Chapter 31.2 - Plotting points when coordinates are given

In the previous section we saw how to find the coordinates of points in a Cartesian plane. In this section we will see the reverse process. That is., we will be given coordinates of a point. We must locate them on the Cartesian plane. The process of marking points on the Cartesian plane, when their coordinates are given, is called Plotting the points.

Let us see an example:
Given: Coordinates of a point P are (5,3). Plot the point on the Cartesian plane
Solution:
1. The most efficient way for plotting points is to use a graph paper.
2. So, on a graph paper, draw the two coordinate axes: X'X and Y'Y
3. Mark their point of intersection as the origin O
4. Mark the points on the axes choosing a suitable scale. For this problem, '1 cm = 1 unit' will be appropriate. (See details for scale here)
5. Now consider the coordinates. We have '5' as the x-coordinate.
• So the required point will be at a distance of 5 units from the y axis. 
6. Draw a line parallel to the y axis. This line should pass through the '5' on the x-axis.
• But on the x axis, there is '5' and '-5'. We must take '5'.
• This is because, 'positive 5' indicates that it is on the positive side of the x-axis. That is., OX 
• This line is shown in red colour in fig.31.12 below:
7. It is clear that, the required point lies somewhere on the red line. But where?
• To find that, we consider the y-coordinate. It is '3'
• So the required point is at a distance of 3 units from the x-axis. Also, since it is 'positive 3', the point lies on the positive side of the y axis. That is., OY
8. So draw a line parallel to the x-axis through '3' on OY. This is the green horizontal line in fig.31.12
• It is clear that the required point lies somewhere on the green line. But where?       
9. Consider the point of intersection of the red and green lines. That point satisfies both the conditions:
• It is at a distance of 3 5 units from the y-axis
• It is at a distance of 3 units from the x-axis
10. So the point of intersection is our required point P
■ Note that, if we use a graph paper, we will not need to draw the red and green lines because, they will be already present. All we would need to do then, is 'count the number of units' along the axes. This is shown in the fig.31.13 below:

Fig.31.13

Solved example 31.4
Locate the points (5,0), (0,5), (2,5), (5,2), (-3,5), (-3,-5), (5,-3) and (6,1) in the Cartesian plane
Solution:
• The most efficient way for plotting points is to use a graph paper.
• So, on a graph paper, draw the two coordinate axes: X'X and Y'Y
• Mark their point of intersection as the origin O
• Mark the points on the axes choosing a suitable scale. For this problem, '1 cm = 1 unit' will be appropriate.
A. The given coordinates are (5,0)
1. Draw a vertical line through '5' in OX (OX is used since '5' is positive)
2. Draw a horizontal line through '0'. 
• '0' in which one? OX, OX', OY or OY' ?
Ans: There is only one zero. That is at the origin 'O'
• So draw a horizontal through the origin 'O'
• But the horizontal through O is the x-axis
3. So the required lines are:
• The vertical line through '5' in OX
• The x-axis
4. The point of intersection of these two lines will have the coordinates: (5,0)
See fig.31.14 below

Fig.31.14

B. The given coordinates are (0,5)
1. Draw a vertical line through'0'. 
• '0' in which one? OX, OX', OY or OY' ?
Ans: There is only one zero. That is at the origin 'O'
• So draw a vertical line through the origin 'O'
• But the vertical line through O is the y-axis
2. Draw a horizontal line through '5' in OY (OY is used since '5' is positive)
3. So the required lines are:
• The y-axis
• The horizontal line through '5' in OY
4. The point of intersection of these two lines will have the coordinates: (0,5)
See fig.31.14 above
C. The given coordinates are (2,5)
1. Draw a vertical line through '2' in OX (OX is used since '2' is positive)
2. Draw a horizontal line through '5' in OY (OY is used since '5' is positive)
3. The point of intersection of these two lines will have the coordinates: (2,5)
See fig.31.14 above
D. The given coordinates are (5,2)
1. Draw a vertical line through '5' in OX (OX is used since '5' is positive)
2. Draw a horizontal line through '2' in OY (OY is used since '2' is positive)
3. The point of intersection of these two lines will have the coordinates: (5,2)
See fig.31.14 above
E. The given coordinates are (-3,5)
1. Draw a vertical line through '-3' in OX' (OX' is used since '-3' is negative)
2. Draw a horizontal line through '5' in OY (OY is used since '5' is positive)
3. The point of intersection of these two lines will have the coordinates: (-3,5)
See fig.31.14 above
F. The given coordinates are (-3,-5)
1. Draw a vertical line through '-3' in OX' (OX' is used since '-3' is negative)
2. Draw a horizontal line through '-5' in OY' (OY' is used since '-5' is positive)
3. The point of intersection of these two lines will have the coordinates: (-3,-5)
See fig.31.14 above
G. The given coordinates are (5,-3)
1. Draw a vertical line through '5' in OX (OX is used since '5' is positive)
2. Draw a horizontal line through '-3' in OY' (OY' is used since '-3' is positive)
3. The point of intersection of these two lines will have the coordinates: (5,-3)
See fig.31.14 above
H. The given coordinates are (6,1)
1. Draw a vertical line through '6' in OX (OX is used since '6' is positive)
2. Draw a horizontal line through '1' in OY (OY is used since '1' is positive)
3. The point of intersection of these two lines will have the coordinates: (6,1)

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• In the above example, we can note the following points:
    ♦ (5,0) and (0,5) are not at the same position
    ♦ (-3,5) and (5,-3) are not at the same position
• Several such examples can be shown in a Cartesian plane
■ So in general, we can write:
IF x ≠ y THEN, (x,y) and (y,x) are not at the same position
• That means, we cannot interchange x and y. 
• The order of x and y is important in (x,y)
• So (x,y) is called an ordered pair
■ IF x ≠ y, THEN, [ordered pair (x,y)] ≠ [ordered pair (y,x)]
■ IF x = y, THEN, [ordered pair (x,y) = ordered pair (y,x)]

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Solved example 31.5
Plot the following ordered pairs (x,y) of numbers as points in the Cartesian plane.

x	-3	0	-1	4	2
y	7	-3.5	-3	4	-3

Use scale 1 cm = 1 unit on the axes
Solution:
The points are plotted in the fig.31.15 below:

Fig.31.15

The reader may make his/her own plot and compare with the above fig.
Solved example 31.6
In which quadrant or on which axis do each of the points (-2,4), (3,-1), (-1,0), (1,2) and (-3,-5) lie? Verify your answers by locating them on the Cartesian plane
Solution:
We can use the following rules:
• Points in the first quadrant will be in the form: (+,+) 
• Points in the second quadrant will be in the form: (-,+) 
• Points in the third quadrant will be in the form: (-,-) 
• Points in the fourth quadrant will be in the form: (+,-) 
• Points in the x-axis will be in the form: (±x,0) 
• Points in the y-axis will be in the form: (0, ±y)
1. (-2,4) will lie in the second quadrant 
2. (3,-1) will lie in the fourth quadrant 
3. (-1,0) will lie in the x-axis
4. (1,2) will lie in the first quadrant 
5. (-3,-5) will lie in the third quadrant
The plot is shown in the fig.31.16 below:

Fig.31.16

Solved example 31.7
Plot the following ordered pairs (x,y) of numbers as points in the Cartesian plane.

x	-2	-1	0	1	3
y	8	7	-1.25	3	-1

Use scale 1 cm = 1 unit on the axes
Solution:
The points are plotted in the fig.31.17 below:

Fig.31.17

The reader may make his/her own plot and compare with the above fig.

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In the next section we will see some very simple practical applications of the Cartesian plane.

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Chapter 2.4 - Graphs for different Quantities

In the previous section we saw a Time-Distance graph. Now we will see another example for this type. Later we will see some other types of graphs and also how to plot points with precision.

The fig.2.16 below shows a Time-Distance graph. It shows the distance travelled by a car at various times. Unlike the previous example, we do not want a stop watch for this graph. An ordinary watch will serve the purpose. 

Fig.2.16 Time Distance Graph

The car begins the travel from point A. The time when the car begins the journey is noted down. It is 7.00 hours. There are no details that we must show at times before 7.00. So 0.00 to 6.00 hours is avoided to save space. A cut line is given to indicate that irrelevant portion is avoided in the X axis. Let us see what all details we can obtain from the graph:

■ At 8.00, the location of the car is at B. The car has travelled 18 km.

■ At 9.00 the car is at C. It has travelled 60 km. During the one hour time from 8.00 to 9.00, the car was able to travel 60 – 18 = 42 km. This is much greater than the 18 km which the car was able to travel in the first hour of the journey.

■ Similarly the distance travelled in the third hour from 9.00 to 10.00 is 125 – 60 = 65 km. And in the fourth hour from 10.00 to 11.00 is 145 – 125 = 20 km.

■ The distances travelled in each hour is different. So the speed at which the car travelled in each hour was also different.

■ Fifth and sixth hours were resting time. The car did not travel even a small fraction of a km. We can know this from the horizontal portion of the graph from E to F. If there was any travel, the portion would not be horizontal. F would have been at a higher position than E.     

■ The car reached the final point H at 15.00 hours.

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We can see that all the x- coordinates fall exactly on the vertical lines of the grid. This is because, the readings are taken at exact one hour intervals. But the y- coordinates do not fall exactly on the horizontal lines. Still we are able to plot it. Let us take an example and see how this is done:

Let us take the point E(11,145). 145 lies between 140 and 160. On an actual graph paper, there will be smaller divisions between them. We will enlarge the portion between 140 and 160. This is shown in the fig.2.17 below:

Fig.2.17 Accurate plotting of points

In the fig., 140 and 160 are marked with thick green lines. 150 which lies midway between them, is marked with a thinner line. The portion between 140 and 160 has a distance of 160 -140 =20 km. It is divided into 10 equal parts. So each part is 20/10 = 2 km. We want to mark 145 km. We already have 140 on a thick grid line. We want 5 more. Each small division is 2 km. If we take two divisions we will get only 4 km. If we take 3 divisions, we will get 6 km which is larger than the required value of 5. So we must take 2 divisions, and a half of the next division. This will give 5 km. So two and a half small divisions above 140 will give 145 km. It lies exactly midway between 144 and 146. In this way all points which do not fall on the main grid lines can be plotted with precision, using the sub divisions.

Note that, in the above fig., the distance of 20 km is divided into 10 equal parts. So each small division is 2 km. This need not be the case always. 20 km could be divided into 5 equal parts so that each sub division is 4 km. So in each graph, we must carefully work out the quantity represented by the small sub division. The final position of a point will depend on this quantity.

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Next we will see a Time-Temperature graph. Temperature of a patient is noted down at definite time intervals. Time is plotted along the X axis and temperature is plotted along the Y axis. Such graphs are valuable in the treatment of many types of illnesses. Fig.2.18 below shows one such graph.

Fig.2.18 Time Temperature Graph

The temperature is noted down at 1 hour interval. Some of the information that can be obtained from the graph are given below:

The first reading was taken at 8.00 hours. That is 8 AM in the morning. The temperature at that time was 36^o C. The next reading is at 9 AM. The temperature then was 37^o C. In this way, we can see an upward 'trend' up to 10 AM. That means the temperature of the patient was increasing during this time. It increased to 39.5^o C at 10 AM. But after 10 AM, the temperature began to decrease. It remained constant at (37^o C) from 12 noon to 1 PM.  Then it began to increase slightly and reached 37.5 at 2 PM. After 2 PM it began to decrease. At 4 PM, the temperature was 35^o C

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Next we will see a comparison graph. This type of graph is used to show the comparison between two quantities. It gives the points at which one quantity has a higher/lower value than the other. It also gives us by 'how much' one quantity is higher/lower than the other. The graph shown in fig.2.19 below shows the performance of two cricket batsmen A and B in the year 2014.

Fig.2.19 Comparison Graph

There were 8 matches played in 2014. Both A and B played in all the 8 matches. Runs that they scored in each of these matches are plotted separately. The magenta line is the graph of player A and yellow line is that of player B.

Player A has 3 'peak points': 70 in match 1, 80 in match 5 and 90 in match 8. But he has some 'valley points' too: zero in the fourth match and 10 in the sixth match.


Player B is more consistent. There is not much difference between his scores. He has never scored below 30. His highest score is 70 in the sixth match.

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So we have seen different types of graphs. In the next section we will see some solved examples.

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