PLEASE HELP!Given triangle ABC with vertices A(−3, 0), B(0, 6), and C(4, 6). Find the equations of the three medians of triangle ABC

Answer:[tex]y=1.2x+3.6\\ \\y=\dfrac{6}{7}x+6\\ \\y=\dfrac{6}{11}x+\dfrac{42}{11}[/tex]Step-by-step explanation:Given triangle ABC with vertices A(−3, 0), B(0, 6), and C(4, 6).First, find midpoints of sides AB, BC and AC:midpoint of AB has coordinates [tex]C_1\left(\dfrac{-3+0}{2},\dfrac{0+6}{2}\right)\rightarrow C_1(-1.5, 3);[/tex]midpoint of BC has coordinates [tex]A_1\left(\dfrac{0+4}{2},\dfrac{6+6}{2}\right)\rightarrow A_1(2,6);[/tex]midpoint of AC has coordinates [tex]B_1\left(\dfrac{-3+4}{2},\dfrac{0+6}{2}\right)\rightarrow B_1(0.5, 3)[/tex]Now find the eqyations of the medians.1. Median at vertex A (line [tex]AA_1[/tex]):[tex]y=\dfrac{6-0}{2-(-3)}(x-(-3))+0\\ \\y=\dfrac{6}{5}(x+3)\\ \\y=1.2x+3.6[/tex]2. Median at vertex B (line [tex]BB_1[/tex]):[tex]y=\dfrac{6-3}{4-0.5}(x-0)+6\\ \\y=\dfrac{6}{7}x+6[/tex]3. Median at vertex C (line [tex]CC_1[/tex]):[tex]y=\dfrac{6-3}{4-(-1.5)}(x-4)+6\\ \\y=\dfrac{6}{11}(x-4)+6\\ \\y=\dfrac{6}{11}x+\dfrac{42}{11}[/tex]