Question

1. model with mathematics the table shows locations of several sites at a high school campus. a landscaper wants to connect two sites with a path perpendicular to the path connecting the cafeteria and the library. which two sites should he connect?

location
cafeteria (5, 5) library (11, 14)
office (4, 12) gym (15, 8)
woodshop (11, 6) art studio (3, 16)

Explanation:

Step1: Find slope of line between cafeteria and library

The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$. For cafeteria $(x_1,y_1)=(5,5)$ and library $(x_2,y_2)=(11,14)$. So $m_{1}=\frac{14 - 5}{11 - 5}=\frac{9}{6}=\frac{3}{2}$.

Step2: Find slope of perpendicular line

The slope of a line perpendicular to a line with slope $m$ is $m_{2}=-\frac{2}{3}$ (since the product of slopes of perpendicular lines is - 1, $m_1\times m_2=-1$).

Step3: Check slopes between other - site pairs

For office $(4,12)$ and gym $(15,8)$: $m_{3}=\frac{8 - 12}{15 - 4}=\frac{-4}{11}$.
For office $(4,12)$ and wood - shop $(11,6)$: $m_{4}=\frac{6 - 12}{11 - 4}=\frac{-6}{7}$.
For office $(4,12)$ and art studio $(3,16)$: $m_{5}=\frac{16 - 12}{3 - 4}=\frac{4}{-1}=-4$.
For gym $(15,8)$ and wood - shop $(11,6)$: $m_{6}=\frac{6 - 8}{11 - 15}=\frac{-2}{-4}=\frac{1}{2}$.
For gym $(15,8)$ and art studio $(3,16)$: $m_{7}=\frac{16 - 8}{3 - 15}=\frac{8}{-12}=-\frac{2}{3}$.
For wood - shop $(11,6)$ and art studio $(3,16)$: $m_{8}=\frac{16 - 6}{3 - 11}=\frac{10}{-8}=-\frac{5}{4}$.

Answer:

Gym and Art Studio