If the vertices of Delta ABC are A(1, 4),B(-3 | vedclass

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(B) For the median $\overline{AD}$,$D$ is the midpoint of $\overline{BC}$,where $B(-3, 2)$ and $C(5, 0)$.The coordinates of $D$ are calculated using t

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$3$

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(B) For the median $\overline{AD}$,$D$ is the midpoint of $\overline{BC}$,where $B(-3, 2)$ and $C(5, 0)$.The coordinates of $D$ are calculated using the midpoint formula: $D = \left(\frac{-3+5}{2}, \frac{2+0}{2}\right) = \left(\frac{2}{2}, \frac{2}{2}\right) = (1, 1)$.Now,we find the length of the median $\overline{AD}$ using the distance formula between $A(1, 4)$ and $D(1, 1)$:$AD = \sqrt{(1-1)^2 + (4-1)^2}$$AD = \sqrt{0^2 + 3^2}$$AD = \sqrt{9} = 3$.Thus,the length of the median $\overline{AD}$ is $3$.

If the vertices of $\Delta ABC$ are $A(1, 4)$,$B(-3, 2)$,and $C(5, 0)$,then the length of the median $\overline{AD}$ is.........

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