Given the coordinates of points [tex]\(A\)[/tex] and [tex]\(B\)[/tex], which are [tex]\(A = (2, -1)\)[/tex] and [tex]\(B = (4, -3)\)[/tex], we need to find the coordinates of a point [tex]\(P\)[/tex] on the directed line segment from [tex]\(A\)[/tex] to [tex]\(B\)[/tex] such that [tex]\(P\)[/tex] is located [tex]\(\frac{2}{3}\)[/tex] of the way along the segment from [tex]\(A\)[/tex] to [tex]\(B\)[/tex].
We use the given formulas for the [tex]\(x\)[/tex]- and [tex]\(y\)[/tex]-coordinates of [tex]\(P\)[/tex]:
[tex]\[ x_P = \left(\frac{m}{m+n}\right) \left(x_2 - x_1 \right) + x_1 \][/tex]
[tex]\[ y_P = \left(\frac{m}{m+n}\right) \left(y_2 - y_1 \right) + y_1 \][/tex]
Here:
- [tex]\(A = (x_1, y_1) = (2, -1)\)[/tex]
- [tex]\(B = (x_2, y_2) = (4, -3)\)[/tex]
- The ratio [tex]\( \frac{m}{m+n} = \frac{2}{3}\)[/tex], so [tex]\( m = 2 \)[/tex] and [tex]\( n = 1 \)[/tex]
Substituting the values into the coordinates equations, we get:
[tex]\[ x_P = \left(\frac{2}{3}\right) \left(4 - 2 \right) + 2 \][/tex]
[tex]\[ y_P = \left(\frac{2}{3}\right) \left(-3 + 1 \right) - 1 \][/tex]
First, calculate [tex]\( x_P \)[/tex]:
[tex]\[ x_P = \left(\frac{2}{3}\right) \cdot 2 + 2 \][/tex]
[tex]\[ x_P = \frac{4}{3} + 2 \][/tex]
[tex]\[ x_P = \frac{4}{3} + \frac{6}{3} \][/tex]
[tex]\[ x_P = \frac{10}{3} \][/tex]
[tex]\[ x_P = 3.3333 \][/tex]
Next, calculate [tex]\( y_P \)[/tex]:
[tex]\[ y_P = \left(\frac{2}{3}\right) \cdot (-2) - 1 \][/tex]
[tex]\[ y_P = \left(-\frac{4}{3}\right) - 1 \][/tex]
[tex]\[ y_P = -\frac{4}{3} - \frac{3}{3} \][/tex]
[tex]\[ y_P = -\frac{7}{3} \][/tex]
[tex]\[ y_P = -2.3333 \][/tex]
Thus, the coordinates of point [tex]\(P\)[/tex] are:
[tex]\[ x_P = 3.3333 \][/tex]
[tex]\[ y_P = -2.3333 \][/tex]
The coordinates of point [tex]\(P\)[/tex] are [tex]\((3.3333, -2.3333)\)[/tex].
