College Algebra, Chapter 2, 2.1, Section 2.1, Problem 50

Find the lengths of the medians of the triangle with vertices $A(1,0), B(3,6)$ and $C(8,2)$
Recall that a median is a line segment from a vertext to the midpoint of the opposite side, so.



First, we must get the midpoint on every sides of the triangle.

$
\begin{equation}
\begin{aligned}
M_{AC} &= \frac{1+8}{2} \text{ and } \frac{0+2}{2} = \left( \frac{9}{2},1 \right)\\
\\
M_{AB} &= \frac{1+3}{2} \text{ and } \frac{0+6}{2} = (2,3)\\
\\
M_{BC} &= \frac{3+8}{2} \text{ and } \frac{6+2}{2} = \left( \frac{11}{2},4 \right)
\end{aligned}
\end{equation}
$


Then, the medians can be computed by using the distance formula from a vertex to the midpoint of the opposite side. So median $b$ can be determined by setting the distance of vertex $B$ to midpoint $AC$

$
\begin{equation}
\begin{aligned}
b &= \sqrt{(6-1)^2 + \left( 3 - \frac{9}{2} \right)^2}\\
\\
&= \sqrt{5^2 + \left( - \frac{3}{2} \right)^2}\\
\\
&= \sqrt{25+ \frac{9}{4}}\\
\\
&= \sqrt{\frac{109}{4}}\\
\\
&= \sqrt{\frac{109}{2}} \text{ units}
\end{aligned}
\end{equation}
$