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 of the line segment from [tex]\(A\)[/tex] to [tex]\(B\)[/tex], we will use the section formula. The section formula is commonly used in coordinate geometry to find the coordinates of a point that divides a line segment in a given ratio.
Given:
- [tex]\(A = (1, 5)\)[/tex]
- [tex]\(B = (0, 3)\)[/tex]
We know that point [tex]\(P\)[/tex] divides the segment [tex]\(AB\)[/tex] in the ratio [tex]\( \frac{1}{3} \)[/tex]. This means that the ratio [tex]\(m:n\)[/tex] is [tex]\(1:2\)[/tex] (where [tex]\(m\)[/tex] is the part of the segment [tex]\(P\)[/tex] from [tex]\(A\)[/tex] and [tex]\(n\)[/tex] from [tex]\(P\)[/tex] to [tex]\(B\)[/tex]).
We use the section formula for the coordinates of point [tex]\(P\)[/tex]:
[tex]\[
\begin{align*}
x_P &= \left( \frac{m}{m+n} \right)(x_2 - x_1) + x_1 \\
y_P &= \left( \frac{m}{m+n} \right)(y_2 - y_1) + y_1
\end{align*}
\][/tex]
Here,
- [tex]\(x_1, y_1 = A = (1, 5)\)[/tex]
- [tex]\(x_2, y_2 = B = (0, 3)\)[/tex]
- [tex]\(m = 1\)[/tex]
- [tex]\(n = 2\)[/tex]
First, let's calculate the [tex]\(x\)[/tex]-coordinate [tex]\(x_P\)[/tex]:
[tex]\[
x_P = \left( \frac{1}{1+2} \right)(0 - 1) + 1 \\
x_P = \left( \frac{1}{3} \right)(-1) + 1 \\
x_P = -\frac{1}{3} + 1 \\
x_P = 1 - \frac{1}{3} \\
x_P = \frac{3}{3} - \frac{1}{3} \\
x_P = \frac{2}{3} \\
x_P = 0.\overline{6}
\][/tex]
Next, we calculate the [tex]\(y\)[/tex]-coordinate [tex]\(y_P\)[/tex]:
[tex]\[
y_P = \left( \frac{1}{1+2} \right)(3 - 5) + 5 \\
y_P = \left( \frac{1}{3} \right)(-2) + 5 \\
y_P = -\frac{2}{3} + 5 \\
y_P = 5 - \frac{2}{3} \\
y_P = \frac{15}{3} - \frac{2}{3} \\
y_P = \frac{13}{3} \\
y_P = 4.\overline{3}
\][/tex]
So, the coordinates of point [tex]\(P\)[/tex] are:
[tex]\[
\boxed{\left( \frac{2}{3}, \frac{13}{3} \right)}
\][/tex]
Numerically, we can write these coordinates approximately as:
[tex]\(\boxed{(0.67, 4.33)}\)[/tex]
