Question

Find the lengths of the median of ∆ABC whose vertices are A(0, −1), B(2, 1) and C(0, 3).

Answer 1

The vertices of ∆ABC are A(0, −1), B(2, 1) and C(0, 3).
Let AD, BE and CF be the medians of ∆ABC.

Let D be the midpoint of BC. So, the coordinates of D are
$D\left(\frac{2+0}{2},\frac{1+3}{2}\right)\mathrm{i}.\mathrm{e}.D\left(\frac{2}{2},\frac{4}{2}\right)\mathrm{i}.\mathrm{e}.D\left(1,2\right)$
Let E be the midpoint of AC. So, the coordinates of E are
$E\left(\frac{0+0}{2},\frac{-1+3}{2}\right)\mathrm{i}.\mathrm{e}.E\left(\frac{0}{2},\frac{2}{2}\right)\mathrm{i}.\mathrm{e}.E\left(0,1\right)$
Let F be the midpoint of AB. So, the coordinates of F are
$F\left(\frac{0+2}{2},\frac{-1+1}{2}\right)\mathrm{i}.\mathrm{e}.F\left(\frac{2}{2},\frac{0}{2}\right)\mathrm{i}.\mathrm{e}.F\left(1,0\right)$
$$\begin{gathered} AD=\sqrt{{\left(1-0\right)}^2+{\left(2-\left(-1\right)\right)}^2}=\sqrt{{\left(1\right)}^2+{\left(3\right)}^2}=\sqrt{1+9}=\sqrt{10}\mathrm{units}. \\ BE=\sqrt{{\left(0-2\right)}^2+{\left(1-1\right)}^2}=\sqrt{{\left(-2\right)}^2+{\left(0\right)}^2}=\sqrt{4+0}=\sqrt{4}=2\mathrm{units}. \\ CF=\sqrt{{\left(1-0\right)}^2+{\left(0-3\right)}^2}=\sqrt{{\left(1\right)}^2+{\left(-3\right)}^2}=\sqrt{1+9}=\sqrt{10}\mathrm{units}. \end{gathered}$$
Therefore, the lengths of the medians: AD = $\sqrt{10}$ units, BE = 2 units and CF = $\sqrt{10}$ units