Answer:
(-0.75, -3.25) and (4.125, -2.125)
Step-by-step explanation:
Let the town, Thayer = point A(-4, -4)
Town Purdy = point D(9, -1).
Let B and C represent the two rest stops that divides the highway (AD) into 3 equal parts, namely: AB = BC = CD.
Distance of point B from A to D = ⅓ of AD.
Point C = the midpoint of point B and D.
First, using the formula for internal division, let's find the coordinates of point B:
Internal division to find the coordinates of point B is given as:
[tex] x = \frac{mx_2 + nx_1}{m + n} [/tex]
[tex] y = \frac{my_2 + ny_1}{m + n} [/tex]
Where,
[tex] A(-4, -4) = (x_1, y_1) [/tex]
[tex] D(9, -1) = (x_2, y_2) [/tex]
[tex] m = 1, n = 3 [/tex]
Plug in the necessary values into the formula stated above to solve for x and y of point B:
[tex] x = \frac{1*9 + 3*(-4)}{1 + 3} [/tex]
[tex] x = \frac{9 - 12}{4} [/tex]
[tex] x = \frac{-3}{4} [/tex]
[tex] x = -\frac{-3}{4} = 0.75 [/tex]
[tex] y = \frac{1*(-1) + 3*(-4)}{m + n} [/tex]
[tex] y = \frac{-1 - 12}{1 + 3} [/tex]
[tex] y = \frac{-13}{4} [/tex]
[tex] y = -3.25 [/tex]
The coordinates of point B = (-0.75, -3.25)
Next, find the coordinates of point C, which is the midpoint of B(-0.75, -3.25) and D(9, -1) using the midpoint formula.
Midpoint (C) of BC, for B(-0.75, -3.25) and D(9, -1) is given as:
[tex] M(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}) [/tex]
Let [tex] B(-0.75, -3.25) = (x_1, y_1) [/tex]
[tex] D(9, -1) = (x_2, y_2) [/tex]
Thus:
[tex] C(\frac{-0.75 + 9}{2}, \frac{-3.25 + (-1)}{2}) [/tex]
[tex] C(\frac{8.25}{2}, \frac{-4.25}{2}) [/tex]
[tex] C(4.125, -2.125) [/tex]
The rest stops should be built at (-0.75, -3.25) and (4.125, -2.125) in the coordinate plane.
