Question

the vertices of triangle abc have coordinates A(4.6) B(5,11) and C(3,y) the length of the median AD drawn from vertex A to side BC is 4.5 units find the value of y

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{mid}\mathrm{point}\mathrm{of}\mathrm{BC}\mathrm{is}=\left(\frac{3+5}{2},\frac{11+\mathrm{y}}{2}\right) \\ =\left(4,\frac{11+\mathrm{y}}{2}\right) \\ \mathrm{Now}\mathrm{distance}\mathrm{between}\mathrm{A}\left(4,6\right)\mathrm{and}\left(4,\frac{11+\mathrm{y}}{2}\right)\mathrm{is}\mathrm{given}\mathrm{as}4.5\mathrm{unit}\mathrm{so} \\ \sqrt{{\left(4-4\right)}^2+{\left(6-\frac{11+\mathrm{y}}{2}\right)}^2}=4.5 \\ 6-\frac{11+\mathrm{y}}{2}=4.5 \\ 1.5=\frac{11+\mathrm{y}}{2} \\ 11+\mathrm{y}=3 \\ \mathrm{y}=-8 \end{gathered}$$