Answer:
[tex]P(\frac{10}{3}, \frac{-2}{3})[/tex]
Step-by-step explanation:
The question is incomplete; However
[tex]A = (-2, -4)[/tex]
[tex]B = (6, 1)[/tex]
Required
Determine coordinates of P
When line segment is divided in ratios, the following formula calculates the coordinates;
[tex]P(x,y) = \{\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\}[/tex]
In this case;
[tex]m:n = 2:1[/tex]
[tex](x_1,y_1) = (-2, -4)[/tex]
[tex](x_2,y_2) = (6, 1)[/tex]
So, the coordinates of P is calculated as thus
[tex]P(x,y) = \{\frac{2 * 6 + 1 * (-2)}{2+1}, \frac{2 * 1 + 1 * (-4)}{2+1}\}[/tex]
[tex]P(x,y) = \{\frac{12 -2}{3}, \frac{2 -4}{3}\}[/tex]
[tex]P(x,y) = \{\frac{10}{3}, \frac{-2}{3}\}[/tex]
Hence, the coordinates of P is
[tex]P(\frac{10}{3}, \frac{-2}{3})[/tex]
The question is incomplete, the points A and B are:
A = (-2, -4)
B = (6, 1)
We want to find a point P in the segment AB such that the ratio of AP to PB is 2:1
We will find that:
[tex]P = (\frac{2}{3} , \frac{7}{3} )[/tex]
The general formula for two points (x₁, y₁) and (x₂, y₂), the coordinates of a point that separates the segment in a ratio j:k is
[tex]P = (\frac{j*x_1 + k*x_2}{j + k} , \frac{j*y_1 + k*y_2}{j + k})[/tex]
So we only need to use that general formula:
[tex]P = P = (\frac{2*(-2) + 1*6}{3} , \frac{2*(-4) + 1*1}{3})\\\\P = (\frac{2}{3} , \frac{7}{3} )[/tex]
If you want to learn more, you can read:
https://brainly.com/question/17623648
