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 $\mathrm{P}\left(x_1,y_1\right)$, $\mathrm{Q}\left(x_2,y_2\right)$ and $\mathrm{R}\left(x_3,y_3\right)$ respectively.
P is thus the mid point of line BC
$$\begin{gathered} \mathrm{P}\left(x_1,y_1\right)=\left(\frac{5+3}{2},\frac{5-3}{2}\right)=\left(4,2\right) \\ \mathrm{AP}=\sqrt{{\left(4+1\right)}^2+{\left(1-1\right)}^2}=\sqrt{25}=5 \end{gathered}$$
Q is the mid point of line AC.
$$\begin{gathered} Q\left(x_2,y_2\right)=\left(\frac{-1+3}{2},\frac{1+5}{2}\right)=\left(1,3\right) \\ \mathrm{BQ}=\sqrt{{\left(5-1\right)}^2+{\left(-3-3\right)}^2}=\sqrt{16+36}=\sqrt{52}=2\sqrt{13} \end{gathered}$$
R is the mid point of AB.
$$\begin{gathered} \mathrm{R}\left(x_3,y_3\right)=\left(\frac{-1+5}{2},\frac{1-3}{2}\right)=\left(2,-1\right) \\ \mathrm{RC}=\sqrt{{\left(3-2\right)}^2+{\left(5+1\right)}^2}=\sqrt{1+36}=\sqrt{37} \end{gathered}$$