Exercise 31 Page 161

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
$D$	$\frac{D}{4}$	$\frac{D}{4}+D$

By dividing the distance from your house to your school by the distance from your house to the movies, we can find the ratio of these distances. 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 slope of the line segment. Knowing the slope, we can write the following equations for finding the $x\text{-}$ and $y\text{-}$coordinates of $P.$ From the coordinates of your house, $(\text{-}4,\text{-}2),$ we have to move $\frac{9}{5}$ steps horizontally in the positive direction, and $\frac{4}{5}$ steps vertically in the positive direction, to arrive at your school. Let's calculate the $x\text{-}$ and $y\text{-}$coordinate of the school: The school is at $\left(\text{-}\frac{11}{5},\text{-}\frac{6}{5}\right).$