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.

Related topics

• Find the coordinates of the point that divides the line segment into three parts
• How to find the ratio in which a point divides a line
• Midpoint of the line segment
• Section formula
• How to find the missing coordinate of a parallelogram

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