Question

Find the equations of the medians of a triangle, the coordinates of whose vertices are (−1, 6), (−3, −9) and (5, −8).

Answer 1

Let A (−1, 6), B (−3, −9) and C (5, −8) be the coordinates of the given triangle.
Let D, E and F be midpoints of BC, CA and AB, respectively.
So, the coordinates of D, E and F are



$$\begin{gathered} D\equiv \left(\frac{-3+5}{2},\frac{-9-8}{2}\right)=\left(1,\frac{-17}{2}\right) \\ E\equiv \left(\frac{-1+5}{2},\frac{6-8}{2}\right)=\left(2,-1\right) \\ F\equiv \left(\frac{-1-3}{2},\frac{6-9}{2}\right)=\left(-2,-\frac{3}{2}\right) \end{gathered}$$

Median AD passes through $A\left(-1,6\right)\mathrm{and}D\left(1,-\frac{17}{2}\right)$.
So, its equation is

$$\begin{gathered} y-6=\frac{-{\displaystyle \frac{17}{2}}-6}{1+1}\left(x+1\right) \\ \Rightarrow 4y-24=-29x-29 \\ \Rightarrow 29x+4y+5=0 \end{gathered}$$

Median BE passes through $B\left(-3,-9\right)\mathrm{and}E\left(2,-1\right)$.
So, its equation is

$$\begin{gathered} y+9=\frac{-1+9}{2+3}\left(x+3\right) \\ \Rightarrow 5y+45=8x+24 \\ \Rightarrow 8x-5y-21=0 \end{gathered}$$

Median CF passes through $C\left(5,-8\right)\mathrm{and}F\left(-2,-\frac{3}{2}\right)$.
So, its equation is

$$\begin{gathered} y+8=\frac{-{\displaystyle \frac{3}{2}}+8}{-2-5}\left(x-5\right) \\ \Rightarrow -14y-112=13x-65 \\ \Rightarrow 13x+14y+47=0 \end{gathered}$$