# Equation of the Medians of a Triangle

To find the equation of the median of a triangle we examine the following example: Consider the triangle having vertices $$A\left( { – 3,2} \right)$$, $$B\left( {5,4} \right)$$ and $$C\left( {3, – 8} \right)$$.

If $$G$$ is the midpoint of side $$AB$$ of the given triangle, then its coordinates are given as $$\left( {\frac{{ – 3 + 5}}{2},\frac{{2 + 4}}{2}} \right) = \left( {\frac{2}{2},\frac{6}{2}} \right) = \left( {1,3} \right)$$.

Since the median $$CG$$ passes through points $$C$$ and $$G$$, using the two-point form of the equation of a straight line, the equation of median $$CG$$ can be found as

\[\begin{gathered} \frac{{y – 3}}{{ – 8 – 3}} = \frac{{x – 1}}{{3 – 1}} \\ \Rightarrow \frac{{y – 3}}{{ – 11}} = \frac{{x – 1}}{2} \\ \Rightarrow 2\left( {y – 3} \right) = – 11\left( {x – 1} \right) \\ \Rightarrow 11x + 2y – 17 = 0 \\ \end{gathered} \]

If $$H$$ is the midpoint of side$$BC$$ of the given triangle, then its coordinates are given as $$\left( {\frac{{3 + 5}}{2},\frac{{ – 8 + 4}}{2}} \right) = \left( {\frac{8}{2},\frac{{ – 4}}{2}} \right) = \left( {4, – 2} \right)$$.

Since the median $$AH$$ passes through points $$A$$ and $$H$$, using the two-point form of the equation of a straight line, the equation of median $$AH$$ can be found as

\[\begin{gathered} \frac{{y – \left( { – 2} \right)}}{{2 – \left( { – 2} \right)}} = \frac{{x – 4}}{{ – 3 – 4}} \\ \Rightarrow \frac{{y + 2}}{4} = \frac{{x – 4}}{{ – 7}} \\ \Rightarrow – 7\left( {y + 2} \right) = 4\left( {x – 4} \right) \\ \Rightarrow 5x + 7y – 2 = 0 \\ \end{gathered} \]

If $$I$$ is the midpoint of side$$AC$$ of the given triangle, then its coordinates are given as $$\left( {\frac{{ – 3 + 3}}{2},\frac{{2 – 8}}{2}} \right) = \left( {0,\frac{{ – 6}}{2}} \right) = \left( {0, – 3} \right)$$.

Since the median $$BI$$ passes through points $$B$$ and $$I$$, using the two-point form of the equation of a straight line, the equation of median $$BI$$ can be found as

\[\begin{gathered} \frac{{y – \left( { – 3} \right)}}{{4 – \left( { – 3} \right)}} = \frac{{x – 0}}{{5 – 0}} \\ \Rightarrow \frac{{y + 3}}{7} = \frac{x}{5} \\ \Rightarrow 5\left( {y + 3} \right) = 7x \\ \Rightarrow 7x – 5y – 15 = 0 \\ \end{gathered} \]