b Interpret the points remembering what x and y represent.
C
c To calculate the slope, substitute the points from Part B into the Slope Formula.
D
d Marisa will be 0 blocks away from Sanjay's home once she reaches it.
A
a Marisa's speed is constant.
B
b x-variable: minutes
y-variable: blocks to Sanjay's house Points on the line: (0,12) and (6,8) Explanation: See solution.
C
cSlope: - 2/3
Interpretation: See solution.
D
dEquation: y-8=- 2/3(x-6)
It will take 18 minutes for Marisa to reach Sanjay's home.
Practice makes perfect
a An equation is linear if its graph is a line. Straight lines have a constant rate of change. Therefore, in order to treat this situation as linear, we need to assume that the rate of change, or slope, is constant. Thus, we should assume that Marisa is walking with a constant speed.
b The number of blocks Marisa walked depends on the number of minutes that have passed. We can use this to define our variables.
& x= time (in minutes)
& y= distance from Sanjay's house (in blocks)
Now, let's identify two points on the line. We know that Marisa started 12 blocks away from Sanjay's home. Because this will be when she has walked for 0 minutes, we have the point (0,12). We also know that, after 6 minutes, she was 8 blocks away from his house. Therefore, a second point is (6,8).
c To find the slope, let's substitute the two points from Part B into the Slope Formula.
The slope is - 23. It tells us that the distance to the Sanjay's home is decreasing by 23 of a block every minute Marisa is walking. We can also think of this as Marisa walking two blocks closer every three minutes.
-2 blocks/3minutes
d Let's write the equation in point-slope form.
y-y_1=m(x-x_1)To do this, we can substitute the value of the slope and one of the points from Part B into this form. Let's use (6,8).
Now, we can use this equation to find the number of minutes it takes Marisa to reach Sanjay's home. We want to solve the equation for x when Marisa is 0 blocks away. Thus, let's substitute 0 for y, and solve for x.
