Let D, E and F are the midpoints of the sides AB, BC and CA respectively

Midpoint of AB = (x₁+x₂)/2 , (y₁+y₂)/2

= (1+0)/2 , (-1+4)/2

= D (1/2, 3/2)

Midpoint of BC = (x₁+x₂)/2 , (y₁+y₂)/2

= (0+(-5))/2 , (4+3)/2

= (0-5/2 , 7/2)

= E (-5/2, 7/2)

Midpoint of CA = (x₁+x₂)/2 , (y₁+y₂)/2

= (-5+1)/2 , (3+(-1))/2

= (-4/2 , 2/2)

= F (-2, 1)

Length of the median AD = √(x₂-x₁)² + (y₂-y₁)²

A (1, -1) and D (1/2,3/2)

= √(1+5/2)² + (-1-7/2)²

= √(7/2)² + (-9/2)²

= √(49/4)+(81/4)

= √(49+81)/4

= √130/4

= √130/2

Length of the median BE = √(x₂-x₁)² + (y₂-y₁)²

B (0, 4) and E (-5/2, 7/2)

= √(-2-0)² + (1-4)²

= √(-2)² + (-3)²

= √4+9

= √13

Length of the median CF = √(x₂-x₁)² + (y₂-y₁)²

C (-5, 3) and F (-2, 1)

= √((1/2)+5)² + ((3/2)-3)²

= √(11/2)² + (-3/2)²

= √(121/4)+(9/4)

= √(121 + 9)/4

= √130/4

= √130/2

Hence length of medians AD, BE and CF are √130/2, √13 and √130/2.