Chapter 31 - Coordinate Geometry

In the previous section we completed a discussion on Trigonometry. In this section we will see Coordinate Geometry. We saw some basic details about graphs in chapter 2. Here we will have a more detailed study.

We have studied about the number lines also in an earlier chapter 22.  Following are the basic features of a number line:
• The distances are marked from a fixed point
    ♦ This fixed point is called the origin
• The distances are marked in equal units
    ♦ If 1 unit from the origin represents the number 1, then 3 units from origin will represent the number 3
    ♦ In the same way, the number 'r' will be 'r units' away from the origin 
• The distances are marked positively in one direction and negatively in the opposite direction
    ♦ The number 'r' will be 'r units' away from the origin in the positive direction
    ♦ The number '-r' will be 'r units' away from the origin in the negative direction
• An example is shown in the fig.31.1 below:

Fig.31.1

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• Consider two such number lines. Place them perpendicular to each other. 

    ♦ But there are infinite possibilities to place 'two lines perpendicular to each other'. The xxx fig.31.2 below shows some of them:

Fig.31.2

• All the three arrangements in the above fig.31.2 satisfies one condition:

    ♦ The lines should be perpendicular to each other. 
• In this chapter, we will be using the arrangement in fig.c. That is:
    ♦ One line will be horizontal and the other line will be vertical

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Let us see how the numbering system can be applied to the two perpendicular lines:

1. Consider any simple horizontal number line as shown in the fig.31.3(a) below. Let us name it X'X
• The positive numbers are marked towards the right from the origin. 
• The negative numbers are marked towards the left from the origin. This is shown in fig.31.3(b)

Fig.31.3

2. Consider another number line. This new number line must be vertical. Let us name it Y'Y. See fig.31.3(b) 
• The positive numbers must be marked upwards from the origin. 
• The negative numbers must be marked downwards from the origin.
3. Now combine the two number lines. 
• The combination should be done in such a way that, the two lines cross each other at their origins.
4. The horizontal line X'X is called the x-axis
• The vertical line Y'Y is called the y-axis 
• The point where X'X and Y'Y cross is called the origin
    ♦ It is denoted by the letter 'O'
5. The positive numbers belonging to the x-axis lies on OX
    ♦ So OX is called the 'positive direction of the x-axis' 
• The negative numbers  belonging to the x-axis lies on OX'
    ♦ So OX' is called the 'negative direction of the x-axis'
• The positive numbers  belonging to the y-axis lies on OY
    ♦ So OY is called the 'positive direction of the y-axis' 
• The negative numbers  belonging to the y-axis lies on OY'
    ♦ So OY' is called the 'negative direction of the y-axis'

6. Consider the fig.31.4 below. Both the x-axis and y-axis are shown. 

Fig.31.4

• The two axes (plural of 'axis' is 'axes') divide the plane of the paper into four parts. 

• These four parts are called quadrants. 

    ♦ In other words, each one of the four parts can be called: A quadrant

(Recall that a polygon with four sides is called a quadrilateral)
7. Each quadrant is given a particular name. The procedure for naming is simple:
• Just number them as I, II, III and IV
• The numbering should be done in the anti clockwise direction starting from OX. So we get the following:
    ♦ Top right: Quadrant I, 
    ♦ Top left: Quadrant II
    ♦ Bottom left: Quadrant III 
    ♦ Bottom right: Quadrant IV
8. Thus the 'plane of the paper' consists of two axes and four quadrants
• This plane is known by three different names. We can use any one of them:
    ♦ The cartesian plane
    ♦ The coordinate plane
    ♦ The xy-plane
9. The axes are called coordinate axes

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Now we will see how the coordinate system can be used to specify the exact location any points on a plane. 

• Consider fig.31.5 below:

Fig.31.5

• Two points P and Q are marked on the cartesian plane. 

• We want to specify the exact locations of P and Q. For that, we use the following steps:

1. Drop the following perpendicular lines:
• perpendicular PM from P on to the x-axis (See fig.31.6 below) 
• perpendicular PN from P on to the y-axis
• perpendicular QR from Q on to the x-axis
• perpendicular QS from Q on to the y-axis

Fig.31.6

2. Dropping the perpendicular lines gives us the following information:
• The perpendicular distance of P from the y-axis is equal to PN. 
• We want the value of this PN. But PN is same as OM (∵ M is the foot of the perpendicular from P)
• We can easily see that OM is 4 units. So PN is 4 units. Thus we get:
• The point P is at a perpendicular distance of 4 units from the  y-axis
    ♦ This 4 units is measured along the positive direction of the x-axis
3. Working in a similar way we get these also:
• The point P is at a perpendicular distance of 3 units from the  x-axis
    ♦ This 3 units is measured along the positive direction of the y-axis
• The point Q is at a perpendicular distance of 6 units from the  y-axis
    ♦ This 6 units is measured along the negative direction of the x-axis
• The point Q is at a perpendicular distance of 2 units from the  x-axis
    ♦ This 2 units is measured along the positive direction of the y-axis
4. Now we have all the required information for specifying the location of the points.
• But there are three rules to be followed. 
• The rules will ensure that the 'final presentation of the results' made by all of us will be in the same form. 
• The 3 rules are:
(i) The x-coordinate of a point, is the perpendicular distance of that point from the y-axis
• This distance should be measured along the x-axis
    ♦ The x-coordinate thus obtained is positive if the measurement is done along OX  
    ♦ The x-coordinate thus obtained is negative if the measurement is done along OX'  
(ii) The y-coordinate of a point, is the perpendicular distance of that point from the x-axis
• This distance should be measured along the y-axis
    ♦ The y-coordinate thus obtained is positive if the measurement is done along OY  
    ♦ The y-coordinate thus obtained is negative if the measurement is done along OY'
(iii) Write the coordinates of a point inside brackets. They must be separated by a comma.
• The x-coordinate is written first and then the y-coordinate
    ♦ Another name for x-coordinate is abscissa
    ♦ Another name for y-coordinate is ordinate
5. Let us apply the rules to point P:
(i) Applying the first rule to find the x-coordinate:
• Perpendicular distance of P from the y-axis (measured along the x-axis) is '4 units'
    ♦ It is measured along OX. So it is positive
    ♦ Thus the x-coordinate of P is '4'
(ii) Applying the second rule to find the y-coordinate:
• Perpendicular distance of P from the x-axis (measured along the y-axis) is '3 units'
    ♦ It is measured along OY. So it is positive
    ♦ Thus the y-coordinate of P is '3'
(iii) Applying the third rule to write down the coordinates:
• x-coordinate first and then the y-coordinate. Inside brackets and separated by comma. We get:
■ The coordinates of P are: (4,3)
6. Let us apply the rules to point Q:
(i) Applying the first rule to find the x-coordinate:
• Perpendicular distance of Q from the y-axis (measured along the x-axis) is '6 units'
    ♦ It is measured along OX'. So it is negative
    ♦ Thus the x-coordinate of P is '-6'
(ii) Applying the second rule to find the y-coordinate:
• Perpendicular distance of Q from the x-axis (measured along the y-axis) is '2 units'
    ♦ It is measured along OY'. So it is negative
    ♦ Thus the y-coordinate of Q is '-2'
(iii) Applying the third rule to write down the coordinates:
• x-coordinate first and then the y-coordinate. Inside brackets and separated by comma. We get:
■ The coordinates of Q are: (-6,-2)
7. In the above example, we drew perpendicular lines PM, PN, QR and QS. If the points P and Q are marked on a graph paper, we will not need to draw those lines. This is shown in the fig.31.7 below:

Fig.31.7

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In the next section we will see some solved examples.

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Chapter 22.6 - Distance as Absolute value - Solved examples - 2

In the previous section we saw solved examples which demonstrate the expression of distance as absolute value. In this section we will see a few more solved examples.

Solved example 22.10
Find the solutions to the equation: |x-1| = |x-3|
Solution:
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is 1
3. We have a point C on the number line. The number corresponding to the point C is 3
4. The difference between the numbers corresponding to A and B = (x-1)
5. Applying theorem 22.5, the distance between A and B is |(x-1)|
6. The difference between the numbers corresponding to A and C = (x-3)
7. Applying theorem 22.5, the distance between A and B is |(x-3)|
8. But these distances are given as equal.
9. That means the distance from B to A is same as the distance from C to A. Now look at the number line in fig.22.26 below:

Fig.22.26

10. If the distance from B toA is same as the distance from C to A, there is only one possibility:
• A is the midpoint between B and C
11. So we are in the following situation:
• We have a point B, whose number is known
• We have a point C, whose number is known
• We have a point A, whose number is not known. But we know that, A is the midpoint between B and C
12. In this situation, we apply theorem 22.3. 
• Sum of the numbers corresponding to B and C = 1+3 = 4
• Number corresponding to the midpoint A = Half of the sum = 4/2 = 2
Thus we get x = 2
13. Check: |x-1| = |x-3| ⇒ |2-1| = |2-3| ⇒ |1| = |-1| ⇒ 1 = 1 Which is true

Solved example 22.11
Find the solutions to the equation: |x-3| = |x-4|
Solution:
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is 3
3. We have a point C on the number line. The number corresponding to the point C is 4
4. The difference between the numbers corresponding to A and B = (x-3)
5. Applying theorem 22.5, the distance between A and B is |(x-3)|
6. The difference between the numbers corresponding to A and C = (x-4)
7. Applying theorem 22.5, the distance between A and B is |(x-4)|
8. But these distances are given as equal.
9. That means, the distance from B to A is same as the distance from C to A. Now look at the number line in fig.22.27 below:

Fig.22.27

10. If the distance from B to A is same as the distance from C to A, there is only one possibility:
• A is the midpoint between B and C
11. So we are in the following situation:
• We have a point B, whose number is known
• We have a point C, whose number is known
• We have a point A, whose number is not known. But we know that, A is the midpoint between B and C
12. In this situation, we apply theorem 22.3. 
• Sum of the numbers corresponding to B and C = 3+4 = 7
• Number corresponding to the midpoint A = Half of the sum = 7/2 = 3.5
Thus we get x = 3.5
13. Check: |x-3| = |x-4| ⇒ |3.5-3| = |3.5-4| ⇒ |0.5| = |-0.5| ⇒ 0.5 = 0.5 Which is true

Solved example 22.12
Find the solutions to the equation: |x+2| = |x-5|
Solution:
• (x+2) can be modified as [x-(-2)]. So we can write the steps as follows: 
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is -2
3. We have a point C on the number line. The number corresponding to the point C is 5
4. The difference between the numbers corresponding to A and B = [x-(-2)]
5. Applying theorem 22.5, the distance between A and B is |[x-(-2)]|
6. The difference between the numbers corresponding to A and C = (x-5)
7. Applying theorem 22.5, the distance between A and B is |(x-5)|
8. But these distances are given as equal.
9. That means, the distance from B to A is same as the distance from C to A. Now look at the number line in fig.22.27 below:

Fig.22.27

10. If the distance from B to A is same as the distance from C to A, there is only one possibility:
• A is the midpoint between B and C
11. So we are in the following situation:
• We have a point B, whose number is known
• We have a point C, whose number is known
• We have a point A, whose number is not known. But we know that, A is the midpoint between B and C
12. In this situation, we apply theorem 22.3. 
• Sum of the numbers corresponding to B and C = -2+5 = 3
• Number corresponding to the midpoint A = Half of the sum = 3/2 = 1.5
Thus we get x = 1.5
13. Check: |x+2| = |x-5| ⇒ |1.5+2| = |1.5-5| ⇒ |3.5| = |-3.5| ⇒ 3.5 = 3.5 Which is true

Solved example 22.13
Find the solutions to the equation: |x| = |x+1|
Solution:
• |x| is the distance of the 'point whose number is x' from zero
• (x+1) can be modified as [x-(-1)]. 
So we can write the steps as follows: 
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is -1
3. The distance of A from zero is |x|
4. The difference between the numbers corresponding to A and B =  [x-(-1)]
5. Applying theorem 22.5, the distance between A and B is | [x-(-1)]|
6. But these distances are given as equal.
7. That means, the distance from B to A is same as the distance from A to zero. Now look at the number line in fig.22.28 below:

Fig.22.28

8. If the distance from B to A is same as the distance from A to zero, there is only one possibility:
• A is the midpoint between B and zero
9. So we are in the following situation:
• We have a point B, whose number is known
• We have the point zero, whose number is '0'
• We have a point A, whose number is not known. But we know that, A is the midpoint between B and zero
10. In this situation, we apply theorem 22.3. 
• Sum of the numbers corresponding to B and zero = -1+0 = -1
• Number corresponding to the midpoint A = Half of the sum = -1/2 = -0.5
Thus we get x = -0.5
11. Check: |x| = |x+1| ⇒ |-0.5| = |-0.5+1| ⇒ |-0.5| = |0.5| ⇒ 0.5 = 0.5 Which is true

Alternate method:
• x can be modified as (x-0)
• (x+1) can be modified as [x-(-1)]. 
So we can write the steps as follows: 
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is 0
3. We have a point C on the number line. The number corresponding to the point C is -1
4. The difference between the numbers corresponding to A and B = (x-0)
5. Applying theorem 22.5, the distance between A and B is |(x-0)|
6. The difference between the numbers corresponding to A and C =  [x-(-1)]
7. Applying theorem 22.5, the distance between A and B is | [x-(-1)]|
8. But these distances are given as equal.
9. That means, the distance from B to A is same as the distance from C to A. Now look at the number line in fig.22.29 below:

Fig.22.29

10. If the distance from B to A is same as the distance from C to A, there is only one possibility:
• A is the midpoint between B and C
11. So we are in the following situation:
• We have a point B, whose number is known
• We have a point C, whose number is known
• We have a point A, whose number is not known. But we know that, A is the midpoint between B and C
12. In this situation, we apply theorem 22.3. 
• Sum of the numbers corresponding to B and C = -1+0 = -1
• Number corresponding to the midpoint A = Half of the sum = -1/2 = -0.5
Thus we get x = -0.5
13. Check: |x| = |x+1| ⇒ |-0.5| = |-0.5+1| ⇒ |-0.5| = |0.5| ⇒ 0.5 = 0.5 Which is true

Solved example 22.14
Prove that if 1 < x < 4 and 1 < y < 4, then |x-y| < 3
Solution:
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is 1
3. We have a point C on the number line. The number corresponding to the point C is 4
4. We have a point D on the number line. The number corresponding to the point D is y
5. Given that 1 < x < 4. So the point A can be any where in between C and D
x can never take a value of exact 1 or exact 4. So hollow circles are shown at B and C in the graph below:

Fig.22.30

6. Given that 1 < y < 4. So the point D can be any where in between B and C
y can never take a value of exact 1 or exact 4.
7. We want the difference between x and y. The 'maximum difference' will be obtained in the following situation:
• When one of them (x or y) is at it's maximum possible value AND
• The other is at it's minimum possible value
8. When one of them is at it's maximum possible value, it will be 'just to the left of 4'
Then the other is at the minimum possible value, which is 'just to the right of 1'
9. Concentrate on the two positions:
• One point is 'just to the left of 4'
• The other point is 'just to the right of 1'
10. In such a situation, the distance between them (calculated as |x-y|) will be less than 3
■ We will get 3 only if one is at 'exact 4' and the other is at 'exact 1'
11. Thus we find that the maximum possible distance is less than 3
We can write: |x-y| < 3

Solved example 22.15
Prove that if x < 3 and y > 7, then |x-y| > 4
Solution:
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is 3
3. We have a point C on the number line. The number corresponding to the point C is 7
4. We have a point D on the number line. The number corresponding to the point D is y
5. Given that x < 3. So the point A can be any where on the red line

x can never take a value of exact 3. So hollow circle is shown at B in the graph below:

Fig.22.31

6. Given that y > 7. So the point D can be any where on the blue line
y can never take a value of exact 7.
7. We want the difference between x and y. The 'minimum difference' will be obtained in the following situation:
• When x is just to the left of B AND
• y is just to the right of C
• If any one of them is at a different position other than the above two, the difference will not be minimum
8. Concentrate on the two positions:
• x is 'just to the left of B'
• y is 'just to the right of C'
10. In such a situation, the distance between them (calculated as |x-y|) will be greater than 4
■ We will get 4 only if x is at 'exact 3' and y is at 'exact 7'
11. Thus we find that the minimum possible distance is greater than 4
We can write: |x-y| > 4


Solved example 22.16
What are the numbers x, for which |x-2| + |x-8| = 6
Solution:
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is 2

3. We have a point C on the number line. The number corresponding to the point C is 8
4. The difference between the numbers corresponding to A and B = (x-2)
5. Applying theorem 22.5, the distance between A and B is |x-2|
6. The difference between the numbers corresponding to A and C =  (x-8)
7. Applying theorem 22.5, the distance between A and B is |x-8|
8. The sum of the two distances is given as 6. Consider the number line shown in fig.21.32 below:

Fig.22.32

9. We can see that the actual distance between B and C is (8-2) = 6 units
10. So A can lie anywhere on the red line. It can also take the exact values 2 and 8
11. So we can write 2 ≤ x ≤ 8
12. Check: • Let x = 2
Then |x-2| + |x-8| = 6 ⇒|2-2| + |2-8| = 6 ⇒ |0| + |-6| = 6 ⇒ 0 + 6 = 6 ⇒6 = 6 which is true 
• Let x = 4
Then |x-2| + |x-8| = 6 ⇒|4-2| + |4-8| = 6 ⇒ |2| + |-4| = 6 ⇒ 2 + 4 = 6 ⇒6 = 6 which is true
• Let x = 8
Then |x-2| + |x-8| = 6 ⇒|8-2| + |8-8| = 6 ⇒ |6| + |0| = 6 ⇒ 6 + 0 = 6 ⇒6 = 6 which is true

Solved example 22.17
What are the numbers x, for which |x-2| + |x-8| = 10
Solution:
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is 2

3. We have a point C on the number line. The number corresponding to the point C is 8
4. The difference between the numbers corresponding to A and B = (x-2)
5. Applying theorem 22.5, the distance between A and B is |x-2|
6. The difference between the numbers corresponding to A and C =  (x-8)
7. Applying theorem 22.5, the distance between A and B is |x-8|
8. The sum of the two distances is given as 10. Consider the number line shown in fig.21.33 below:

9. The two possible positions of A are shown in blue colour. The corresponding numbers are 0 and 10
10. Check: • Let x = 0
Then |x-2| + |x-8| = 10 ⇒|0-2| + |0-8| = 10 ⇒ |-2| + |-8| = 10 ⇒ 2 + 8 = 10 ⇒10 = 10 which is true   
• Let x = 10
Then |x-2| + |x-8| = 10 ⇒|10-2| + |10-8| = 10 ⇒ |8| + |2| = 10 ⇒ 8 + 2 = 10 ⇒10 = 10 which is true
Another method:
We will consider all the possible values that x can take.
Case 1: x < 2
1. Then (x-2) will be a negative quantity. This is same as (2-x) is a positive quantity
2. Whether we take (x-2) or (2-x), both are 'difference'. And when we take the absolute value, we are dropping the '-' sign.
3. So, when x < 2, |x-2| = (2-x)
4. When x is less than 2, it will be less than 8 also. So |x-8| = (8-x)
5. Substituting these values in the given equation we get:
(2-x) + (8-x) = 10 ⇒ 10 -2x = 10 ⇒ 2x = 0 ⇒ x = 0
Case 2: x = 2
1. Then (x-2) = 0 and |0| = 0
2. |x-8| = |(2-8)| = |-6| = 6
3. Substituting these values in the given equation we get:
0 + 6 = 10. This is not true. So we cannot consider case 2
Case 3: x > 2 AND x < 8
1. That means x is greater than 2, but at the same time, x is less than 8
2. Another way of writing this is: 2 <  x < 8
3. Graphically, this will be a line from 2 to 8, with hollow circles at 2 and 8
4. Let us see whether any of such values of x satisfies the given equation:
5. When x > 2, (x-2) is positive. So |x-2| = (x-2)
6. When x is less than 8, (x-8) is negative. So |x-8| = (8-x)
7. Substituting these values in the given equation we get:
(x-2) + (8-x) = 10 ⇒ 6 = 10. This is not true. So we cannot consider case 3
Case 4: x = 8
1. Then (x-2) = 6 and |6| = 6
2. |x-8| = 0 and |0| = 0
3. Substituting these values in the given equation we get:
6 + 0 = 10. This is not true. So we cannot consider case 4
Case 5: x > 8
1. Then x will be greater than 2 also. So |x-2| = (x-2)    
2. (x-8) will be positive. So |x-8| = (x-8)
3. Substituting these values in the given equation we get:    
4. (x-2) + (x-8) = 10 ⇒ 2x-10 = 10 ⇒ 2x = 20 ⇒ x = 10

■ So we considered 5 cases. Those 5 cases cover all possible numbers on the number line. But only the cases 1 and 5 will give us a valid result.
• The valid results are: x = 0 and x = 10

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In the next section we will see Solids.

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Chapter 22.5 - Distance as Absolute value - Solved examples

In the previous section we saw the 'distance between any two points' expressed as absolute value. We also saw some practical cases. In this section we will see a few more cases.

The equations that we saw were:
• |x-1| = 3 
• |x+1| = 3 
Now we will see some inequalities:
Solved example 22.6
Find the solutions to the inequality |x-1| ≤ 3 
Solution:
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is 1
• The difference between the numbers = (x-1)
3. Applying theorem 22.5, the distance between A and B is |(x-1)|
4. But |(x-1)| is less than or equal to 3. So the distance between A and B should never exceed 3 units
5. That means A should always be within a distance of 3 units from B. Now look at the number line in fig.22.22 below:

Fig.22.22

6. There are two possibilities:
• A can be within a distance of 3 units towards the left from B
• A can be within a distance of 3 units towards the right from B
7. Let us consider the first possibility. We are in the following situation:
• We have a point B, whose number is known
• We have a point A, whose number is not known
• We have the 'maximum allowed distance that A can be away' from B
8. In this situation, we apply theorem 22.2. A is on the left side of B. So A is lesser than B
The number corresponding to A = Number corresponding to B - Distance between A and B
= (1 - 3) = -2
• So A is not allowed to take any position to the left of -2. If it does, it's distance from B will exceed 3 units
9. Let us consider the second possibility. We are in the following situation:
• We have a point B, whose number is known
• We have a point A, whose number is not known
• We have the 'maximum allowed distance that A can be away' from B
10. In this situation, we apply theorem 22.2. A is on the right side of B. So A is larger than B
The number corresponding to A = Number corresponding to B + Distance between A and B
= (1 + 3) = 4
• So A is not allowed to take any position to the right of 4. If it does, it's distance from B will exceed 3 units
11. All the points that lie in between -2 and 4 will satisfy the given inequality. If we join all such points, we will get a graph of the given inequality. This graph is the red line shown in the upper number line in fig.22.22 above.
12. Note that, the inequality is 'less than OR equal to'. Because of the presence of 'equal to', the points -2 and 4 also qualify to be part of the graph
13. If the inequality is |x-1| < 3, then only those points whose distances are less than 3 from B should be included in the graph. In such a situation, the points at a distance 'exact 3' will be shown in 'hollow circles'.
14. This is shown in the lower number line in fig.22.22 above. Such hollow circles indicate that, those points are not part of the graph.
15. Check: -1.5 is a number in the red line
Put x = -1.5 in the inequality |x-1| ≤ 3
We get: |-1.5-1| ≤ 3 ⇒ |-2.5| ≤ 3 ⇒ 2.5 ≤ 3 Which is true

Solved example 22.7
Find the solutions to the inequality |x-1| ≥ 3 
Solution:
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is 1
• The difference between the numbers = (x-1)
3. Applying theorem 22.5, the distance between A and B is |(x-1)|
4. But |(x-1)| is greater than or equal to 3. So the distance between A and B should never be less than 3 units
5. That means A should always be away from B by a distance of 3 units or more. In other words, A should never come closer to B by a distance less than 3 units. Now look at the number line in fig.22.23 below:

Fig.22.23

6. There are two possibilities:
• A can be away by a distance of 3 or more units towards the left from B
• A can be away by a distance of 3 or more units towards the right from B
7. Let us consider the first possibility. We are in the following situation:
• We have a point B, whose number is known
• We have a point A, whose number is not known
• We have the 'minimum required distance that A should be away' from B
8. In this situation, we apply theorem 22.2. A is on the left side of B. So A is lesser than B
The number corresponding to A = Number corresponding to B - Distance between A and B
= (1 - 3) = -2
• So A is not allowed to take any position to the right of -2. If it does, it's distance from B will become less than 3 units
9. Let us consider the second possibility. We are in the following situation:
• We have a point B, whose number is known
• We have a point A, whose number is not known
• We have the 'minimum required distance that A should be away' from B
10. In this situation, we apply theorem 22.2. A is on the right side of B. So A is larger than B
The number corresponding to A = Number corresponding to B + Distance between A and B
= (1 + 3) = 4
• So A is not allowed to take any position to the left of 4. If it does, it's distance from B will become less than 3 units
11. ♦ All the numbers that lie to the left of -2 will satisfy the given inequality.
      ♦ All the numbers that lie to the right of 4 will satisfy the given inequality.
If we join all such points, we will get a graph of the given inequality. This graph is the red line shown in the upper number line in fig.22.22 above.
• But the red line is broken. This is because, the points which lie in between -2 and 4 does not satisfy the given inequality
• Arrows are given at the left and right ends of the line. This is to show that the graph extends upto infinity on both sides.
    ♦ All numbers on the left of -2 upto infinity satisfies the inequality
    ♦ All numbers on the right of 4 upto infinity satisfies the inequality
12. Note that, the inequality is 'greater than OR equal to'. Because of the presence of 'equal to', the points -2 and 4 also qualify to be part of the graph
13. If the inequality is |x-1| > 3, then only those points whose distances are greater than 3 should be included in the graph. In such a situation, the points at a distance 'exact 3' will be shown in 'hollow circles'.
14. This is shown in the lower number line in fig.22.23 above. Such hollow circles indicate that, those points are not part of the graph.
15. Check: -9.3 is a number in the red line
Put x = -9.3 in the inequality |x-1| ≥ 3

We get: |-9.3-1| ≥ 3 ⇒ |-10.3| ≥ 3 ⇒ 10.3 ≥ 3 Which is true

Solved example 22.8
Find the solutions to the inequality |x+1| ≤ 3
Solution:
• (x+1) can be written as [x-(-1)]
• So  |x+1| ≤ 3 ⇒ |[x-(-1)]| ≤ 3
• Based on this modification we can write:
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is -1
• The difference between the numbers = [x-(-1)]
3. Applying theorem 22.5, the distance between A and B is |[x-(-1)]|
4. But |[x-(-1)]| is less than or equal to 3. So the distance between A and B should never exceed 3 units
5. That means A should always be within a distance of 3 units from B. Now look at the number line in fig.22.24 below:

Fig.22.24

6. There are two possibilities:
• A can be within a distance of 3 units towards the left from B
• A can be within a distance of 3 units towards the right from B
7. Let us consider the first possibility. We are in the following situation:
• We have a point B, whose number is known
• We have a point A, whose number is not known
• We have the 'maximum allowed distance that A can be away' from B
8. In this situation, we apply theorem 22.2. A is on the left side of B. So A is lesser than B
The number corresponding to A = Number corresponding to B - Distance between A and B
= (-1 - 3) = -4
• So A is not allowed to take any position to the left of -4. If it does, it's distance from B will exceed 3 units
9. Let us consider the second possibility. We are in the following situation:
• We have a point B, whose number is known
• We have a point A, whose number is not known
• We have the 'maximum allowed distance that A can be away' from B
10. In this situation, we apply theorem 22.2. A is on the right side of B. So A is larger than B
The number corresponding to A = Number corresponding to B + Distance between A and B
= (-1 + 3) = 2
• So A is not allowed to take any position to the right of 2. If it does, it's distance from B will exceed 3 units
11. All the points that lie in between -4 and 2 will satisfy the given inequality. If we join all such points, we will get a graph of the given inequality. This graph is the red line shown in the upper number line in fig.22.24 above.
12. Note that, the inequality is 'less than OR equal to'. Because of the presence of 'equal to', the points -4 and 2 also qualify to be part of the graph
13. If the inequality is |x+1| < 3, then only those points whose distances are less than 3 from B should be included in the graph. In such a situation, the points at a distance 'exact 3' will be shown in 'hollow circles'.
14. This is shown in the lower number line in fig.22.24 above. Such hollow circles indicate that, those points are not part of the graph.
15. Check: -3.75 is a number in the red line
Put x = -3.75 in the inequality |x+1| ≤ 3

We get: |-3.75+1| ≤ 3 ⇒ |-2.75| ≤ 3 ⇒ 2.75 ≤ 3 Which is true

Solved example 22.9
Find the solutions to the inequality |x+1| ≥ 3
Solution:
• (x+1) can be written as [x-(-1)]
• So  |x+1| ≥ 3 ⇒ |[x-(-1)]| ≥ 3
• Based on this modification we can write:
1. We have a point A on the number line. The number corresponding to the point A is x
2. We have a point B on the number line. The number corresponding to the point B is -1
• The difference between the numbers = [x-(-1)]
3. Applying theorem 22.5, the distance between A and B is |[x-(-1)]|
4. But |(x-1)| is greater than or equal to 3. So the distance between A and B should never be less than 3 units
5. That means A should always be away from B by a distance of 3 units or more. In other words, A should never come closer to B by a distance less than 3 units. Now look at the number line in fig.22.25 below:

Fig.22.25

6. There are two possibilities:
• A can be away by a distance of 3 or more units towards the left from B
• A can be away by a distance of 3 or more units towards the right from B
7. Let us consider the first possibility. We are in the following situation:
• We have a point B, whose number is known
• We have a point A, whose number is not known
• We have the 'minimum required distance that A should be away' from B
8. In this situation, we apply theorem 22.2. A is on the left side of B. So A is lesser than B
The number corresponding to A = Number corresponding to B - Distance between A and B
= (-1 - 3) = -4
• So A is not allowed to take any position to the right of -4. If it does, it's distance from B will become less than 3 units
9. Let us consider the second possibility. We are in the following situation:
• We have a point B, whose number is known
• We have a point A, whose number is not known
• We have the 'minimum required distance that A should be away' from B
10. In this situation, we apply theorem 22.2. A is on the right side of B. So A is larger than B
The number corresponding to A = Number corresponding to B + Distance between A and B
= (-1 + 3) = 2
• So A is not allowed to take any position to the left of 2. If it does, it's distance from B will become less than 3 units
11. ♦ All the numbers that lie to the left of -4 will satisfy the given inequality.
      ♦ All the numbers that lie to the right of 2 will satisfy the given inequality.
If we join all such points, we will get a graph of the given inequality. This graph is the red line shown in the upper number line in fig.22.25 above.
• But the red line is broken. This is because, the points which lie in between -4 and 2 does not satisfy the given inequality
• Arrows are given at the left and right ends of the line. This is to show that the graph extends upto infinity on both sides.
    ♦ All numbers on the left of -4 upto infinity satisfies the inequality
    ♦ All numbers on the right of 2 upto infinity satisfies the inequality
12. Note that, the inequality is 'greater than OR equal to'. Because of the presence of 'equal to', the points -4 and 2 also qualify to be part of the graph
13. If the inequality is |x+1| > 3, then only those points whose distances are greater than 3 should be included in the graph. In such a situation, the points at a distance 'exact 3' will be shown in 'hollow circles'.
14. This is shown in the lower number line in fig.22.23 above. Such hollow circles indicate that, those points are not part of the graph.
15. Check: -20.5 is a number in the red line
Put x = -20.5 in the inequality |x+1| ≥ 3
We get: |-20.5+1| ≥ 3 ⇒ |-19.5| ≥ 3 ⇒ 19.5 ≥ 3 Which is true

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In the next section we will see more solved examples.

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