To 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] the length from [tex]\( A \)[/tex] to [tex]\( B \)[/tex], we can use the section formula:
Given:
- Point [tex]\( A = (1, 5) \)[/tex]
- Point [tex]\( B = (-4, -5) \)[/tex]
- The ratio is [tex]\( m : n = 1 : 2 \)[/tex] (since [tex]\( P \)[/tex] is [tex]\( \frac{1}{3} \)[/tex] of the way, the remaining segment is [tex]\( \frac{2}{3} \)[/tex])
The section formula for coordinates [tex]\( (x,y) \)[/tex] 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]
1. Calculate the [tex]\( x \)[/tex]-coordinate of [tex]\( P \)[/tex]:
[tex]\[
\begin{array}{l}
x_1 = 1 \\
x_2 = -4 \\
m = 1 \\
n = 2
\end{array}
\][/tex]
[tex]\[
x = \left( \frac{m}{m+n} \right)(x_2 - x_1) + x_1
\][/tex]
Substitute the values:
[tex]\[
x = \left( \frac{1}{1+2} \right)(-4 - 1) + 1
\][/tex]
[tex]\[
x = \left( \frac{1}{3} \right)(-5) + 1
\][/tex]
[tex]\[
x = -\frac{5}{3} + 1
\][/tex]
Convert 1 into thirds:
[tex]\[
x = -\frac{5}{3} + \frac{3}{3}
\][/tex]
[tex]\[
x = -\frac{5}{3} + \frac{3}{3} = -\frac{2}{3}
\][/tex]
2. Calculate the [tex]\( y \)[/tex]-coordinate of [tex]\( P \)[/tex]:
[tex]\[
\begin{array}{l}
y_1 = 5 \\
y_2 = -5 \\
m = 1 \\
n = 2
\end{array}
\][/tex]
[tex]\[
y = \left( \frac{m}{m+n} \right)(y_2 - y_1) + y_1
\][/tex]
Substitute the values:
[tex]\[
y = \left( \frac{1}{1+2} \right)(-5 - 5) + 5
\][/tex]
[tex]\[
y = \left( \frac{1}{3} \right)(-10) + 5
\][/tex]
[tex]\[
y = -\frac{10}{3} + 5
\][/tex]
Convert 5 into thirds:
[tex]\[
y = -\frac{10}{3} + \frac{15}{3}
\][/tex]
[tex]\[
y = -\frac{10}{3} + \frac{15}{3} = \frac{5}{3}
\][/tex]
Thus, the coordinates of point [tex]\( P \)[/tex] are [tex]\( \left( -\frac{2}{3}, \frac{5}{3} \right) \)[/tex] which is approximately [tex]\( (-0.6667, 1.6667) \)[/tex].
