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$A(3, 0)$, $B(6, 4)$ and $C(-1, 3)$ are vertices of a triangle $ABC$. Find length of its median $BE$.
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Mid-point of $AC$ is $E (\frac{3-1}{2}, \frac{0+3}{2}) = E(1, \frac{3}{2})$
Length of median $BE = \sqrt{(6-1)^2 + (4-\frac{3}{2})^2} = \sqrt{5^2 + (\frac{5}{2})^2} = \sqrt{25 + \frac{25}{4}} = \sqrt{\frac{100+25}{4}} = \sqrt{\frac{125}{4}}$ or $\frac{5\sqrt{5}}{2}$
Length of median $BE = \sqrt{(6-1)^2 + (4-\frac{3}{2})^2} = \sqrt{5^2 + (\frac{5}{2})^2} = \sqrt{25 + \frac{25}{4}} = \sqrt{\frac{100+25}{4}} = \sqrt{\frac{125}{4}}$ or $\frac{5\sqrt{5}}{2}$