Question

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

Answer 1

Let the medians meet the lines BC, AC and AB at points be 

\[P\left( x_1 , y_1 \right)\]

\[Q\left( x_2 , y_2 \right)\] and 

\[R\left( x_3 , y_3 \right)\]respectively. 
P is thus the mid point of line BC

\[P\left( x_1 , y_1 \right) = \left( \frac{5 + 3}{2}, \frac{5 - 3}{2} \right) = \left( 4, 1 \right)\]

\[AP = \sqrt{\left( -4 - 1 \right)^2 + \left( 1 - 1 \right)^2} = \sqrt{25} = 5\]

Q is the mid point of line AC. 

\[Q\left( x_2 , y_2 \right) = \left( \frac{- 1 + 3}{2}, \frac{1 + 5}{2} \right) = \left( 1, 3 \right)\]

\[BQ = \sqrt{\left( 5 - 1 \right)^2 + \left( - 3 - 3 \right)^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}\]

R is the mid point of AB. 

\[R\left( x_3 , y_3 \right) = \left( \frac{- 1 + 5}{2}, \frac{1 - 3}{2} \right) = \left( 2, - 1 \right)\]

\[RC = \sqrt{\left( 3 - 2 \right)^2 + \left( -1 - 5 \right)^2} = \sqrt{1 + 36} = \sqrt{37}\]