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]$\frac{1}{3}$[/tex] the length of the line segment from [tex]$A$[/tex] to [tex]$B$[/tex]?

[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]

(1.5)

[tex]$(0,3)$[/tex]

[tex]$(-4,-5)$[/tex]

[tex]$(-5,-7)$[/tex]



Answer :

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]