The point P(–4, 4) that is [tex]\frac{2}{5}[/tex] of the way from A to B on the directed line segment AB.
Solution:
The points of the line segment are A(–8, –2) and B(6, 19).
P is the point that bisect the line segment in [tex]\frac{2}{5}[/tex].
So, m = 2 and n = 5.
[tex]x_1=-8, y_1=-2, x_2=6, y_2=19[/tex]
By section formula:
[tex]$P(x, y)=\left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right)[/tex]
[tex]$P(x, y)=\left(\frac{2\times 6+5\times (-8)}{2+5}, \frac{2\times 19+5\times (-2)}{2+5}\right)[/tex]
[tex]$P(x, y)=\left(\frac{12-40}{7}, \frac{38-10}{7}\right)[/tex]
[tex]$P(x, y)=\left(\frac{-28}{7}, \frac{28}{7}\right)[/tex]
P(x, y) = (–4, 4)
Hence the point P(–4, 4) that is [tex]\frac{2}{5}[/tex] of the way from A to B on the directed line segment AB.
