Showing posts with label midpoint. Show all posts
Showing posts with label midpoint. Show all posts

Wednesday, February 7, 2018

Chapter 34.4 - Solved examples on Section formula

In the previous section we saw the section formula. We also saw some examples. In this section we will see a few more solved examples.

 Solved example 34.6
The coordinates of the vertices of a quadrilateral are (2,1), (5,3), (8,7) and (4,9) in order.
(i) Find the coordinates of the midpoints of all sides
(ii) Prove that the quadrilateral with these midpoints as vertices is a parallelogram
Solution:
• Let us name the coordinates:
A(2,1), B(5,3), C(8,7), D(4,9)
1. Now we will calculate the midpoints:
(i) Midpoint of AB (Let us call it P):
[(x1+x2)⁄2,(y1+y2)⁄2] = [(2+5)⁄2,(1+3)⁄2] = [7⁄2,4⁄2] = (3.5,2)
(ii) Midpoint of BC (Let us call it Q):
[(5+8)⁄2,(3+7)⁄2] = [13⁄2,10⁄2] = (6.5,5)
(iii) Midpoint of CD (Let us call it R):
[(8+4)⁄2,(7+9)⁄2] = [12⁄2,16⁄2] = (6,8)
(iv) Midpoint of DA (Let us call it S):
[(4+2)⁄2,(9+1)⁄2] = [6⁄2,10⁄2] = (3,5)
2. We took the vertices A, B C and D in order.
• So P, Q, R and S are the midpoints taken in order:
P(3.5,2), Q(6.5,5), R(6,8), S(3,5)
• We have to prove that these points are vertices of a parallelogram
■ If both pairs of opposite sides are equal, then the quadrilateral is a parallelogram. 
3. In our case:
• One pair of opposite sides is [PQ, RS]
• The other pair of opposite sides is [QR,SP]
(A rough sketch of a quadrilateral will help to identify opposite sides)
4. We will use the distance formula to calculate the distances
• Length of PQ = √[(6.5-3.5)2 + (5-2)2] = √[(3)2 + (3)2] = √[2×(3)2] = 3√2
• Length of RS = √[(3-6)2 + (5-8)2] = √[(-3)2 + (-3)2] = √[2×(-3)2] = 3√2
• Length of QR = √[(6-6.5)2 + (8-5)2] = √[(-0.5)2 + (3)2] = √[0.25 + 9] = √9.25
• Length of SP = √[(3.5-3)2 + (2-5)2] = √[(0.5)2 + (-3)2] = √[0.25 + 9] = √9.25
■ Thus we find that PQ = RS AND QR = SP
So both pairs of opposite sides are equal. Thus it is a parallelogram
• The actual points in the Cartesian plane are shown in fig.34.13 below:
Fig.34.13
Solved example 34.7
In the fig.34.14(a) below, the midpoints of the sides of the large quadrilateral ABCD are joined to draw the small quadrilateral PQRS inside
Fig.34.14
(i) Find the coordinates of the fourth vertex P of the small quadrilateral
(ii) Find the coordinates of the other three vertices B, C and D of the large quadrilateral
Solution:
• We have seen in an earlier chapter that, the quadrilateral obtained by joining the midpoints of any other quadrilateral will always be a parallelogram
    ♦ See solved example 18.13 in section 18.7
• So PQRS is a parallelogram. We have seen the method to find the unknown vertex of a parallelogram when all the other three vertices are given
1.So the first step is to find the coordinates of P
(i) Group the vertices into two: [P,Q] and [R,S]
(The members of a group should not be diagonally opposite)
• Coordinates of both vertices in the second group are known. So we can write the details of the travel within that group:
(ii) To reach S from R:
• First travel 3 units horizontally to the left [∵ (3-6) = -3]
• Then travel 3 unit vertically downwards [∵ (3-6) = -3]
(iii) The same procedure of travel must be followed for the travel from Q to P
• First travel 3 units horizontally to the left
    ♦ At the end of this travel, the coordinates will be (6,5) [∵ (9-3) = 6]
• Then travel 3 unit vertically downwards
    ♦ At the end of this final lap, the coordinates will be (6,2) [∵ (5-3) = 2]
■ So the coordinates of P are (6,2)
2. P is the midpoint of AB
(i) Let the coordinates of A be (x1,y1)
    ♦ They are already given as (2,1)
(ii) Let the coordinates of B be (x2,y2)
(iii) Applying midpoint formula, we have:
• x coordinate of the midpoint P = (x1+x2)⁄2 = (2+x2)⁄2 = 6
⟹ (2+x2) = 12 ⟹ x2 = 10
• y coordinate of the midpoint P = (y1+y2)⁄2 = (1+y2)⁄2 = 2
⟹ (1+y2) = 4 ⟹ y2 = 3
(iv) Thus coordinates of B are (10,3)
3. Q is the midpoint of BC
(i) Let the coordinates of B be (x1,y1)
    ♦ They are already obtained as (10,3)
(ii) Let the coordinates of C be (x2,y2)
(iii) Applying midpoint formula, we have:
• x coordinate of the midpoint Q = (x1+x2)⁄2 = (10+x2)⁄2 = 9
⟹ (10+x2) = 18 ⟹ x2 = 8
• y coordinate of the midpoint Q = (y1+y2)⁄2 = (3+y2)⁄2 = 5
⟹ (3+y2) = 10 ⟹ y2 = 7
(iv) Thus coordinates of C are (8,7)
4. R is the midpoint of CD
(i) Let the coordinates of C be (x1,y1)
    ♦ They are already obtained as (8,7)
(ii) Let the coordinates of D be (x2,y2)
(iii) Applying midpoint formula, we have:
• x coordinate of the midpoint R = (x1+x2)⁄2 = (8+x2)⁄2 = 6
⟹ (8+x2) = 12 ⟹ x2 = 4
• y coordinate of the midpoint R = (y1+y2)⁄2 = (7+y2)⁄2 = 6
⟹ (7+y2) = 12 ⟹ y2 = 5
(iv) Thus coordinates of D are (4,5)
• The actual points in the Cartesian plane are shown in fig.34.14(b) above

 Solved example 34.8
The vertices of a triangle ABC are points with coordinates (3,5), (9,13) and (10,6). Prove that it is an isosceles triangle. Calculate it's area.
Solution:
1. Fig.34.15(a) below shows a rough sketch
Fig.34.15
• We will use the distance formula to calculate the distances
• Length of AB = √[(9-3)2 + (13-5)2] = √[(6)2 + (8)2] = √[36 + 64] = √[100] = 10
• Length of BC = √[(10-9)2 + (6-13)2] = √[(1)2 + (7)2] = √[1 + 49] = √[50] = √[2×(5)2] = 5√2
• Length of AC = √[(10-3)2 + (6-5)2] = √[(7)2 + (1)2] = √[49 + 1] = √[50] = √[2×(5)2] = 5√2
2. So two sides AC and BC are equal. It is an isosceles triangle
• The meeting point of the two equal sides will be the apex. 
• And the third side will be the base
• So C is the apex and AB is the base
3. In an isosceles triangle, if we drop a perpendicular from the apex to the base, then the foot of the perpendicular will be the midpoint of the base. 
• Let this midpoint be D
• So we must find the coordinates of D. We can use the midpoint formula:
• x coordinate of the midpoint D = (x1+x2)⁄2 = (3+9)⁄2 = 6
• y coordinate of the midpoint D = (y1+y2)⁄2 = (5+13)⁄2 = 9
Thus the coordinates of D are: (6,9)
4. Now, CD is the altitude. We can use the distance formula to find it's length:
• Length of CD = √[(6-10)2 + (9-6)2] = √[(-4)2 + (3)2] = √[16 + 9] = √[25] = 5
5. Area of the triangle = 1⁄2 × base × altitude 
= 1⁄2 × AB × CD = 1⁄2 × 10 × 5 = 25 square units
• The fig.34.15(b) above shows the actual positions of the points in the Cartesian plane

 Solved example 34.9
The coordinates of the vertices of a triangle are (-1,5), (3,7), (3,1). Find the coordinates of it's centroid
Solution:
Fig.34.16(a) below shows a rough sketch.
Fig.34.16
• D is the midpoint of AB. This midpoint is joined to the opposite vertex C. So CD is a median
• E is the midpoint of AC. This midpoint is joined to the opposite vertex B. So BE is a median
• The point of intersection of the two medians will give the centroid of the triangle ABC (Details here)
• Now we can write the steps:
1. Coordinates of D using the midpoint formula:
• x coordinate = (x1+x2)⁄2 = (-1+3)⁄2 = 1
• y coordinate = (y1+y2)⁄2 = (5+1)⁄2 = 3
2. Coordinates of E using the midpoint formula:
• x coordinate = (x1+x2)⁄2 = (-1+3)⁄2 = 1
• y coordinate = (y1+y2)⁄2 = (7+5)⁄2 = 6
3. The centroid will divide the median in the ratio 2:1 measured from the vertex
• Let us divide the median CD in the ratio 2:1
    ♦ This ratio 2:1 is measured from the vertex. 
    ♦ So the segment corresponding to 2 will lie near the vertex C of the triangle
    ♦ And the segment corresponding to 1 will lie near the base side AB
4. We will use the section formula:
• m:n is 2:1
• k = 2+1 = 3
• The segment corresponding to 'm' lies near (x1,y1)
    ♦ So the segment corresponding to '2' lies near (x1,y1)
    ♦ Thus coordinates of C must be taken as (x1,y1) and coordinates of D must be taken as (x2,y2)  
5. Then x coordinate of the centroid = x1 + {[m⁄k]×(x2-x1)} 
= 3 + {[2⁄3]×(1-3)} 
= 3 + {[2⁄3]×(-2)} 
= 3 - 4⁄3 = 5⁄3
6. y coordinate of the centroid = y1 + {[m⁄k]×(y2-y1)} 
= 7 + {[2⁄3]×(3-7)}
= 7 + {[2⁄3]×(-4)} 
= 7 - 8⁄3 = 13⁄3
• So coordinates of the centroid are (5⁄3,13⁄3)
Check:
We can check the result by using the other median BE:
• Let us divide the median BE in the ratio 2:1
    ♦ This ratio 2:1 is measured from the vertex. 
    ♦ So the segment corresponding to 2 will lie near the vertex B of the triangle
    ♦ And the segment corresponding to 1 will lie near the base side AC
1. We will use the section formula:
• m:n is 2:1
• k = 2+1 = 3
• The segment corresponding to 'm' lies near (x1,y1)
    ♦ So the segment corresponding to '2' lies near (x1,y1)
    ♦ Thus coordinates of B must be taken as (x1,y1) and coordinates of E must be taken as (x2,y2)  
2. Then x coordinate of the centroid = x1 + {[m⁄k]×(x2-x1)} 
= 3 + {[2⁄3]×(1-3)} 
= 3 + {[2⁄3]×(-2)} 
= 3 - 4⁄3 = 5⁄3
3. y coordinate of the centroid = y1 + {[m⁄k]×(y2-y1)} 
= 1 + {[2⁄3]×(6-1)}
= 1 + {[2⁄3]×(5)} 
= 1 + 10⁄3 = 13⁄3
• So coordinates of the centroid are (5⁄3,13⁄3)
• These are the same coordinates that we obtained earlier
• The actual positions of the points in the Cartesian plane are shown in fig.34.16(b) above 

 Solved example 34.10
The centre of a circle is (1,2) and (3,2) is a point on it. Find the coordinates of the other end of the diameter through this point
Solution:
1. Imagine a small horizontal line 
• Horizontal because y coordinates are same 
2. Put O(1,2) at it's center
3. Put the coordinates B(3,2) at the right end 
• B is towards right of O because x coordinate of B is greater than that of O
4. Let the left end be A
• We have to find the coordinates of A
5. AB is the diameter and O is the centre
• So OA = OB
6. Since OB is horizontal, it's length can be calculated from x coordinates alone. 
• That is., OB = (3-1) = 2 units
7. So OA = 2 units.
• So A is 2 units to the left of O
• Thus it's x coordinate will be (1-2) = -1
• y coordinate will be same as that of O and B, which is 2
• So the coordinates of A are (-1,2)

 Solved example 34.11
Find the ratio in which the line segment joining the points (-3,10) and (6,-8) is divided by (-1,6)
Solution:
• Let A(x1,y1) = (-3,10)
• Let B(x2,y2) = (6,-8)
• Let P(-1,6) divide AB in the ratio m:n
1. Then x coordinate of P =  x1 + {[m⁄k]×(x2-x1)} = -1
⟹ -3 + {[m⁄k]×(6-(-3))} = -1
⟹ -3 + {[m⁄k]×(9)} = -1
⟹ {[m⁄k]×(9)} = 2 ⟹ {[m⁄k]} = 2⁄9
⟹ {[m⁄(m+n)]} = 2⁄9 ⟹ 9m = 2m + 2n ⟹ 7m = 2n
⟹ m⁄n = 2⁄7. That is., m:n = 2:7
Check:
Considering the y coordinates:
y coordinate of P =  y1 + {[m⁄k]×(y2-y1)} = 6
⟹ 10 + {[m⁄k]×(-8-10)} = 6
⟹ 10 + {[m⁄k]×(-18)} = 6
⟹ {[m⁄k]×(-18)} = -4 ⟹ {[m⁄k]} = 2⁄9
⟹ {[m⁄(m+n)]} = 2⁄9 ⟹ 9m = 2m + 2n ⟹ 7m = 2n
⟹ m⁄n = 2⁄7. That is., m:n = 2:7
• This is the same ratio as above.
• The actual positions of the points in the Cartesian plane are shown in the fig.34.17 below:
Fig.34.17
Another method of check:
1. Distance AB = √[(6-(-3))2 + (-8-10)2] = √[(9)2 + (-18)2] = √[81 + 324] = √[405] = √[5×9×9] = 9√5
2. Distance AP = √[(-1-(-3))2 + (6-10)2] = √[(2)2 + (-4)2] = √[4 + 16] = √[20] = √[5×2×2] = 2√5  
3. If AP is 2⁄9 of AB, our calculations are correct
2⁄9 of AB = 2⁄9 × 9√5 = 2√5
4. From (2) we see that AP is indeed 2√5
• So our calculations are correct

In the next section, we will see Lines.


PREVIOUS      CONTENTS       NEXT

                        Copyright©2018 High school Maths lessons. blogspot.in - All Rights Reserved
Posted by at No comments:
Labels: , , , , , ,

Monday, January 9, 2017

Chapter 22.3 - Absolute value of a Number

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. 
Fig.22.13
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:
Fig.22.14
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.
Fig.22.15
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.


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
• 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()'

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
■ 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:
Fig.22.18
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:
Fig.22.19
• 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
So we have seen the distance between a point and zero. In the next section we will see the distance between two points in terms of absolute value.


PREVIOUS      CONTENTS       NEXT

                         Copyright©2016 High school Maths lessons. blogspot.in - All Rights Reserved
Posted by at No comments:
Labels: , , ,
Subscribe to: Posts (Atom)