Question

Find the lengths of the medians of $∆$ABC whose vertices are A(7, $-$3) B(5,3) and C(3,$-$1)

Answer 1

$$\begin{gathered} \mathrm{Given}:\mathrm{A}\equiv \left(7,-3\right),\mathrm{B}\equiv \left(5,3\right),\mathrm{C}\equiv \left(3,-1\right) \\ \mathrm{Let}\mathrm{mid}-\mathrm{point}\mathrm{of}\mathrm{AB}\equiv \mathrm{E}\left(x_1,y_1\right) \\ \mathrm{mid}-\mathrm{point}\mathrm{of}\mathrm{BC}\equiv \mathrm{D}\left(x_2,y_2\right) \\ \mathrm{and}\mathrm{mid}-\mathrm{point}\mathrm{of}\mathrm{AC}\equiv \mathrm{F}\left(x_3,y_3\right) \\ \therefore x_1=\frac{7+5}{2}=6\mathrm{and}{y}_1=\frac{-3+3}{2}=0 \\ \therefore x_2=\frac{5+3}{2}=4\mathrm{and}{y}_2=\frac{3-1}{2}=1 \\ \therefore x_3=\frac{7+3}{2}=5\mathrm{and}{y}_3=\frac{-3-1}{2}=-2 \\ \mathrm{AD}=\sqrt{{\left(7-4\right)}^2+{\left(-3-1\right)}^2} \\ =\sqrt{{\left(3\right)}^2+{\left(-4\right)}^2} \\ =\sqrt{9+16} \\ =5\mathrm{units} \\ \mathrm{BF}=\sqrt{{\left(5-5\right)}^2+{\left[3-\left(-2\right)\right]}^2} \\ =\sqrt{{\left(0\right)}^2+{\left(5\right)}^2} \\ =\sqrt{0+25} \\ =5\mathrm{units} \\ \mathrm{CE}=\sqrt{{\left(3-6\right)}^2+{\left(-1-0\right)}^2} \\ =\sqrt{{\left(-3\right)}^2+{\left(-1\right)}^2} \\ =\sqrt{9+1} \\ =\sqrt{10}\mathrm{units} \\ \therefore \text{L}\mathrm{engths}\mathrm{of}\text{m}\mathrm{edians}\mathrm{are}5\mathrm{units},5\mathrm{units}\mathrm{and}\sqrt{10}\mathrm{units}. \end{gathered}$$