In the previous section we saw the distance between two points. In this section we will see the midpoint between two points. Later in this section, we will learn about Absolute value.
Consider the fig.22.13 below. Two points A and B are marked in red. C is the midpoint between A and B.
Our aim is to find the number corresponding to the midpoint C. The following steps can be used:
1. The numbers corresponding to A and B are x1 and x2.
2. Point A is on the left side of point B. So x1 is less than x2.
3. According to theorem 22.1 that we saw in the previous section, the distance between A and B is (x2-x1)
4. Half of this distance is (x2-x1)⁄2
• This is the distance between A and C
• This is also the distance between B and C
5. So, we have the following information:
• The number corresponding to the left side point A is x1
• The distance between A and C is (x2-x1)⁄2
• The number corresponding to the right side point B is x2
• The distance between B and C is (x2-x1)⁄2
6. To apply theorem 22.2, we can use the 'first two' or 'last two' of the above four information
7. Let us take the first two. Applying theorem 22.2,
The number corresponding to C = number corresponding to A + distance between A and C
= x1 + (x2-x1)⁄2 = (x1)⁄2 + (x2)⁄2 = (x1+x2)⁄2
8. Let us check the above result by using the last two information in (5)
Applying theorem 22.2,
The number corresponding to C = number corresponding to B - distance between A and C
= x2 - (x2-x1)⁄2 = (2x2)⁄2 - (x2)⁄2 + (x1)⁄2 = (x2)⁄2 + (x1)⁄2 = (x1+x2)⁄2
This is the same result in (7). So we have an easy method to find the midpoint between two points. We can write it in the form of a theorem.
Theorem 22.3:
• We are given two points
• The numbers corresponding to those points are also given
• Take the sum of those numbers
• Half of that sum will be the number corresponding to the midpoint between the given points
Let us see a sample calculation:
A and B are two points on the number line. The number corresponding to A is -21⁄2. The number corresponding to B is 43⁄4. Find the number corresponding to the midpoint C between A and B
Solution: Sum of the numbers is (-21⁄2 + 43⁄4 ) = 21⁄4.
The number corresponding to the midpoint = half of the above sum = 1⁄2 × 21⁄4 = 11⁄8. This is shown in the fig.22.14 below:
Now we will see some solved examples
Solved example 22.1
Find the distance between two points on the number line denoted by each pair of numbers given below:
(i) 1, -5 (ii) 1⁄2 , 2⁄3 (iii) -1⁄2 , -1⁄3 (iv) -1⁄2 , 3⁄4 (v) -√2 , -√3
Solution:
We can find the distance by applying theorem 22.1
In each taking the smaller number as x1 and the larger number as x2 , the distance is (x1-x2)
(i) Distance = 1-(-5) = 1+5 = 6
(ii) Distance = 2⁄3 - 1⁄2 = 4⁄6 - 3⁄6 = 1⁄6
(iii) Distance = -1⁄3 -(- 1⁄2) = -1⁄3 + 1⁄2 = -2⁄6 + 3⁄6 = 1⁄6
(iv) Distance = 3⁄4 -(- 1⁄2) = 3⁄4 + 1⁄2 = 3⁄4 + 2⁄4 = 5⁄4 = 11⁄4
(v) Distance = -√2 -(-√3) = -√2 + √3 = (√3 - √2)
Solved example 22.2
Find the midpoint of each pair of points in the first problem
Solution:
We can use theorem 22.3 that we saw above in this section. We have to take half of the sum
(i) Midpoint = [1+(-5)]⁄2 = -4⁄2 = -2
(ii) Midpoint = 1⁄2 × (1⁄2 + 2⁄3) = 1⁄2 × (3⁄6 + 4⁄6) = 1⁄2 × (7⁄6) = 7⁄12.
(iii) Midpoint = 1⁄2 × [(-1⁄2 )+(-1⁄3)] = 1⁄2 × [(-3⁄6 )+(-2⁄6)] = 1⁄2 × (-5⁄6) = -5⁄12.
(iv) Midpoint = 1⁄2 × [(-1⁄2 )+(3⁄4)] = 1⁄2 × [(-2⁄4 )+(3⁄4)] = 1⁄2 × (1⁄4) = 1⁄8.
(v) Midpoint = [(-√2)+(-√3)]⁄2 = -(√2+√3)⁄2
Solved example 22.3
The part of the number line between the points denoted by numbers 1⁄3 and 1⁄2 is divided into four equal parts. Find the numbers denoting the ends of each such part
Solution:
1. It is better to draw a rough sketch as shown in fig.22.15 below.
1⁄3 is less than 1⁄2 . So 1⁄3 (marked as A) will be on the left of 1⁄2 (marked as B)
2. The distance between the two points A and B is (1⁄2 - 1⁄3) = (3⁄6 - 2⁄6) = 1⁄6.
3. This distance is divided into 4 equal parts. The division is done by three points C, D and E. Each of the 4 equal parts will be (1⁄4 × 1⁄6 ) = 1⁄24.
4. So, applying theorem 22.3, the number corresponding to C
= Number corresponding to A + distance AC
= (1⁄3 + 1⁄24) = (8⁄24 + 1⁄24) = 9⁄24 = 3⁄8.
5. Number corresponding to D = (9⁄24 + 1⁄24) = 10⁄24 = 5⁄12
6. Number corresponding to E = (10⁄24 + 1⁄24) = 11⁄24
7. Check: Number corresponding to B = (11⁄24 + 1⁄24) = 12⁄24 = 1⁄2.
1⁄2 is the number corresponding to B. Thus we reach point B. The calculations are correct.
♦ The distance of A from zero is 4 units
• Consider point B. The number corresponding to B is 3.
♦ The distance of B from zero is 3 units
In general:
• Consider any point on the right side of zero.
♦ It's distance from zero is same as it's number
♦ All we need to do is, add the word 'units' to the number. This is to indicate that it is a distance.
♦ The unit may be cm, m or any other appropriate unit
• Consider any point on the left side of zero
♦ It's distance from zero is same as it's 'number with out the negative sign'
♦ All we need to do is, add the word 'units' to the number. This is to indicate that it is a distance.
♦ The unit may be cm, m or any other appropriate unit
We know that, if we multiply a negative number by '-1', the number will become positive. So, for the points on the left we can say this:
• Consider any point on the left side of zero
♦ It's distance from zero is same as it's 'number multiplied by -1'
♦ All we need to do is, add the word 'units' to the number. This is to indicate that it is a distance.
♦ The unit may be cm, m or any other appropriate unit
For example, if the number corresponding to a point is -3, the distance of that point from zero is:
-1× -3 = 3 units
Note that if the point under consideration is zero, it's distance from zero is '0'. We can write the three situations together:
Consider any point on the number line. Let the number corresponding to the point be 'x'. Then:
• Distance of the point from zero = x (if x > 0)
• Distance of the point from zero = 0 (if x = 0)
• Distance of the point from zero = -1 x (if x < 0)
This distance of a point from zero is called Absolute value of that number. It is written using symbol as |x|. It is read as 'Absolute value of x'.
We can write the above points in the form of a theorem.
Theorem 22.4
• We have a point A on the number line.
• We know the number corresponding to the point A
• The distance of point A from zero, is the absolute value of the number
Example:
1. A point is marked on the number line. The number corresponding to that point is 5.
2. Then the distance of that point from zero = |5| = 5
Another example:
1. A point is marked on the number line. The number corresponding to that point is -2.
2. Then the distance of that point from zero = |-2| = 2
A spread sheet showing the absolute values of some numbers can be seen here. The formula used is '=ABS()'
• We want the value of x that will satisfy the above equation
Solution:
1. 'x' is a number on the number line. If we put a particular value for 'x', the left side of the equation will become equal to the right side. We want this 'particular value'
2. The left side is: Absolute value of x
• We know that absolute value of a number is the distance of the 'corresponding point' from zero
3. A point whose number is '2' will be at a distance of 2 units from zero.
4. So the number that we want is 2. We can write: |2| = 2. So x = 2
5. But there is another point also.
We know that, the point whose number is '-2' will also be at a distance of 2 units from zero
We can write |-2| = 2. So x = -2
6. Thus, there are two values for 'x' that will solve the given equation. They are x = 2 and -2
7. This is shown in the fig.22.17 below:
■ We started out to find 'the value' of x which will satisfy the equation |x| = 2. When all steps are completed, we find that there are 'two values' of x that will satisfy the equation.
Another problem:
• Consider the inequality |x| ≤ 2
• We want the value of x that will satisfy the above inequality
Solution:
1. 'x' is a number on the number line. If we put a particular value for 'x', the left side of the inequality will become less than or equal to the right side. We want this 'particular value'
2. The left side is: Absolute value of x
• We know that absolute value of a number is the distance of the 'corresponding point' from zero
3. A point whose number is '2' will be at a distance of 2 units from zero.
4. So the number that we want is 2. We can write: |2| = 2. So x = 2
5. But there is another point also.
We know that, the point whose number is '-2' will also be at a distance of 2 units from zero
We can write |-2| = 2. So x = -2
6. But 2 and -2 are not the only points. Look at the inequality symbol '≤'. It is less than OR equal to. Points whose distances are less than 2 will also satisfy the inequality.
7. For example consider 1.5 and -1.5. Both have a distance less than 2. There are numerous such points
8. All such points which satisfy the given inequality lies between -2 and 2. 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.22.18 below:
9. Note that, the inequality is 'less than OR equal to'. Because of the presence of 'equal to', the points 2 and -2 also qualify to be part of the graph
• We know that, on the red line, there are numerous points. Every point on the red line will satisfy the inequality |x| ≤ 2
♦For example weird integers like (-1182⁄2574) fall on the red line. They will also satisfy the inequality |x| ≤ 2
10. If the inequality is |x| < 2, then only those points whose distances are less than 2 should be included in the graph. In such a situation, the points at a distance 'exact 2' will be shown in 'hollow circles'.
11. This is shown in the lower number line in fig.22.18 above. Such hollow circles indicate that, those points are not part of the graph.
■ We started out to find 'the value' of x which will satisfy the inequality |x| ≤ 2. When all steps are completed, we find that there are 'numerous values' of x that will satisfy the inequality
One more problem:
• Consider the inequality |x| ≥ 2
• We want the value of x that will satisfy the above inequality
Solution:
1. 'x' is a number on the number line. If we put a particular value for 'x', the left side of the inequality will become greater than or equal to the right side. We want this 'particular value'
2. The left side is: Absolute value of x
• We know that absolute value of a number is the distance of the 'corresponding point' from zero
3. A point whose number is '2' will be at a distance of 2 units from zero.
4. So the number that we want is 2. We can write: |2| = 2. So x = 2
5. But there is another point also.
We know that, the point whose number is '-2' will also be at a distance of 2 units from zero
We can write |-2| = 2. So x = -2
6. But 2 and -2 are not the only points. Look at the inequality symbol '≥'. It is greater than OR equal to. Points whose distances are greater than 2 will also satisfy the inequality.
7. For example consider 2.5 and -2.5. Both have a distance greater than 2. There are numerous such points
8. • All points which lie to the left of -2 are at a distance greater than 2 from zero. They satisfy the inequality
• All points which lie to the right of 2 are at a distance greater than 2 from zero. They satisfy the 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.22.19 below:
• But the red line is broken. This is because, the points which lie in between -2 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 -2 upto infinity satisfies the inequality
♦ All numbers on the right of 2 upto infinity satisfies the inequality
9. Note that, the inequality is 'greater than OR equal to'. Because of the presence of 'equal to', the points 2 and -2 also qualify to be part of the graph
• We know that, on the red line, there are numerous points. Every point on the red line will satisfy the inequality |x| ≥ 2
♦For example weird integers like (-201742⁄2574) fall on the red line. They will also satisfy the inequality |x| ≥ 2
10. If the inequality is |x| > 2, then only those points whose distances are greater than 2 should be included in the graph. In such a situation, the points at a distance 'exact 2' will be shown in 'hollow circles'.
11. This is shown in the lower number line in fig.22.19 above. Such hollow circles indicate that, those points are not part of the graph.
■ We started out to find 'the value' of x which will satisfy the inequality |x| ≥ 2. When all steps are completed, we find that there are 'numerous values' of x that will satisfy the inequality
Consider the fig.22.13 below. Two points A and B are marked in red. C is the midpoint between A and B.
 |
| Fig.22.13 |
1. The numbers corresponding to A and B are x1 and x2.
2. Point A is on the left side of point B. So x1 is less than x2.
3. According to theorem 22.1 that we saw in the previous section, the distance between A and B is (x2-x1)
4. Half of this distance is (x2-x1)⁄2
• This is the distance between A and C
• This is also the distance between B and C
5. So, we have the following information:
• The number corresponding to the left side point A is x1
• The distance between A and C is (x2-x1)⁄2
• The number corresponding to the right side point B is x2
• The distance between B and C is (x2-x1)⁄2
6. To apply theorem 22.2, we can use the 'first two' or 'last two' of the above four information
7. Let us take the first two. Applying theorem 22.2,
The number corresponding to C = number corresponding to A + distance between A and C
= x1 + (x2-x1)⁄2 = (x1)⁄2 + (x2)⁄2 = (x1+x2)⁄2
8. Let us check the above result by using the last two information in (5)
Applying theorem 22.2,
The number corresponding to C = number corresponding to B - distance between A and C
= x2 - (x2-x1)⁄2 = (2x2)⁄2 - (x2)⁄2 + (x1)⁄2 = (x2)⁄2 + (x1)⁄2 = (x1+x2)⁄2
This is the same result in (7). So we have an easy method to find the midpoint between two points. We can write it in the form of a theorem.
Theorem 22.3:
• We are given two points
• The numbers corresponding to those points are also given
• Take the sum of those numbers
• Half of that sum will be the number corresponding to the midpoint between the given points
Let us see a sample calculation:
A and B are two points on the number line. The number corresponding to A is -21⁄2. The number corresponding to B is 43⁄4. Find the number corresponding to the midpoint C between A and B
Solution: Sum of the numbers is (-21⁄2 + 43⁄4 ) = 21⁄4.
The number corresponding to the midpoint = half of the above sum = 1⁄2 × 21⁄4 = 11⁄8. This is shown in the fig.22.14 below:
 |
| Fig.22.14 |
Solved example 22.1
Find the distance between two points on the number line denoted by each pair of numbers given below:
(i) 1, -5 (ii) 1⁄2 , 2⁄3 (iii) -1⁄2 , -1⁄3 (iv) -1⁄2 , 3⁄4 (v) -√2 , -√3
Solution:
We can find the distance by applying theorem 22.1
In each taking the smaller number as x1 and the larger number as x2 , the distance is (x1-x2)
(i) Distance = 1-(-5) = 1+5 = 6
(ii) Distance = 2⁄3 - 1⁄2 = 4⁄6 - 3⁄6 = 1⁄6
(iii) Distance = -1⁄3 -(- 1⁄2) = -1⁄3 + 1⁄2 = -2⁄6 + 3⁄6 = 1⁄6
(iv) Distance = 3⁄4 -(- 1⁄2) = 3⁄4 + 1⁄2 = 3⁄4 + 2⁄4 = 5⁄4 = 11⁄4
(v) Distance = -√2 -(-√3) = -√2 + √3 = (√3 - √2)
Solved example 22.2
Find the midpoint of each pair of points in the first problem
Solution:
We can use theorem 22.3 that we saw above in this section. We have to take half of the sum
(i) Midpoint = [1+(-5)]⁄2 = -4⁄2 = -2
(ii) Midpoint = 1⁄2 × (1⁄2 + 2⁄3) = 1⁄2 × (3⁄6 + 4⁄6) = 1⁄2 × (7⁄6) = 7⁄12.
(iii) Midpoint = 1⁄2 × [(-1⁄2 )+(-1⁄3)] = 1⁄2 × [(-3⁄6 )+(-2⁄6)] = 1⁄2 × (-5⁄6) = -5⁄12.
(iv) Midpoint = 1⁄2 × [(-1⁄2 )+(3⁄4)] = 1⁄2 × [(-2⁄4 )+(3⁄4)] = 1⁄2 × (1⁄4) = 1⁄8.
(v) Midpoint = [(-√2)+(-√3)]⁄2 = -(√2+√3)⁄2
Solved example 22.3
The part of the number line between the points denoted by numbers 1⁄3 and 1⁄2 is divided into four equal parts. Find the numbers denoting the ends of each such part
Solution:
1. It is better to draw a rough sketch as shown in fig.22.15 below.
 |
| Fig.22.15 |
2. The distance between the two points A and B is (1⁄2 - 1⁄3) = (3⁄6 - 2⁄6) = 1⁄6.
3. This distance is divided into 4 equal parts. The division is done by three points C, D and E. Each of the 4 equal parts will be (1⁄4 × 1⁄6 ) = 1⁄24.
4. So, applying theorem 22.3, the number corresponding to C
= Number corresponding to A + distance AC
= (1⁄3 + 1⁄24) = (8⁄24 + 1⁄24) = 9⁄24 = 3⁄8.
5. Number corresponding to D = (9⁄24 + 1⁄24) = 10⁄24 = 5⁄12
6. Number corresponding to E = (10⁄24 + 1⁄24) = 11⁄24
7. Check: Number corresponding to B = (11⁄24 + 1⁄24) = 12⁄24 = 1⁄2.
1⁄2 is the number corresponding to B. Thus we reach point B. The calculations are correct.
Absolute value of numbers
• Consider the point A in the number line shown in fig.22.16 below. The number corresponding to A is -4.♦ The distance of A from zero is 4 units
 |
| Fig.22.16 |
♦ The distance of B from zero is 3 units
In general:
• Consider any point on the right side of zero.
♦ It's distance from zero is same as it's number
♦ All we need to do is, add the word 'units' to the number. This is to indicate that it is a distance.
♦ The unit may be cm, m or any other appropriate unit
• Consider any point on the left side of zero
♦ It's distance from zero is same as it's 'number with out the negative sign'
♦ All we need to do is, add the word 'units' to the number. This is to indicate that it is a distance.
♦ The unit may be cm, m or any other appropriate unit
We know that, if we multiply a negative number by '-1', the number will become positive. So, for the points on the left we can say this:
• Consider any point on the left side of zero
♦ It's distance from zero is same as it's 'number multiplied by -1'
♦ All we need to do is, add the word 'units' to the number. This is to indicate that it is a distance.
♦ The unit may be cm, m or any other appropriate unit
For example, if the number corresponding to a point is -3, the distance of that point from zero is:
-1× -3 = 3 units
Note that if the point under consideration is zero, it's distance from zero is '0'. We can write the three situations together:
Consider any point on the number line. Let the number corresponding to the point be 'x'. Then:
• Distance of the point from zero = x (if x > 0)
• Distance of the point from zero = 0 (if x = 0)
• Distance of the point from zero = -1 x (if x < 0)
This distance of a point from zero is called Absolute value of that number. It is written using symbol as |x|. It is read as 'Absolute value of x'.
We can write the above points in the form of a theorem.
Theorem 22.4
• We have a point A on the number line.
• We know the number corresponding to the point A
• The distance of point A from zero, is the absolute value of the number
Example:
1. A point is marked on the number line. The number corresponding to that point is 5.
2. Then the distance of that point from zero = |5| = 5
Another example:
1. A point is marked on the number line. The number corresponding to that point is -2.
2. Then the distance of that point from zero = |-2| = 2
A spread sheet showing the absolute values of some numbers can be seen here. The formula used is '=ABS()'
Let us see some common situations where we use the absolute value:
• Consider the equation |x| = 2• We want the value of x that will satisfy the above equation
Solution:
1. 'x' is a number on the number line. If we put a particular value for 'x', the left side of the equation will become equal to the right side. We want this 'particular value'
2. The left side is: Absolute value of x
• We know that absolute value of a number is the distance of the 'corresponding point' from zero
3. A point whose number is '2' will be at a distance of 2 units from zero.
4. So the number that we want is 2. We can write: |2| = 2. So x = 2
5. But there is another point also.
We know that, the point whose number is '-2' will also be at a distance of 2 units from zero
We can write |-2| = 2. So x = -2
6. Thus, there are two values for 'x' that will solve the given equation. They are x = 2 and -2
7. This is shown in the fig.22.17 below:
 |
| Fig.22.17 |
Another problem:
• Consider the inequality |x| ≤ 2
• We want the value of x that will satisfy the above inequality
Solution:
1. 'x' is a number on the number line. If we put a particular value for 'x', the left side of the inequality will become less than or equal to the right side. We want this 'particular value'
2. The left side is: Absolute value of x
• We know that absolute value of a number is the distance of the 'corresponding point' from zero
3. A point whose number is '2' will be at a distance of 2 units from zero.
4. So the number that we want is 2. We can write: |2| = 2. So x = 2
5. But there is another point also.
We know that, the point whose number is '-2' will also be at a distance of 2 units from zero
We can write |-2| = 2. So x = -2
6. But 2 and -2 are not the only points. Look at the inequality symbol '≤'. It is less than OR equal to. Points whose distances are less than 2 will also satisfy the inequality.
7. For example consider 1.5 and -1.5. Both have a distance less than 2. There are numerous such points
8. All such points which satisfy the given inequality lies between -2 and 2. 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.22.18 below:
 |
| Fig.22.18 |
• We know that, on the red line, there are numerous points. Every point on the red line will satisfy the inequality |x| ≤ 2
♦For example weird integers like (-1182⁄2574) fall on the red line. They will also satisfy the inequality |x| ≤ 2
10. If the inequality is |x| < 2, then only those points whose distances are less than 2 should be included in the graph. In such a situation, the points at a distance 'exact 2' will be shown in 'hollow circles'.
11. This is shown in the lower number line in fig.22.18 above. Such hollow circles indicate that, those points are not part of the graph.
■ We started out to find 'the value' of x which will satisfy the inequality |x| ≤ 2. When all steps are completed, we find that there are 'numerous values' of x that will satisfy the inequality
One more problem:
• Consider the inequality |x| ≥ 2
• We want the value of x that will satisfy the above inequality
Solution:
1. 'x' is a number on the number line. If we put a particular value for 'x', the left side of the inequality will become greater than or equal to the right side. We want this 'particular value'
2. The left side is: Absolute value of x
• We know that absolute value of a number is the distance of the 'corresponding point' from zero
3. A point whose number is '2' will be at a distance of 2 units from zero.
4. So the number that we want is 2. We can write: |2| = 2. So x = 2
5. But there is another point also.
We know that, the point whose number is '-2' will also be at a distance of 2 units from zero
We can write |-2| = 2. So x = -2
6. But 2 and -2 are not the only points. Look at the inequality symbol '≥'. It is greater than OR equal to. Points whose distances are greater than 2 will also satisfy the inequality.
7. For example consider 2.5 and -2.5. Both have a distance greater than 2. There are numerous such points
8. • All points which lie to the left of -2 are at a distance greater than 2 from zero. They satisfy the inequality
• All points which lie to the right of 2 are at a distance greater than 2 from zero. They satisfy the 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.22.19 below:
 |
| Fig.22.19 |
• 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 2 upto infinity satisfies the inequality
9. Note that, the inequality is 'greater than OR equal to'. Because of the presence of 'equal to', the points 2 and -2 also qualify to be part of the graph
• We know that, on the red line, there are numerous points. Every point on the red line will satisfy the inequality |x| ≥ 2
♦For example weird integers like (-201742⁄2574) fall on the red line. They will also satisfy the inequality |x| ≥ 2
10. If the inequality is |x| > 2, then only those points whose distances are greater than 2 should be included in the graph. In such a situation, the points at a distance 'exact 2' will be shown in 'hollow circles'.
11. This is shown in the lower number line in fig.22.19 above. Such hollow circles indicate that, those points are not part of the graph.
■ We started out to find 'the value' of x which will satisfy the inequality |x| ≥ 2. When all steps are completed, we find that there are 'numerous values' of x that will satisfy the inequality
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