In the previous section we saw that the distance between any number and zero is the absolute value of that number. In this section we will see the 'distance between any two points' expressed as absolute value.
We have seen the method to calculate the distance between any two points on the number line:
• The smaller number is subtracted from the larger number.
We saw it in theorem 22.1. Let us analyse it again:
Calculation of the distance between a few pairs of points is shown below. Calculations are based on theorem 22.1.
Example 1
Number corresponding to Point A = 5
Number corresponding to Point B = 3
Smaller of the two is 3
So distance between A and B = 5 – 3 = 2 units
Example 2
Number corresponding to Point A = 4
Number corresponding to Point B = -2
Smaller of the two is -2
So distance between A and B = 4 – (-2) = 4 + 2 = 6 units
Example 3
Number corresponding to Point A = -8
Number corresponding to Point B = -5
Smaller of the two is -8
So distance between A and B = -5 – (-8) = -5 + 8 = 3 units
Let us check for the above examples:
• In example 1, we got the result as 2 units.
If we subtract the larger from the smaller, the result will be 3 – 5 = -2 units
• In example 2, we got the result as 6 units.
If we subtract the larger from the smaller, the result will be -2 – 4 = -6 units
• In example 3, we got the result as 3 units.
If we subtract the larger from the smaller, the result will be -8 – (-5) = -8 + 5 = -3 units
So we find that the signs have reversed. But for specifying distance, we need only the magnitude. Sign does not have a role to play. For example, the petrol consumed by a car for travelling 10 km will be the same whether it travels towards the east, or towards the west, if the road conditions are the same.
That means, we do not have to strictly follow the rule:
Subtract smaller number from the larger number
All we need is the difference. If we take the absolute value of that difference, we will get the required distance without any sign. Let us see the above 3 examples again:
• In example 1, we may get the difference as 2 or -2, whatever be the difference, we must take it's absolute value:
|2| = 2 AND |-2| = 2
• In example 2, we may get the difference as 6 or -6, whatever be the difference, we must take it's absolute value:
|6| = 6 AND |-6| = 6
• In example 3, we may get the difference as 3 or -3, whatever be the difference, we must take it's absolute value:
|3| = 3 AND |-3| = 3
■ So, if we decide to always take the absolute value of the difference, we will get the required distance, regardless of whether we subtract the smaller from the larger or vice versa.
We can write this in the form of a theorem:
Theorem 22.5
• We have a point A on the number line. We know the number corresponding to the point A
• We have a point B on the number line. We know the number corresponding to the point B
• Calculate the difference between the numbers
• Take the absolute value of the difference
• This absolute value is the 'distance between the points A and B'
Example:
• We have a point A on the number line. The number corresponding to the point A is 2
• We have a point B on the number line. The number corresponding to the point B is 5
• Calculate the difference between the numbers: (2 - 5) = -3
• Take the absolute value of the difference: |-3| = 3
• This absolute value '3' is the distance between the points A and B
Now we will see some solved examples:
Solved example 22.4
• Consider the equation |x-1| = 3
• Find the value of x that will satisfy the above equation
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. 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 given as 3. So the distance between A and B is 3 units
5. That means A is 3 units away from B. Now look at the number line in fig.22.20 below:
6. There are two possibilities:
• A can be at a distance of 3 units towards the left from B
• A can be at 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 distance between A and 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
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 distance between A and 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
11. • From (8), we have A = -2
• From (10), we have A = 4
• Both values will satisfy the equation |x-1| = 3
• They are marked by red 'filled circles' in the number line in fig.22.20
12. Check: Put x = -2 in the equation |x-1| = 3
We get: |-2-1| = 3 ⇒ |-3| = 3 ⇒ 3 = 3 Which is true
Put x = 4 in the equation |x-1| = 3
We get: |4-1| = 3 ⇒ |3| = 3 ⇒ 3 = 3 Which is true
■ We started out to find 'the value' of x which will satisfy the equation |x-1| = 3. When all steps are completed, we find that there are 'two values' of x that will satisfy the equation
■ The process of finding the value or values of the variable (x in our case) can be written in two ways:
♦ Finding the solutions to the equation
♦ Solving the equation
Solved example 22.5
• Consider the equation |x+1| = 3
• Find the value of x that will satisfy the above equation
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
3. 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 given as 3. So the distance between A and B is 3 units
5. That means A is 3 units away from B. Now look at the number line in fig.22.21 below:
6. There are two possibilities:
• A can be at a distance of 3 units towards the left from B
• A can be at 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 distance between A and 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
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 distance between A and 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
11. • From (8), we have A = -4
• From (10), we have A = 2
• Both values will satisfy the equation |[x-(-1)]| = 3
• They are marked by red 'filled circles' in the number line in fig.22.21
12. Check: Put x = -4 in the equation |x+1| = 3
We get: |-4+1| = -3 ⇒ |-3| = 3 ⇒ 3 = 3 Which is true
Put x = 2 in the equation |x+1| = 3
We get: |2+1| = 3 ⇒ |3| = 3 ⇒ 3 = 3 Which is true
■ We started out to find 'the value' of x which will satisfy the equation |x+1| = 3. When all steps are completed, we find that there are 'two values' of x that will satisfy the equation
■ The process of finding the value or values of the variable (x in our case) can be written in two ways:
♦ Finding the solutions to the equation
♦ Solving the equation
We have seen the method to calculate the distance between any two points on the number line:
• The smaller number is subtracted from the larger number.
We saw it in theorem 22.1. Let us analyse it again:
Calculation of the distance between a few pairs of points is shown below. Calculations are based on theorem 22.1.
Example 1
Number corresponding to Point A = 5
Number corresponding to Point B = 3
Smaller of the two is 3
So distance between A and B = 5 – 3 = 2 units
Example 2
Number corresponding to Point A = 4
Number corresponding to Point B = -2
Smaller of the two is -2
So distance between A and B = 4 – (-2) = 4 + 2 = 6 units
Example 3
Number corresponding to Point A = -8
Number corresponding to Point B = -5
Smaller of the two is -8
So distance between A and B = -5 – (-8) = -5 + 8 = 3 units
■ Note that, when we subtract the smaller from the larger, we are calculating the 'difference' between the two numbers.
■ If we subtract the larger number from the smaller, then also, we will get the 'difference'. In that case, we will get a negative answer.
Let us check for the above examples:
• In example 1, we got the result as 2 units.
If we subtract the larger from the smaller, the result will be 3 – 5 = -2 units
• In example 2, we got the result as 6 units.
If we subtract the larger from the smaller, the result will be -2 – 4 = -6 units
• In example 3, we got the result as 3 units.
If we subtract the larger from the smaller, the result will be -8 – (-5) = -8 + 5 = -3 units
So we find that the signs have reversed. But for specifying distance, we need only the magnitude. Sign does not have a role to play. For example, the petrol consumed by a car for travelling 10 km will be the same whether it travels towards the east, or towards the west, if the road conditions are the same.
That means, we do not have to strictly follow the rule:
Subtract smaller number from the larger number
All we need is the difference. If we take the absolute value of that difference, we will get the required distance without any sign. Let us see the above 3 examples again:
• In example 1, we may get the difference as 2 or -2, whatever be the difference, we must take it's absolute value:
|2| = 2 AND |-2| = 2
• In example 2, we may get the difference as 6 or -6, whatever be the difference, we must take it's absolute value:
|6| = 6 AND |-6| = 6
• In example 3, we may get the difference as 3 or -3, whatever be the difference, we must take it's absolute value:
|3| = 3 AND |-3| = 3
■ So, if we decide to always take the absolute value of the difference, we will get the required distance, regardless of whether we subtract the smaller from the larger or vice versa.
We can write this in the form of a theorem:
Theorem 22.5
• We have a point A on the number line. We know the number corresponding to the point A
• We have a point B on the number line. We know the number corresponding to the point B
• Calculate the difference between the numbers
• Take the absolute value of the difference
• This absolute value is the 'distance between the points A and B'
Example:
• We have a point A on the number line. The number corresponding to the point A is 2
• We have a point B on the number line. The number corresponding to the point B is 5
• Calculate the difference between the numbers: (2 - 5) = -3
• Take the absolute value of the difference: |-3| = 3
• This absolute value '3' is the distance between the points A and B
Now we will see some solved examples:
Solved example 22.4
• Consider the equation |x-1| = 3
• Find the value of x that will satisfy the above equation
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. 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 given as 3. So the distance between A and B is 3 units
5. That means A is 3 units away from B. Now look at the number line in fig.22.20 below:
 |
| Fig.22.20 |
• A can be at a distance of 3 units towards the left from B
• A can be at 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 distance between A and 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
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 distance between A and 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
11. • From (8), we have A = -2
• From (10), we have A = 4
• Both values will satisfy the equation |x-1| = 3
• They are marked by red 'filled circles' in the number line in fig.22.20
12. Check: Put x = -2 in the equation |x-1| = 3
We get: |-2-1| = 3 ⇒ |-3| = 3 ⇒ 3 = 3 Which is true
Put x = 4 in the equation |x-1| = 3
We get: |4-1| = 3 ⇒ |3| = 3 ⇒ 3 = 3 Which is true
■ We started out to find 'the value' of x which will satisfy the equation |x-1| = 3. When all steps are completed, we find that there are 'two values' of x that will satisfy the equation
■ The process of finding the value or values of the variable (x in our case) can be written in two ways:
♦ Finding the solutions to the equation
♦ Solving the equation
Solved example 22.5
• Consider the equation |x+1| = 3
• Find the value of x that will satisfy the above equation
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
3. 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 given as 3. So the distance between A and B is 3 units
5. That means A is 3 units away from B. Now look at the number line in fig.22.21 below:
 |
| Fig.22.21 |
• A can be at a distance of 3 units towards the left from B
• A can be at 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 distance between A and 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
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 distance between A and 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
11. • From (8), we have A = -4
• From (10), we have A = 2
• Both values will satisfy the equation |[x-(-1)]| = 3
• They are marked by red 'filled circles' in the number line in fig.22.21
12. Check: Put x = -4 in the equation |x+1| = 3
We get: |-4+1| = -3 ⇒ |-3| = 3 ⇒ 3 = 3 Which is true
Put x = 2 in the equation |x+1| = 3
We get: |2+1| = 3 ⇒ |3| = 3 ⇒ 3 = 3 Which is true
■ We started out to find 'the value' of x which will satisfy the equation |x+1| = 3. When all steps are completed, we find that there are 'two values' of x that will satisfy the equation
■ The process of finding the value or values of the variable (x in our case) can be written in two ways:
♦ Finding the solutions to the equation
♦ Solving the equation
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