The coordinates of a triangle ABC are A(1, 5), B(-2, 3), and C(0,-4); find the equation of the median AD? ·

  Given that $$\triangle$$ABC

Since AD is the median to BC, D will be mid point of BC. So coordinates of D = ($$\frac{-2+0}{2} ,\frac{3-4}{2}$$) = (-1 ,$$\frac{-1}{2}$$)

Equation of line passing through points A(1,5) and D(-1,$$\frac{-1}{2}$$) will be :

                $$(y-x_{1})=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})$$                   (Here $$(x_{1},y_{1})=(1,5)$$ ,  $$(x_{2},y_{2})=(-1,\frac{-1}{2})$$)
                
                $$(y-5)=\frac{(\frac{-1}{2})-(5)}{(-1)-(1)}(x-1)$$
                
                $$4(y-5)=11(x-1)$$
                
                $$11x-4y+9=0$$