Chapter 2 Integers And Rational Numbers
Section 2.5: Find Distances On The Coordinate Plane
Page 95 Exercise 1 Answer
Plot the points (3,3), (0,0), (−4,−4), (−9,0), (−4,4), (0,0), (3,−3), and (3,3).
When the points are connected in order they make a fish.
You can use the structure of the grid to find a pair of points that are the same distance from the x-axis. The absolute value of the y-coordinate represents the distance of the point from the x-axis. Opposites have the same absolute value so points with opposite y-coordinates will be the same distance from the x-axis.
A pair of points that are the same distance from the x-axis are for example (−4,4) and (−4,−4), because their y-coordinates are opposites.
∣4∣ = ∣−4∣ = 4
When the points are connected in order they make a fish. The points (−4,4) and (−4,−4) are the same distance from the x-axis because their y-coordinates are opposites.
Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions
Page 95 Exercise 1 Answer
To find the total length of the picture graphed in the Solve & Discuss It! problem, we need to look at the point that is farthest to the left and the point that is farthest to the right.
The point farthest to the left is (−9,0) and the points farthest to the right are (3,−3) and (3,−3).
The length of the picture is the horizontal distance between these two points so we can look at the x-coordinates of these points.
For (−9,0), the horizontal distance to the origin is 9 since ∣−9∣ = 9.
For (3,−3) and (3,3), the horizontal distance to the origin is 3 since ∣3∣ = 3.
The total length of the picture is then 9 + 3 = 12 units.
Result
The total length of the picture is the horizontal distance between the point farthest on the left and the points farthest on the right. Since the x-coordinates of these points are −9 and 3, then the distance between the points is ∣−9∣ + ∣3∣ = 12 units.
Page 96 Exercise 1 Answer
The school is located at (−4,2) and the playground is located at (−4,5). Since the playground and the school have the same x-coordinates, we can use their y-coordinates to find the distance between them. They are in the same quadrant, Quadrant II, so to find the distance between them we must calculate the difference between their y-coordinates, which is ∣5∣ − ∣2∣ = 5 − 2 = 3.
Result
The distance from the school to the playground is 3 miles.
Page 97 Exercise 2 Answer
∣−85.5∣ + ∣65.5∣ = 85.5 + 65.5 = 151
The total distance of the Coulter’s return trip is 151 miles.
If we place the Water Park and the Coulter’s Home in a coordinate plane as points they are not in the same quadrants, the Water Park is in Quadrant III and the Coulter’s Home is in Quadrant IV, so we take the absolute value of their x-coordinates, since their y-coordinates are the same, and add them to find the total distance of their return trip.
Result
The total distance of the Coulter’s return trip is 151 miles.
Page 97 Exercise 3 Answer
The coordinates of point A are (5,7) and of point B (5,0) so the distance from point A to point B is 7, as seen in the Example 3. Point D is the same distance from point B as point A, so if we add the absolute values of the y-coordinates of point B and point D we must get the same number as if we added the absolute values point A and point B.
∣7∣ + ∣0∣ = 7 + 0 = 7
Since point D is in Quadrant IV it has a positive x-coordinate and a negative y-coordinate so the y-coordinate of point D is −7, and the x-coordinate is the same as that of points A and B.
Result
The coordinates of point D are (5,−7).
Page 98 Exercise 1 Answer
To find the distance between two points on a coordinate plane we can use absolute value by adding or subtracting the absolute values of their x-coordinates or y-coordinates.
For example, if A(4,−3) and B(4,8), then to the distance between A and B we must add the absolute values of their y-coordinates since their x-coordinates are the same.
∣−3∣ + ∣8∣ = −(−3) + 8 = 3 + 8 = 11
Thus, the distance from A to B is 11.
Another example, if C(2,7) and D(2,5), then to find the distance from C to D we must subtract their y-coordinates.
∣7∣ − ∣5∣ = 7 − 5 = 2
The distance from C to D is 2.
Result
To find the distance between two points on a coordinate plane we can use absolute value by adding or subtracting the absolute values of their x-coordinates or y-coordinates.
Page 98 Exercise 2 Answer
When two points are in the same quadrant we subtract their absolute values and when they are not in the same quadrant we add their absolute values to find the distance between them.
For example, point A(3,4) and B(7,4) are both in the first quadrant. They have the same y-coordinates, however, their x-coordinates differ. To find the distance from A to B we must subtract the absolute values of their x-coordinates.
∣7∣ − ∣3∣ = 7 − 3 = 4
Thus, the distance from A to B is four.
A different example, C(-3,4) and D(-3,-9) are not in the same quadrant, C is in the second quadrant and D in the third. They have the same x-coordinates and different y-coordinates. To find the distance between them we must add the absolute values of their y-coordinates.
∣4∣ + ∣−9∣ = 4 + (−(−9)) = 4 + 9 = 13
Thus, the distance from C to D is 13.
Result
When two points are in the same quadrant we subtract their absolute values and when they are not in the same quadrant we add their absolute values to find the distance between them.
Page 98 Exercise 3 Answer
From Example 1, Li’s house has coordinates of (−4,−3) and Tammy’s house has coordinates of (0,0).
Li’s house and Tammy’s house do not have the same x-coordinates or the same y-coordinates. Therefore, we cannot use absolute values to find the distance between Li’s house and Tammy’s house.
Result
No, we cannot use absolute values to find the distance because they don’t have the same x-coordinates or the same y-coordinates.
Page 98 Exercise 4 Answer
To find the distance from (-5,2) to (-5,6), since they have the same x-coordinates and are in the same quadrant, the second quadrant, we must subtract the absolute values of their y-coordinates.
∣6∣ − ∣2∣ = 6 − 2 = 4
Result
The distance between (−5,2) and (−5,6) is four.
Page 98 Exercise 5 Answer
To find the distance from (4.5, -3.3) to (4.5, 5.5), since they have the same x-coordinates and are not in the same quadrant, (4.5, -3.3) is in the second quadrant and (4.5, 5.5) in the first, so we must add the absolute values of their y-coordinates.
∣−3.3∣ + ∣5.5∣ = −(−3.3) + 5.5 = 3.3 + 5.5 = 8.8
Result
The distance between (4.5,−3.3) and (4.5,5.5) is 8.8.
Page 98 Exercise 6 Answer
To find the distance from (\(5 \frac{1}{2}\), –\(7 \frac{1}{2}\)) to (\(5 \frac{1}{2}\), –\(1 \frac{1}{2}\)), since they have the same x-coordinates and are in the same quadrant, the fourth quadrant, we must subtract the absolute values of their y-coordinates.
\(\left|-7 \frac{1}{2}\right|-\left|-1 \frac{1}{2}\right|=7 \frac{1}{2}-1 \frac{1}{2}=\frac{15}{2}-\frac{3}{2}=\frac{15-3}{2}=\frac{12}{2}=6\)Result
The distance between (\(5 \frac{1}{2}\), –\(7 \frac{1}{2}\)) and (\(5 \frac{1}{2}\), –\(1 \frac{1}{2}\)) is six.
Page 98 Exercise 7 Answer
To find the distance from (-\(2 \frac{1}{4}\),−8) to (\(7 \frac{3}{4}\),−8), since they have the same y-coordinates and are not in the same quadrant, (-\(2 \frac{1}{4}\),−8) is in the third quadrant and (\(7 \frac{3}{4}\),−8) in the fourth, so we must add the absolute values of their x-coordinates.
\(\left|-2 \frac{1}{4}\right|+\left|7 \frac{3}{4}\right|=2 \frac{1}{4}+7 \frac{3}{4}=\frac{9}{4}+\frac{31}{4}=\frac{9+31}{4}=\frac{40}{4}=10\)Result
The distance between (-\(2 \frac{1}{4}\),−8) and (\(7 \frac{3}{4}\),−8) is ten.
Page 98 Exercise 8 Answer
To find the distance from (\(5 \frac{1}{4}\), –\(\frac{1}{4}\)) to (\(5 \frac{1}{4}\), –\(6 \frac{1}{4}\)), since they have the same x-coordinates and are in the same quadrant, the fourth quadrant, we must subtract the absolute values of their y-coordinates.
\(\left|-6 \frac{1}{4}\right|-\left|-3 \frac{1}{4}\right|=6 \frac{1}{4}-3 \frac{1}{4}=\frac{25}{4}-\frac{13}{4}=\frac{25-13}{4}=\frac{12}{4}=3\)Result
The distance between (\(5 \frac{1}{4}\), –\(\frac{1}{4}\)) and (\(5 \frac{1}{4}\), –\(6 \frac{1}{4}\)) is three.
Page 98 Exercise 9 Answer
To find the distance from (-\(1 \frac{1}{2}\), –\(6 \frac{1}{2}\)) to (-\(2 \frac{1}{2}\), –\(6 \frac{1}{2}\)), since they have the same y-coordinates and are in the same quadrant, the third quadrant, we must subtract the absolute values of their x-coordinates.
\(\left|-2 \frac{1}{2}\right|-\left|-1 \frac{1}{2}\right|=2 \frac{1}{2}-1 \frac{1}{2}=\frac{5}{2}-\frac{3}{2}=\frac{5-3}{2}=\frac{2}{2}=1\)Result
The distance between (-\(1 \frac{1}{2}\), –\(6 \frac{1}{2}\)) and (-\(2 \frac{1}{2}\), –\(6 \frac{1}{2}\)) is one.
Page 99 Exercise 10 Answer
Since their y-coordinates are the same and (−2,8) is in Quadrant II and (7,8) in Quadrant I, to find the distance between (−2,8) and (7,8), we need to calculate the sum of absolute values of their x-coordinates.
∣−2∣ + ∣7∣ = 2 + 7 = 9
Result
The distance between (−2,8) and (7,8) is 9 units.
Page 99 Exercise 11 Answer
Since their x-coordinates are the same and both (−6.1,−8.4) and (−6.1,−4.2) are in Quadrant III, to find the distance between (−6.1,−8.4) and (−6.1,−4.2), we need to calculate the difference of absolute values of their y-coordinates.
∣−8.4∣ − ∣−4.2∣ = 8.4 − 4.2 = 4.2
Result
The distance between (−6.1,−8.4) and (−6.1,−4.2) is 4.2 units.
Page 99 Exercise 12 Answer
Since their y-coordinates are the same and (\(12 \frac{1}{2}\), \(3 \frac{3}{4}\)) is in Quadrant I and (-\(4 \frac{1}{2}\), \(3 \frac{3}{4}\)) in Quadrant II, to find the distance between (\(12 \frac{1}{2}\), \(3 \frac{3}{4}\)) and (-\(4 \frac{1}{2}\), \(3 \frac{3}{4}\)), we need to calculate the sum of absolute values of their x-coordinates.
\(\left|12 \frac{1}{2}\right|+\left|-4 \frac{1}{2}\right|=12 \frac{1}{2}+4 \frac{1}{2}=\frac{25}{2}+\frac{9}{2}=\frac{34}{2}=17\)Result
The distance between (\(12 \frac{1}{2}\), \(3 \frac{3}{4}\)) and (-\(4 \frac{1}{2}\), \(3 \frac{3}{4}\)) is 17 units.
Page 99 Exercise 13 Answer
Since their x-coordinates are the same and both (−5,−3) and (−5,−6) are in Quadrant III, to find the distance between (−5,−3) and (−5,−6), we need to calculate the difference of absolute values of their y-coordinates.
∣−6∣ − ∣−3∣ = 6 − 3 = 3
Result
The distance between (−5,−3) and (−5,−6) is 3 units.
Page 99 Exercise 14 Answer
Since their y-coordinates are the same and (−5.4,4.7) is in Quadrant II and (0.6,4.7) in Quadrant I, to find the distance between (−5.4,4.7) and (0.6,4.7), we need to calculate the sum of absolute values of their x-coordinates.
∣0.6∣ + ∣−5.4∣ = 0.6 + 5.4 = 6
Result
The distance between (−5.4,4.7) and (0.6,4.7) is 6 units.
Page 99 Exercise 15 Answer
Since their x-coordinates are the same and both (\(7 \frac{1}{2}\), –\(5 \frac{3}{4}\)) and (\(7 \frac{1}{2}\), –\(1 \frac{1}{4}\)) is in Quadrant IV, to find the distance between (\(7 \frac{1}{2}\), –\(5 \frac{3}{4}\)) and (\(7 \frac{1}{2}\), –\(1 \frac{1}{4}\)), we need to calculate the difference of absolute values of their y-coordinates.
\(\left|-5 \frac{3}{4}\right|-\left|-1 \frac{1}{4}\right|=5 \frac{3}{4}-1 \frac{1}{4}=\frac{23}{4}-\frac{5}{4}=\frac{23-5}{4}=\frac{18}{4}=\frac{16}{4}+\frac{2}{4}=4 \frac{2}{4}=4 \frac{1}{2}\)Result
The distance between (\(7 \frac{1}{2}\), –\(5 \frac{3}{4}\)) and (\(7 \frac{1}{2}\), –\(1 \frac{1}{4}\)) is \(4 \frac{1}{2}\) units.
Page 99 Exercise 16 Answer
First, we need to find the coordinates which represent the location of roller coaster 1 and the swings. Follow the grid lines directly to the x-axis to find the x-coordinates. Follow the grid lines directly to the y-axis to find the y-coordinates. The coordinates of roller coaster 1 are (−6,7) and the coordinates of the swings are (1,7). Since (−6,7) is in Quadrant II and (1,7) is Quadrant I and their y-coordinates are the same, to find the distance from roller coaster 1 to the swings the sum of absolute values of their x-coordinate.
∣1∣ + ∣−6∣ = 1 + 6 = 7
Result
The distance from roller coaster 1 to the swings is 7 units.
Page 99 Exercise 17 Answer
First, we need to find the coordinates of the Ferris wheel and the roller coaster 3. Follow the grid lines directly to the x-axis to find the x-coordinates. Follow the grid lines directly to the y-axis to find the y-coordinates. The coordinates of the Ferris wheel (-6,2) and the coordinatess of the roller coaster 3 (-6,-8).
Since (-6,2) is in Quadrant II and (-6,-8) is in Quadrant III and their x-coordinates are the same, to find the distance between (-6,2) and (-6,-8) we need to calculate the sum of their y-coordinates.
∣2∣ + ∣−8∣ = 2 + 8 = 10
Result
The distance between the Ferris wheel and the roller coaster 3 is 10 units.
Page 99 Exercise 18 Answer
The total distance from roller coaster 2 to roller coaster 3 and then to the water slide is the sum of the distance from roller coaster 2 to roller coaster 3 and from roller coaster 3 to the water slide. First we must find their coordinates. Follow the grid line directly to the x-axis to find the x-coordinates. Follow the grid line directly to the y-axis to find the y-coordinates.
The coordinates of roller coaster 2 are (-6,-3), of roller coaster 3 are (-6,-8), and of the water slide are (8, -8). Both (-6,-3) and (-6,-8) are in Quadrant III and (8, -8) is in Quadrant IV.
To find the distance between roller coaster 2 and roller coaster 3, since their x-coordinates and they are in the same quadrant, we need to calculate the difference between their y-coordinates.
∣−8∣ − ∣−3∣ = 8 − 3 = 5
The distance between roller coaster 2 and roller coaster 3 is 5 units.
To find the distance between roller coaster 3 and the water slide, since their y-coordinates and (-6,-8) is in Quadrant III and (8, -8) is in Quadrant IV, we need to calculate the sum between their x-coordinates.
∣−6∣ + ∣8∣ = 6 + 8 = 14
The distance between roller coaster 3 and the water slide is 14 units.
The total distance from roller coaster 2 to roller coaster 3 and then to the water slide is the sum
5 + 14 = 19.
Result
The total distance from roller coaster 2 to roller coaster 3 and then to the water slide is 19 units.
Page 99 Exercise 19 Answer
Since distance is defined as the positive length between two points, the distance between two points is always the same no matter which point is chosen as the starting point.
This means the distance between the merry-go-round and the water slide is the same as the distance between the water slide and merry-go-round since the two distances represent the same positive length between the two locations.
Result
The distances are the same since distance between two points is always the same no matter which point is chosen as the starting point.
Page 100 Exercise 20 Answer
H is graphed at (n,−6) and J at (n,−3). Since H and J have the same x-coordinate and are in the same quadrant, to find the distance between them we must caculate the difference of the absolute values of their y-coordinates.
∣−6∣ − ∣−3∣ = 6 − 3 = 3
Result
The distance between H and J is 3 units.
Page 100 Exercise 21 Answer
From the graph we can see that points H and G have the same y-coordinates, which are both equal to -6. Since point G is on the y-axis, its x-coordinate is zero. So, point G is graphed at (0,-6).
The distance between point H and point G is equal to the distance of point H to point J which is 3 units as calculated in the Exercise 20 on page 100.
To find the value of n, we must calcualte the following:
∣n∣ − ∣0∣ = 3
n − 0 = 3
n = 3
Result
n = 3
Page 100 Exercise 22 Answer
Since a, b, and c are all negative numbers, both points (a,b) and (a,c) are in Quadrant III. Their x-coordinates are the same so to find the distance between (a,b) and (a,c) we need to calculate the difference of the absolute values of their y-coordinates.
We don’t know whether b is greater than c or if c is greater than b. Since the distance is always represented by a positive number, we will take the absolute value of the difference.
∣∣b∣ − ∣c∣∣ = ∣b−c∣
Suppose b is greater then c, then the distance is b − c. However if c is greater than b, then the distance is c − b.
Result
If b is greater than c, then the distance is b − c and if c is greater than b, then the distance is c − b.
Page 100 Exercise 23 Answer
To answer the question, we must find the distance between points (-1,8) and (85,8). When placed in a coordinate plane, the epicenter (-1,8) is in Quadrant II and the farthest location reported to have felt the earthquake (85,8) is in Quadrant I. Since their y-coordinates are the same, to find the distance between (-1,8) and (85,8) we need to calculate the sum of their x-coordinates.
∣−1∣ + ∣85∣ = 1 + 85 = 86
Result
The earthquake was felt 86 miles from the epicenter.
Page 100 Exercise 24 Answer
To find the dimensions of the rectangular piece of land we need to find the distance between points A and B, and the distance between points A and D. From the graph we can determine the coordinates of points A, B, and D, which are in order (0,0), (0, \(4 \frac{1}{2}\)), and (\(5 \frac{1}{2}\),0).
Both points A and B are on the y-axis so their x-coordinates are zero. The distance between them is the difference of absolute values of their y-corodinates.
\(\left|4 \frac{1}{2}\right|-|0|=4 \frac{1}{2}-0=4 \frac{1}{2}\)The distance between points A and B is \(4 \frac{1}{2}\) units.
\(4 \frac{1}{2} \times 1,000=\frac{9}{2} \times \frac{1000}{1}=\frac{9 \times 1000}{2 \times 1}=\frac{9000}{2}=4500\)Since each unit represents 1,000 feet, the distance between points A and B is 4,500 feet.
Both points A and D are on the x-axis so their y-coordinates are zero. The distance between them is the difference of absolute values of their x-corodinates.
\(\left|5 \frac{1}{2}\right|-|0|=5 \frac{1}{2}-0=5 \frac{1}{2}\) \(5 \frac{1}{2} \times 1,000=\frac{11}{2} \times \frac{1000}{1}=\frac{11 \times 1000}{2 \times 1}=\frac{11000}{2}=5500\)The distance between points A and D is 5,500 feet.
Result
The width of the rectangular piece of land is 4,500 feet and the length is 5,500 feet.
Page 100 Exercise 25 Answer
To find two ordered pairs that are 4.5 units apart, we can choose from the ones which have either the same x-coordinates or the same y-coordinates, and from those which are in the same quadrant or those that are not. For example, (1,1) and (1,5.5) are both in Quadrant I, have the same x-coordinates, and are 4.5 units apart.
∣5.5∣ − ∣1∣ = 5.5 − 1 = 4.5
Points (3.5,−1) and (−1,−1) have the same y-quadrant, but point (3.5,−1) is in Quadrant IV and point (−1,−1) is in Quadrant III, so the distance among them is the sum of absolute values of their x-coordinates.
∣3.5∣ + ∣−1∣ = 3.5 + 1 = 4.5
Result
(1,1) and (1,5.5), (3.5,−1) and (−1,−1)