Let's find the 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]\(\frac{1}{3}\)[/tex] of the way from [tex]\( A \)[/tex] to [tex]\( B \)[/tex].
Given points:
- [tex]\( A = (0, 3) \)[/tex]
- [tex]\( B = (-4, -5) \)[/tex]
The formula to find the coordinates of such a point is:
[tex]\[
\begin{array}{l}
x = \left(\frac{m}{m+n}\right) \left(x_2 - x_1\right) + x_1 \\
y = \left(\frac{m}{m+n}\right) \left(y_2 - y_1\right) + y_1
\end{array}
\][/tex]
Where [tex]\( x_1 \)[/tex] and [tex]\( y_1 \)[/tex] are the coordinates of point [tex]\( A \)[/tex], [tex]\( x_2 \)[/tex] and [tex]\( y_2 \)[/tex] are the coordinates of point [tex]\( B \)[/tex], [tex]\( m \)[/tex] is the portion towards point [tex]\( B \)[/tex], and [tex]\( n \)[/tex] is the remaining portion.
Here, [tex]\( P \)[/tex] is [tex]\(\frac{1}{3}\)[/tex] along the line from [tex]\( A \)[/tex] to [tex]\( B \)[/tex]. Thus, [tex]\( m = 1 \)[/tex] and [tex]\( m + n = 3 \)[/tex]. Hence, [tex]\( n = 2 \)[/tex].
Now, applying the given coordinate values:
For the x-coordinate:
[tex]\[
x = \left(\frac{1}{1 + 2}\right) \left(-4 - 0\right) + 0
\][/tex]
[tex]\[
x = \left(\frac{1}{3}\right) \left(-4\right) + 0
\][/tex]
[tex]\[
x = - \frac{4}{3}
\][/tex]
For the y-coordinate:
[tex]\[
y = \left(\frac{1}{1+2}\right) \left(-5 - 3\right) + 3
\][/tex]
[tex]\[
y = \left(\frac{1}{3}\right) \left(-8\right) + 3
\][/tex]
[tex]\[
y = -\frac{8}{3} + 3
\][/tex]
Converting 3 to [tex]\(\frac{9}{3}\)[/tex] for easy subtraction:
[tex]\[
y = -\frac{8}{3} + \frac{9}{3}
\][/tex]
[tex]\[
y = \frac{1}{3}
\][/tex]
Therefore, the coordinates of point [tex]\( P \)[/tex] are:
[tex]\[
\left( -\frac{4}{3}, \frac{1}{3} \right)
\][/tex]
In decimal form, the coordinates are:
[tex]\[
\left( -1.3333333333333333, 0.3333333333333333 \right)
\][/tex]
