What are the [tex]$x$[/tex]- and [tex]$y$[/tex]-coordinates of point [tex]$P$[/tex] on the directed line segment from [tex]$A$[/tex] to [tex]$B$[/tex] such that [tex]$P$[/tex] is [tex]$\frac{2}{3}$[/tex] the length of the line segment from [tex]$A$[/tex] to [tex]$B$[/tex]?

[tex]\[
\begin{array}{l}
x = \left(\frac{2}{3}\right) \left(x_2 - x_1\right) + x_1 \\
y = \left(\frac{2}{3}\right) \left(y_2 - y_1\right) + y_1
\end{array}
\][/tex]

A. [tex]$(2, -1)$[/tex]
B. [tex]$(4, -3)$[/tex]
C. [tex]$(-1, 2)$[/tex]
D. [tex]$(3, -2)$[/tex]



Answer :

To determine the coordinates of point [tex]\(P\)[/tex] on the directed line segment from point [tex]\(A\)[/tex] (with coordinates [tex]\((2, -1)\)[/tex]) to point [tex]\(B\)[/tex] (with coordinates [tex]\((4, -3)\)[/tex]), such that [tex]\(P\)[/tex] divides the segment [tex]\(AB\)[/tex] in the ratio [tex]\( \frac{2}{3} \)[/tex], we can use the section formula for internal division.

Given:
- Coordinates of [tex]\(A\)[/tex]: [tex]\( (x_1, y_1) = (2, -1) \)[/tex]
- Coordinates of [tex]\(B\)[/tex]: [tex]\( (x_2, y_2) = (4, -3) \)[/tex]
- Ratio [tex]\( m : n = 2 : 1 \)[/tex]

The formula for finding the coordinates of point [tex]\( P \)[/tex] that divides the line segment [tex]\(AB\)[/tex] in the ratio [tex]\(m : n\)[/tex] is:

[tex]\[ P \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) \][/tex]

Substituting [tex]\( m = 2 \)[/tex] and [tex]\( n = 1 \)[/tex]:

1. Calculate [tex]\(P_x\)[/tex] (the x-coordinate of [tex]\(P\)[/tex]):
[tex]\[ P_x = \frac{m x_2 + n x_1}{m + n} = \frac{2 \cdot 4 + 1 \cdot 2}{2 + 1} = \frac{8 + 2}{3} = \frac{10}{3} = 3.333333333333333 \][/tex]

2. Calculate [tex]\(P_y\)[/tex] (the y-coordinate of [tex]\(P\)[/tex]):
[tex]\[ P_y = \frac{m y_2 + n y_1}{m + n} = \frac{2 \cdot (-3) + 1 \cdot (-1)}{2 + 1} = \frac{-6 - 1}{3} = \frac{-7}{3} = -2.333333333333333 \][/tex]

Thus, the coordinates of point [tex]\(P\)[/tex] are:

[tex]\[ P \left( 3.333333333333333, -2.333333333333333 \right) \][/tex]