Chapter 2 Integers And Rational Numbers
Section 2.6: Represent Polygons On The Coordinate Plane
Page 101 Exercise 1 Answer
Since we know the vertices of the polygon we can graph them in a coordinate plane and then calculate the distance from one vertice to another to find the perimeter of the polygon.
The distance from point A to point B is the difference of absolute values of their x-coordinates, since they have the same y-coordinates and are both in Quadrant II.
∣−7∣ − ∣−1∣ = 7 − 1 = 6
The distance from point A to point B is 6 units.
The distance from point B to point C is the sum of absolute values of their y-coordinates, since they have the same x-coordinates and are not in the same quadrant.
∣6∣ + ∣−3∣ = 6 + 3 = 9
Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions
The distance from point B to point C is 9 units.
We can see from the graph that the polygon is a rectangle, so side AB is of equal length as side CD, and side BC is of equal length as side AD. However, we can check.
The distance from point C to point D is the difference of absolute values of their x-coordinates, since they have the same y-coordinates and are both in Quadrant III, thus the distance from point C to point D is 6 units.
The distance from point A to point D is the sum of absolute values of their y-coordinates, since they have the same x-coordinates and are not in the same quadrant, thus the distance from point B to point C is 9 units.
The perimeter is the sum of the lengths of all sides. 6 + 9 + 6 + 9 = 30
Result
The perimeter of the polygon is 30 units.
Page 101 Exercise 1 Answer
The polygon drawn in the Solve & Discuss It! exercise is a rectangle since it is a quadrilateral with four right angles.
Points A(−1,6), B(−7,6), C(−7,−3), and D(−1,−3) are vertices of this polygon. Since lines segments AB and CD are both parallel to the x-axis, and line segments BC and AD are both parallel to the y-axis, angles ∠ABC, ∠BCD, ∠CDA, and ∠DAB are all right angles which means this polygon is a rectangle.
Result
Rectangle
Page 102 Exercise 1 Answer
First we need to find the length of each side of the new rectangle.
The vertices of the new rectangle are A(−4,6), B(2,6), Working Tent, and Food Tent. To find the coordinates of the Working Tent and the Food Tent 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 Working Tent is at (−4,−2) and the Food Tent is at (2,−2).
A to B: ∣−4∣ + ∣2∣ = 4 + 2 = 6
B to food tent: ∣6∣ + ∣−2∣ = 6 + 2 = 8
Food tent to working tent: ∣2∣ + ∣−4∣ = 2 + 4 = 6
Working tent to A: ∣−2∣ + ∣6∣ = 2 + 6 = 8
6 + 8 + 6 + 8 = 28
The sum of the lengths of all sides is 28, so the archaeologist needs 28 meters of rope.
To find the perimeter of the larger rectangle we could have used just the distance from A to B and the distance from A to the Working Tent. Since points A and the Working Tent, and points B and the Food Tent have the same x-coordinates, and points A and B, and points the Working Tent and the Food tent have the same y-coordinates, the distance from A to the Working Tent and the distance from B to the Food Tent are the same, so is the distance from A to B, and from the Working Tent to the Food Tent, are the same as well.
Result
A to B = 6 m Food tent to working tent = 6 m
B to food tent: 8 m Working tent to A = 28 m
The archeologist needs 28 meters of rope.
Page 103 Exercise 2 Answer
Find the side lengths. For all the points which have the same x-coordinates and are in the same quadrant, find the difference of the absolute values of their y-coordinates. For all the points which have the same y-coordinates and are in the same quadrant, find the difference of the absolute values of their x-coordinates. If they are in different quadrants, find the sum instead of the difference.
AB = ∣15∣ − ∣4.5∣ = 15 − 4.5 = 10.5
BC = ∣14∣ − ∣8∣ = 14 − 8 = 6
DC = ∣15∣ − ∣10∣ = 15 − 10 = 5
DE = ∣8∣ − ∣3∣ = 8 − 3 = 5
EF = ∣15∣ − ∣10∣ = 15 − 10 = 5
FG = ∣3∣ + ∣−2.25∣ = 3 + 2.25 = 5.25
HG = ∣15∣ − ∣4.5∣ = 15 − 4.5 = 10.5
AH = ∣14∣ + ∣−2.25∣ = 14 + 2.25 = 16.25
10.5 + 6 + 5 + 5 + 5 + 5.25 + 10.5 + 16.25 = 63.5
The perimeter is the sum of all the lengths which is 63.5.
Result
The rancher needs 63.5 yards of fencing.
Page 103 Exercise 3 Answer
If quadrilateral ADEF is a square, then all of its sides have an equal length. Let’s check.
The length of side AD = ∣−5∣ + ∣7∣ = 5 + 7 = 12 units.
The length of side AF = ∣2∣ + ∣−8∣ = 2 + 8 = 10 units.
The length of side DE = ∣2∣ + ∣−8∣ = 2 + 8 = 10 units.
The length of side EF = ∣−5∣ + ∣7∣ = 5 + 7 = 12 units.
Not all the sides have an equal length, so quadrilateral ADEF is not a square.
Result
Quadrilateral ADEF is not a square since not all the sides have an equal length.
Page 104 Exercise 1 Answer
When a polygon is placed in a coordinate plane, its vertices are all assigned specific unique coordinates.
The lengths of the sides can be calculated by adding or subtracting the absolute values of the coordinates to find the distances between the vertices. When we know the lengths of sides of polygons we can further calculate its perimeter and its area.
For example, if we have a rectangle whose vertices are A(−2,7), B(5,7), C(5,−3), and D(−2,−3). The lengths of its side are then:
AB = ∣−2∣ + ∣5∣ = 2 + 5 = 7
BC = ∣7∣ + ∣−3∣ = 7 + 3 = 10
CD = ∣5∣ + ∣−2∣ = 5 + 2 = 7
AD = ∣7∣ + ∣−3∣ = 7 + 3 = 10
Its area is 7 × 10 = 70 units, and its perimeter is 7 + 10 + 7 + 10 = 34 units.
Result
When a polygon is placed in a coordinate plane, its vertices are all assigned specific unique coordinates.
The lengths of the sides can be calculated by adding or subtracting the absolute values of the coordinates to find the distances between the vertices. When we know the lengths of sides of polygons we can further calculate its perimeter and its area.
Page 104 Exercise 2 Answer
In Example 1, points A and B are in different quadrants. If we imagine walking the distance straight from A to B, first we must cross four units to get to the y-axis, since A is at (-4, 6), and then another two units to get to point B, which is at (2, 6).
Thus, to find the distance from A to B we need to add the units, to add the distance from A to the y-axis and the distance from B to the y-axis, which are actually the x-coordinates of the points.
However, points B and C are in the same quadrant, Quadrant I. If we imagin wlaking a straight line from B to C, we won’t cross either of the axes. Since B is at (2, 6) and C at (2, 1), we need to cross five units to get from B to C.
Thus, to find the distance from B to C we need to subtract the units, to subtract the distance from B to the x-axis and the distance from C to the y-axis, which are actually the y-coordinates of the points.
Result
A and B are in different quadrants so we need to add to find the distance. However, B and C are in the same quadrant so we need to subtract to find the distance.
Page 104 Exercise 3 Answer
To find the length of the diagonal AC we can neither add nor subtract the absolute values of coordinates since points A and C have different both the x-coordinates and the y-coordinates.
Result
To find the length of the diagonal AC we can neither add nor subtract the absolute values of coordinates since points A and C have different both the x-coordinates and the y-coordinates.
Page 104 Exercise 4 Answer
The vertices of rectangle MNOP are M(−2,5), N(−2,−4), O(3,−4), and P(3,5). To find the lengths of sides we need to calculate the distances from one vertice to another.
Points M and N are in different quadrants and have the same x-coordinates, so to find the distance from M to N we need to add the absolute values of their y-coordinates. The same is true for points O and P.
MN = ∣5∣ + ∣−4∣ = 5 + 4 = 9
OP = ∣5∣ + ∣−4∣ = 5 + 4 = 9
Points N and O are in different quadrants and have the same y-coordinates, so to find the distance from N to O we need to add the absolute values of their x-coordinates. The same is true for points P and M.
NO = ∣−2∣ + ∣3∣ = 2 + 3 = 5
MP = ∣−2∣ + ∣3∣ = 2 + 3 = 5
The perimeter is equal to the sum of the lengths of all four sides of rectangle MNOP.
9 + 9 + 5 + 5 = 28
Result
The perimeter of rectangle MNOP is 28 units.
Page 104 Exercise 5 Answer
If a polygon EFGH is a square, then all of its sides are equal length. To check if they are equal length, we need to check if the distance between E and F, F and G, G and H, and the distance between E and H, are same.
Points E and F are in different quadrants and have the same y-coordinates. To find the distance from E to F we need to add the absolute values of their x-coordinates. The same is true for points G and H.
EF = ∣−2∣ + ∣3∣ = 2 + 3 = 5
GH = ∣−2∣ + ∣3∣ = 2 + 3 = 5
Points F and G are in different quadrants and have the same x-coordinates. To find the distance from F to G we need to add the absolute values of their y-coordinates. The same is true for points E and H.
FG = ∣3.5∣ + ∣−1.5∣ = 3.5 + 1.5 = 5
EH = ∣3.5∣ + ∣−1.5∣ = 3.5 + 1.5 = 5
All the side of a rectangle EFGH are the same length, which means it is a square.
Result
EFGH is a square.
Page 104 Exercise 6 Answer
To find the area of square ABCD we must first find the length of one of its side.
Points A and B have the same y-coordinates and are in different quadrants, to find the distance from A to B we need to calculate the sum of absolute values of their x-coordinates.
AB = ∣−4.5∣ + ∣3.5∣ = 4.5 + 3.5 = 8
The area of square ABCD is equal to 8 × 8 = 64.
Result
The area of square ABCD is 64 units.
Page 105 Exercise 7 Answer
Points J and K have the same x-coordinates and are in different quadrants, so to find the distance between J and K we need to calculate the sum of absolute values of their y-coordinates.
JK = ∣8∣ + ∣−1∣ = 8 + 1 = 9
Points K and L have the same y-coordinates and are in different quadrants, so to find the distance between K and L we need to calculate the sum of absolute values of their x-coordinates.
KL = ∣−3∣ + ∣4∣ = 3 + 4 = 7
The rectangle has two sides with length 9 and two side with length 7.
The perimeter is equal to 2 × 9 + 2 × 7 = 18 + 14 = 32 units.
Result
The perimeter is equal to 32 units.
Page 105 Exercise 8 Answer
Points W and X have the same y-coordinates and are in different quadrants, so to find the distance between W and X we need to calculate the sum of absolute values of their x-coordinates.
WX = ∣−3∣ + ∣4∣ = 3 + 4 = 7
Points X and Y have the same x-coordinates and are in the same quadrant, Quadrant IV, so to find the distance between X and Y we need to calculate the difference of absolute values of their y-coordinates.
XY = ∣−5∣ − ∣−2∣ = 5 − 2 = 3
The rectangle has two sides with length 7 and two side with length 3.
The perimeter is equal to 2 × 7 + 2 × 3 = 14 + 6 = 20 units.
Result
The perimeter is 20 units.
Page 105 Exercise 9 Answer
If triangle JKL is equilateral, then the lengths of all of its sides are equal. To find the length of its side we need to calculate the distance from one vertice to another.
Since point J is the origin and point K is a point on the x-axis, the distance from J to K is equal to the absolute value of the x-coordinate of point K. Thus, the distance from J to K is 5 units.
Again, point J is the origin and point L is a point on the y-axis, the distance from J to L is equal to the absolute value of the y-coordinate of point L. Thus, the distance from J to L is 3 units.
Sides JK and JL are not of equal length, so triangle JKL is not equilateral.
Result
Triangle JKL is not equilateral.
Page 105 Exercise 10 Answer
Polygon WXYZ with vertices W(−1.5,1.5), X(6,1.5), Y(6,−4.5), and Z(−1.5,−4.5) is a rectangle if it is a quadrilateral with four right angles. Graph the vertices and then connect them with segments to draw the polygon.
From the graph, the polygon has four sides so it is a quadrilateral.
W and Z have the same x-coordinates and X and Y have the same x-coordinates. Sides WZ and XY are then both vertical.
W and X have the same y-coordinates and Z and Y have the same y-coordinates. Sides WX and ZY are then both horizontal.
Since horizontal lines and vertical lines intersect to form right angles, then quadrilateral WXYZ has four right angles. The polygon is then a rectangle.
Result
WXYZ is a rectangle since it has four sides and four right angles.
Page 105 Exercise 11 Answer
First we need to find the coordinates of the vertices of rectangle ABCD. For each point, follow the grid lines directly to the x-axis to find the x-coordinates, and follow the grid lines directly to the y-axis to find the y-coordinates.
The coordinates of the vertices are as follows: A(−3,−7), B(4,−7), C(4,6), D(−3,6).
To find the perimeter of rectangle ABCD, we need to calculate the sum of the lengths of all of its sides.
Points A and B have the same y-coordinates and are in different quadrants, so the distance between them is the sum of absolute values of their x-coordinates. The same is true for points D and C.
AB = ∣−3∣ + ∣4∣ = 3 + 4 = 7
DC = ∣−3∣ + ∣4∣ = 3 + 4 = 7
Points A and D have the same x-coordinates and are in different quadrants, so the distance between them is the sum of absolute values of their y-coordinates. The same is true for points B and C.
AD = ∣−7∣ + ∣6∣ = 7 + 6 = 13
BC = ∣−7∣ + ∣6∣ = 7 + 6 = 13
The perimeter of rectangle ABCD is
2 × 7 + 2 × 13 = 14 + 26 = 40.
The area of rectangle ABCD is
7 × 13 = 91.
Result
The perimeter of rectangle ABCD is 40, and the area is 91.
Page 105 Exercise 12 Answer
The perimeter of the patio is the sum of the lengths of all sides of the patio.
All the vertices are in the same quadrant, Quadrant I. If the points have the same x-coordinates, the distance between them is the difference of absolute values of their y-coordinates. If the points have the same y-coordinates, the distance between them is the difference of absolute values of their x-coordinates.
AB = ∣4∣ − ∣2∣ = 4 − 2 = 2
BC = ∣7∣ − ∣3∣ = 7 − 3 = 4
CD = ∣7∣ − ∣4∣ = 7 − 4 = 3
DE = ∣9∣ − ∣7∣ = 9 − 7 = 2
EF = ∣7∣ − ∣2∣ = 7 − 2 = 5
AF = ∣9∣ − ∣3∣ = 9 − 3 = 6
The perimeter of the patio is
AB + BC + CD + DE + EF + AF = 2 + 4 + 3 + 2 + 5 + 6 = 22.
Result
The perimeter of the patio is 22 yards.
Page 106 Exercise 13 Answer
The total distance that Jordan ran is the perimeter of polygon HBLCS.
HB = \(\left|-\frac{3}{4}\right|+|1|=\frac{3}{4}+1=1 \frac{3}{4}\)
BL = \(\left|\frac{1}{2}\right|+\left|-\frac{1}{4}\right|=\frac{1}{2}+\frac{1}{4}=\frac{2}{4}+\frac{1}{4}=\frac{2+1}{4}=\frac{3}{4}\)
LP = \(|1|-\left|\frac{1}{2}\right|=1-\frac{1}{2}=\frac{2}{2}-\frac{1}{2}=\frac{2-1}{2}=\frac{1}{2}\)
PC = \(\left|-\frac{3}{4}\right|-\left|-\frac{1}{4}\right|=\frac{3}{4}-\frac{1}{4}=\frac{3-1}{4}=\frac{2}{4}=\frac{1}{2}\)
SC = \(\left|-\frac{3}{4}\right|+\left|\frac{1}{2}\right|=\frac{3}{4}+\frac{1}{2}=\frac{3}{4}+\frac{2}{4}=\frac{3+2}{4}=\frac{5}{4}=\frac{4}{4}+\frac{1}{4}=1 \frac{1}{4}\)
HS = \(\left|\frac{1}{2}\right|+\left|-\frac{3}{4}\right|=\frac{1}{2}+\frac{3}{4}=\frac{2}{4}+\frac{3}{4}=\frac{2+3}{4}=\frac{5}{4}=\frac{4}{4}+\frac{1}{4}=1 \frac{1}{4}\)
Thee perimeter of polygon HBLCS is
\(H B+B L+L P+P C+S C+H S=1 \frac{3}{4}+\frac{3}{4}+\frac{1}{2}+\frac{1}{2}+1 \frac{1}{4}+1 \frac{1}{4}=6 .\)Result
The total distance that Jordan ran is 6 miles.
Page 106 Exercise 14 Answer
Three vertices of a rectangle are known, (−1.2,−3.5), (−1.2,4.4), and (5.5,4.4).
Point (−1.2,−3.5) is in Quadrant III, point (−1.2,4.4) in Quadrant II, and point (5.5,4.4) in Quadrant I. We can now conclude that points (5.5,4.4) and (−1.2,−3.5) are adjecent points to the fourth vertex. We can also graph the known points to check this.
From the graph, we can also conclude that the fourth vertex is in Quadrant IV. Its x-coordinate must be the same as the x-coordinate of point (5.5,4.4) and its y-coordinate must be the same as the y-coordinate of point (−1.2,−3.5).
Result
The coordinates of the fourth vertex are (5.5,−3.5).
Page 106 Exercise 15 Answer
To calculate the area of the backyard needed for the pool we need to first find the lengths of sides of the rectangular pool.
Points A and B have the same y-coordinates and are in different quadrants, so the distance from A to B is the sum of absolute values of their x-coordinates.
∣−5∣ + ∣1∣ = 5 + 1 = 6
The lengths of one of the sides is 6 yards.
Points B and C have the same x-coordinates and are in different quadrants, so the distance from B to C is the sum of absolute values of their y-coordinates.
∣7∣ + ∣−1∣ = 7 + 1 = 8
The lengths of one of the sides is 8 yards.
The area of the rectangular pool is the product
AB × BC = 6 × 8 = 48.
Result
The area of the backyard needed for the pool is 48 square yards.
Page 106 Exercise 16 Answer
Each point in a coordinate plane has its unique coordinates, real numbers which describe its location. Coordinates can be negative, zero, or positive. The distance between points can be calculated using their coordinates, however, coordinates can be negative, but distance is always positive, so we use absolute values of coordinates to calculate the distances.
The absolute value of the x-coordinate of a point represents the distance between that point and the x-axis, however the sign of the x-coordinate tells us in which direction from the origin the point is located. The same is true for the y-axis as well.
Result
The distance between points can be calculated using their coordinates, however, coordinates can be negative, but distance is always positive, so we use absolute values of coordinates to calculate the distances.
Page 106 Exercise 17 Answer
Since it is a square, all of its sides are of equal length. Perimeter is 10 units, to find the length of one side we can divide the perimeter by four.
10 ÷ 4 = 2.5
The length of one side is 2.5.
One vertex is (−0.5,−2), one of the adjecent vertices will have the same x-coordinate, and the other adjecent vertex will have the same y-coordinates.
If the x-coordinate is the same as that of the given vertex, than the other is either the difference −2 + 2.5 or −2 − 2.5. Since the difference −2 + 2.5 is equal 0.5, which is positive, so that point wouldn’t be in Quadrant III. So the other coordinate must be −2 − 2.5, which is equal to −4.5.
Second vertex is (−0.5,−4.5).
If the y-coordinate is the same as that of the given vertex, than the other is either the difference −0.5 + 2.5 or −0.5 − 2.5. Since the difference −0.5 + 2.5 is equal 2, which is positive, so that point wouldn’t be in Quadrant III. So the other coordinate must be −0.5 − 2.5, which is equal to −3.
Third vertex is (−3,−2).
To determine the coordinates of the fourth vertex we can graph the three known vertices and simply read them from the graph.
From the graph we can see that the fourth vertex has the same x-coordinates as point (−3,−2) and the same y-coordinates as point (−0.5,−4.5). The fourth vertex is (−3,−4.5).
Result
The coordinates of the other vertices are (−3,−2), (−0.5,−4.5), and (−3,−4.5).
Page 106 Exercise 18a Answer
Points A and B have the same x-coordinates and are in the same quadrant, Quadrant III, which we know because both of their coordinates are negative. To find the distance between points A and B we need to calculate the difference between their y-coordinates.
\(|-3|-\left|-\frac{1}{2}\right|=3-\frac{1}{2}=\frac{6}{2}-\frac{1}{2}=\frac{6-1}{2}=\frac{5}{2}=\frac{4}{2}+\frac{1}{2}=2 \frac{1}{2}\)Result
The distance between points A and B is \(2 \frac{1}{2}\) units.
Page 106 Exercise 18b Answer
From point C we can move 8 units parallel to the x-axis in two directions, and also parallel to the y-axis in two directions.
If we move parallel to the x-axis, the y-coordinate is the same as that of point C. If we move parallel to the y-axis, the x-coordinate is the same as that of point C.
−3 + 8 = 5
−3 − 8 = −11
4 − 8 = −4
4 + 8 = 12
Two points which are 8 units from point C are (4,5) and (-4,-3).
Result
The coordinates of two points which are 8 units from point C are (4,5) and (−4,−3).