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Chapter 2: Problem 6

Plot each pair of points and calculate the slope of the line that passes through them. a. (3,5) and (8,15) d. (-2,6) and (2,-6) b. (-1,4) and (7,0) e. (-4,-3) and (2,-3) c. (5,9) and (-5,9)

Short Answer

Expert verified
The slopes are 2, -3, -1/2, 0, and 0.

Step by step solution

01

Plot Points for Pair a

Plot the points (3,5) and (8,15) on a graph.
02

Calculate the Slope for Pair a

Use the slope formula \(\frac{y_2 - y_1}{x_2 - x_1}\) where \(x_1 = 3\), \(y_1 = 5\), \(x_2 = 8\), and \(y_2 = 15\). So, \frac{15 - 5}{8 - 3} = 2\
03

Plot Points for Pair d

Plot the points (-2,6) and (2,-6) on a graph.
04

Calculate the Slope for Pair d

Use the slope formula \(\frac{y_2 - y_1}{x_2 - x_1}\) where \(x_1 = -2\), \(y_1 = 6\), \(x_2 = 2\), and \(y_2 = -6\). So, \frac{-6 - 6}{2 - (-2)} = -3\
05

Plot Points for Pair b

Plot the points (-1,4) and (7,0) on a graph.
06

Calculate the Slope for Pair b

Use the slope formula \(\frac{y_2 - y_1}{x_2 - x_1}\) where \(x_1 = -1\), \(y_1 = 4\), \(x_2 = 7\), and \(y_2 = 0\). So, \frac{0 - 4}{7 - (-1)} = -\frac{1}{2}\
07

Plot Points for Pair e

Plot the points (-4,-3) and (2,-3) on a graph.
08

Calculate the Slope for Pair e

Use the slope formula \(\frac{y_2 - y_1}{x_2 - x_1}\) where \(x_1 = -4\), \(y_1 = -3\), \(x_2 = 2\), and \(y_2 = -3\). So, \frac{-3 - (-3)}{2 - (-4)} = 0\
09

Plot Points for Pair c

Plot the points (5,9) and (-5,9) on a graph.
10

Calculate the Slope for Pair c

Use the slope formula \(\frac{y_2 - y_1}{x_2 - x_1}\) where \(x_1 = 5\), \(y_1 = 9\), \(x_2 = -5\), and \(y_2 = 9\). So, \frac{9 - 9}{-5 - 5} = 0\

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Plotting Points
Before we dive into calculating the slope, we need to understand how to plot points on a coordinate plane. Each point is a pair of coordinates, written as \((x, y)\). The first number, \x\, tells you how far to move horizontally from the origin (the point where the x-axis and y-axis intersect). Move right if \x\ is positive and left if it's negative. The second number, \y\, tells you how far to move vertically. Move up if \y\ is positive and down if it's negative.

Let's plot the points given in your exercise:
  • (3,5) and (8,15): Move right 3 and up 5 for the first point, then right 8 and up 15 for the second point.
  • (-2,6) and (2,-6): Move left 2 and up 6 for the first point, then right 2 and down 6 for the second point.
  • (-1,4) and (7,0): Move left 1 and up 4 for the first point, then right 7 and don't move vertically for the second point.
  • (-4,-3) and (2,-3): Move left 4 and down 3 for the first point, then right 2 and down 3 for the second point.
  • (5,9) and (-5,9): Move right 5 and up 9 for the first point, then left 5 and up 9 for the second point.

Plotting the points correctly is essential as it sets the foundation for calculating the slope. Once our points are plotted, we can visually see the trend of the line.
Slope Formula
The slope of a line indicates how steep the line is. It's calculated using the slope formula:
\(\text{slope} = \frac{y_2 - y_1}{x_2 - x_1}\). Here, \(x_1\) and \(y_1\) are the coordinates of the first point, and \(x_2\) and \(y_2\) are the coordinates of the second point. This formula finds the rate of change between two points and gives you the 'rise over run.'

Let's calculate the slope for each pair:
  • For (3,5) and (8,15): \(\frac{15 - 5}{8 - 3} = 2\) means the line rises 2 units for each 1 unit it runs.
  • For (-2,6) and (2,-6): \(\frac{-6 - 6}{2 - (-2)} = -3\) means the line falls 3 units for every unit it runs.
  • For (-1,4) and (7,0): \(\frac{0 - 4}{7 - (-1)} = -\frac{1}{2}\) means the line falls 0.5 units for each 1 unit it runs.
  • For (-4,-3) and (2,-3): \(\frac{-3 - (-3)}{2 - (-4)} = 0\), there's no rise or fall since both \(y\) coordinates are the same.
  • For (5,9) and (-5,9): \(\frac{9 - 9}{-5 - 5} = 0\) also has no rise or fall as both \- coordinates remain the same.
Understanding the slope formula helps us grasp how steep or flat a line is between two points.
Graphing Lines
Finally, let's talk about graphing lines using the points and slopes. Once you have calculated the slope, you can draw the line connecting the points by following the 'rise over run' rule. Start from one point and apply the slope to find another point on the line.

Here's how we can graph each pair of points:
  • For (3,5) and (8,15) with a slope of 2: From point (3,5), move up 2 units and right 1 unit repeatedly until you reach (8,15).
  • For (-2,6) and (2,-6) with a slope of -3: From point (-2,6), move down 3 units and right 1 unit repeatedly to reach (2,-6).
  • For (-1,4) and (7,0) with a slope of -0.5: From point (-1,4), move down 1 unit and right 2 units repeatedly to reach (7,0).
  • For (-4,-3) and (2,-3) with a slope of 0: Plot a horizontal line passing through both points since there's no vertical change.
  • For (5,9) and (-5,9) with a slope of 0: Also, plot a horizontal line since there's no vertical change.

Drawing accurate lines helps visualize the relationship between points on a graph and emphasizes the importance of the slope in determining the direction and steepness of a line.

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Most popular questions from this chapter

Construct a linear equation for each of the following conditions. a. A negative slope and a positive \(y\) -intercept b. A positive slope and a vertical intercept of -10.3 c. A constant rate of change of \(\$ 1300 /\) year

The \(y\) -axis, the \(x\) -axis, the line \(x=6,\) and the line \(y=12\) determine the four sides of a 6 -by- 12 rectangle in the first quadrant (where \(x>0\) and \(y>0\) ) of the \(x y\) plane. Imagine that this rectangle is a pool table. There are pockets at the four corners and at the points (0,6) and (6,6) in the middle of each of the longer sides. When a ball bounces off one of the sides of the table, it obeys the "pool rule": The slope of the path after the bounce is the negative of the slope before the bounce. (Hint: It helps to sketch the pool table on a piece of graph paper first.) a. Your pool ball is at (3,8) . You hit it toward the \(y\) -axis, along the line with slope \(2 .\) i. Where does it hit the \(y\) -axis? ii. If the ball is hit hard enough, where does it hit the side of the table next? And after that? And after that? iii. Show that the ball ultimately returns to \((3,8) .\) Would it do this if the slope had been different from \(2 ?\) What is special about the slope 2 for this table? b. A ball at (3,8) is hit toward the \(y\) -axis and bounces off it at \(\left(0, \frac{16}{3}\right) .\) Does it end up in one of the pockets? If so, what are the coordinates of that pocket? c. Your pool ball is at (2,9) . You want to shoot it into the pocket at \((6,0) .\) Unfortunately, there is another ball at (4,4.5) that may be in the way. i. Can you shoot directly into the pocket at (6,0)\(?\) ii. You want to get around the other ball by bouncing yours off the \(y\) -axis. If you hit the \(y\) -axis at \((0,7),\) do you end up in the pocket? Where do you hit the line \(x=6 ?\) iii. If bouncing off the \(y\) -axis at (0,7) didn't work, perhaps there is some point \((0, b)\) on the \(y\) -axis from which the ball would bounce into the pocket at (6,0) Try to find that point.

Complete the table for each of the linear functions, and then sketch a graph of each function. Make sure to choose an appropriate scale and label the axes. a. $$ \begin{array}{rl} \hline x & f(x)=0.10 x+10 \\ \hline-100 & \\ 0 & \\ 100 & \\ \hline \end{array} $$ b. $$ \begin{array}{ll} \hline x & h(x)=50 x+100 \\ \hline-0.5 & \\ 0 & \\ 0.5 & \end{array} $$

Which pairs of points produce a line with a negative slope? a. (-5,-5) and (-3,-3) d. (4,3) and (12,0) h. (-2,6) and (-1,4) e. (0,3) and (4,-10) c. (3,7) and (-3,-7) f. (4,2) and (6,2)

(Graphing program optional.) The accompanying table indicates the number of juvenile arrests (in thousands) in the United States for aggravated assault. $$ \begin{array}{ccc} \hline & \begin{array}{c} \text { Juvenile } \\ \text { Arrests } \\ \text { Year } \end{array} & \begin{array}{c} \text { Annual Average } \\ \text { (thousands) } \end{array} & \begin{array}{c} \text { Rate of Change } \\ \text { over Prior 5 Years } \end{array} \\ \hline 1985 & 36.8 & \text { n.a. } \\ 1990 & 54.5 & \\ 1995 & 68.5 & \\ 2000 & 49.8 & \\ 2005 & 36.9 & \\ \hline \end{array} $$ a. Fill in the third column in the table by calculating the annual average rate of change. b. Graph the annual average rate of change versus time. c. During what 5 -year period was the annual average rate of change the largest? d. Describe the change in aggravated assault cases during these years by referring both to the number and to the annual average rate of change.
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