Solution:
Given:
[tex]\begin{gathered} Point\text{ P }\frac{2}{5}\text{ of the way to line AB} \\ A=(-8,-2) \\ B=(6,19) \end{gathered}[/tex]Using the section formula;
[tex](\frac{m_1x_2+m_2x_1}{m_2+m_1},\frac{m_1y_2+m_2y_1}{m_2+m_1})[/tex]where;
[tex]\begin{gathered} the\text{ ratio of }m_1:m_2=2:3 \\ \\ Hence, \\ m_1=2 \\ m_2=3 \\ x_1=-8 \\ y_1=-2 \\ x_2=6 \\ y_2=19 \end{gathered}[/tex]Substituting the values into the formula;
[tex]\begin{gathered} (\frac{(2\times6)+(3\times-8)}{3+2},\frac{(2\times19)+(3\times-2)}{3+2}) \\ =(\frac{12-24}{5},\frac{38-6}{5}) \\ =(\frac{-12}{5},\frac{32}{5}) \\ =(-2.4,6.4) \end{gathered}[/tex]Therefore, the point P that is 2/5 of the way from A to B on the directed line segment AB is (-2.4, 6.4)
