If we call the distance from the school to the movie theater D, then the distance from your house to your school will be one fourth of this. Let's show this in a table.
school to movies
|
house to school
|
house to movies
|
D
|
D/4
|
D/4+D
|
By dividing the distance from your house to your school by the distance from your house to the movies, we can find the of these distances.
.D/4 /(D/4+d).
.D/4 /(D/4+4D/4).
.D/4 /5D/4.
D/4* 4/5D
4D/20D
1/5
The distance from your house to the school is one-fifth of the distance between your house and the movie theater. Because we are working with fifths, we have to divide the distance D into five congruent pieces. Let's call the length of these d.
To arrive at your school, you have to go one-fifth of the
run
, as well as one-fifth of the
rise
, when starting from your house. Thus, we first have to find the of the .
m=y_2-y_1/x_2-x_1
m=2-( - 2)/5-( - 4)
m=2+2/5+4
m=4/9
Knowing the slope, we can write the following equations for finding the x- and y-coordinates of P.
Rise: &1/5* 4= 4/5 [0.8em]
Run: &1/5* 9= 9/5
From the coordinates of your house, (-4, - 2), we have to move 95 steps horizontally in the positive direction, and 45 steps vertically in the positive direction, to arrive at your school. Let's calculate the x- and y-coordinate of the school:
y-coordinate: & - 2+ 4/5=- 6/5 [0.8em]
x-coordinate: &- 4+ 9/5= - 11/5
The school is at (- 115,- 65).