Question

Find the point P that is 2/5 of the way from A to B on the directed line segment AB

Answer 1

The point P(–4, 4) that is $\frac{2}{5}$ 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 $\frac{2}{5}$.

So, m = 2 and n = 5.

$x_1=-8, y_1=-2, x_2=6, y_2=19$

By section formula:

$P(x, y)=\left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right)$

$P(x, y)=\left(\frac{2\times 6+5\times (-8)}{2+5}, \frac{2\times 19+5\times (-2)}{2+5}\right)$

$P(x, y)=\left(\frac{12-40}{7}, \frac{38-10}{7}\right)$

$P(x, y)=\left(\frac{-28}{7}, \frac{28}{7}\right)$

P(x, y) = (–4, 4)

Hence the point P(–4, 4) that is $\frac{2}{5}$ of the way from A to B on the directed line segment AB.