oint A is located at (2, 6) and point B is located at (18, 12).What point partitions the directed line segment ​ AB¯¯¯¯¯ ​ into a 2:3 ratio?825, 825)(1012, 1012)(1139, 935)(1412, 12)

Answer:[tex](\frac{42}{5},\frac{42}{5})[/tex]Step-by-step explanation:We are given that Point A is at (2,6) and  point B is at  (18,12).We have to find the point which partitions the directed line segment AB into a ratio 2:3.We have [tex]x_1=2, y_1=6,x_2=18,y_2=12,m_1=2,m_2=3[/tex]Section formula:[tex]x=\frac{m_1x_1+m_2x_2}{m_1+m_2}, y=\frac{m_1y_1+m_2y_2}{m_1+m_2}[/tex]Substitute the values in the given formula then, we get The coordinates of point which partitions the segment AB is given by[tex]x=\frac{2(18)+3(2)}{2+3}, y=\frac{2(12)+3(6)}{2+3}[/tex][tex]x=\frac{36+6}{5},y=\frac{24+18}{5}[/tex][tex]x=\frac{42}{5}, y=\frac{42}{5}[/tex]Hence, the coordinated of the point which partitions the line segment AB =[tex](\frac{42}{5},\frac{42}{5})[/tex]