\(|a-b|\;\text{or}\;|b-a|\)
Note: Distance is always positive.
Step 1: Count the number of units from one endpoint to another.
Step 2: There are \(\text{17 units}\) between the endpoints, \(-13\) and \(4\).
Thus, the distance between \(-13\) and \(4\)
\(=17\) units
By using the distance formula, let's find out the distance between \(-13\) and \(4\).
\(|a-b|=|-13-4|=|-17|=17\) units
or
\(|b-a|=|4-(-13)|=|4+13|=|17|=17\) units
A \(m=-8,\;n=12,\;\text{Distance}=20\,\text{units}\)
B \(m=17,\;n=21,\;\text{Distance}=10\,\text{units}\)
C \(m=-16,\;n=-24,\;\text{Distance}=-40\,\text{units}\)
D \(m=-22,\;n=17,\;\text{Distance}=-30\,\text{units}\)
Example: Cody wants to go for a picnic with his friends. He gets \($35\) from his father for the picnic. Show the increment in the amount of money Cody has, considering the fact that he had nothing before that.
To understand it, consider an example.
Model \(2\cdot(-3)\) on the number line.
Step 1: Check whether the first factor is positive or negative.
Here, \(2\) is a positive number.
Step 2: There are \(2\) possible cases:
(a) If the first factor is positive then we move in the direction of the sign of the second factor.
(b) If the first factor is negative then we move in the opposite direction of the sign of the second factor.
Here, the first factor is positive, so we move in the direction of the sign of the second factor i.e. negative direction.
Step 3: Start from zero and make the groups in which each group equals to the second factor, that is \((-3)\) in this case.
Step 4: Number of moves is equal to the value of the first factor, that is \(2\) in this case.
Here, the groups(or number of units in a move) that are equal to the second factor and the moves which are equal to the first factor are representing that \(-3\) is multiplied \(2\) times i.e. \(2\cdot(-3)=-6\)
Note: When we make an expression from the given number line model, we follow the steps in the reverse order.
A \(-4\cdot2\)
B \(-4\cdot(-2)\)
C \(4\cdot2\)
D Both (B) and (C)
\('+'\longrightarrow\)
\(\longleftarrow\,'-'\)
\('-(-)'\longrightarrow\)
\(\longleftarrow\,'-(+)'\).
For example: Subtract \(-7\) from \(-10\) on the number line.
\(-10-(-7)=\Box\)
First, draw a number line from \(-10\) to \(0.\)
Start from zero and move \(10\) points backward to show \(-10\).
Now, start from \(-10\) and move \(7\) points forward to subtract \(-7\) from \(-10.\)
We reach the number \(-3\).
Thus, \(-10-(-7)=-3\)
We can subtract positive and negative integers with the help of a number line.
To show positive integers, move in the forward direction.
\('+'\longrightarrow\)
\(\longleftarrow\,'-'\)
\('-(-)'\longrightarrow\)
\(\longleftarrow\,'-(+)'\)
For example: Subtract \(-2\) from \(4\).
\(4-(-2)=\Box\)
First, draw a number line from \(0\) to \(4.\)
Start from zero and move \(4\) points forward to show positive \(4.\)
Now, to subtract \(-2\) from \(4,\) we move in the forward direction.
We reach the number \(6.\)
Thus, \(4-(-2)=6\)
Case-I: When dividend and divisor, both are positive.
To understand it, consider an example.
\(4\div 2\)
Step 1: Plot the dividend on the number line.
Here, the dividend is \(4.\)
Step 2: Make the groups of units between zero and the end point (dividend) such that each group has number of units equal to the value of the divisor.
Here, the divisor is \(2.\)
Step 3: Start from zero, face to the positive direction and count the number of groups by moving forward.
Since the number of moves is \(2,\) therefore, the quotient is \(2.\)
So, \(4\div2=2\)
Case-II: When dividend and divisor, both are negative.
Example: \(-6\div(-2)\)
Step 1: Plot the dividend on the number line.
Here, the dividend is \(-6\).
Step 2: Make the groups of units between zero and end point (dividend) which should be equal to the divisor, i.e. \(-2\).
Step 3: Start from zero, face to the positive direction and count the number of groups by moving backward.
Since the number of moves is \(3,\) therefore, the quotient is \(3.\)
Note:
Case-III: When the dividend is positive and the divisor is negative.
Example: \(8\div(-4)\)
Step 1: Plot the dividend on the number line.
Step 2: Make the groups of units between zero and end point (dividend) which should be equal to the divisor.
Here, the divisor is \(-4\).
Step 3: Start from zero, face to the negative direction and count the number of groups by moving backward.
Since the number of moves is \(2\) therefore the quotient is \(2\) but with negative sign because of facing to the negative direction.
So, the quotient is \(-2\).
Case-IV: When the dividend is negative and the divisor is positive.
Example: \(-9\div3\)
Step 1: Plot the dividend on the number line.
Step 2: Make the groups of units between zero and end point (dividend) which should be equal to the divisor.
Here, the divisor is \(3.\)
Step 3: Start from zero, face to the negative direction and count the number of groups by moving forward.
Since the number of moves is \(3,\) therefore, the quotient is \(3\) but with negative sign because of facing to the negative direction.
So, the quotient is \(-3\).
Note:
Example: Carl plans a birthday celebration for his sister's birthday. It will cost him \($33\) to buy a chocolate cake, \($29\) for a gift to his sister and \($40\) for snacks. Every month, he saves some money from his pocket money. Till now he has saved \($95\) . How much more dollars Carl needs to save so that he has exactly as much as he plans to spend?
Carl will spend \($33\) on a chocolate cake. Spending represents a negative amount.
Carl will spend \($29\) for a gift. As spending means losing money, so we will subtract \(29\) from \(–33\).
\(-33-29=-62\)
Thus, he is in the deficit of \($62\).
He will spend \($40\) on snacks.
Therefore,
\(–62–40=\;–102\)
So, up till now, he is in the deficit of \($102\).
He has saved \($95\) from his pocket money. Saving represents a positive amount so we will add \($95\) to \(-$102\).
\(-102+95=7\)
Thus, in the end, he is in the deficit of \($7\).
Now, Carl needs to save exactly as much as he plans to spend, so let that amount be \(x\).
\(\therefore\) \(–7+x=0\)
\(\Rightarrow x=7\)
Thus, Carl needs \($7\) more so that he has exactly as much as he plans to spend.
(1) For positive integers
\('+'\longrightarrow\)
(2) For adding positive integers
\('+(+)'\longrightarrow\)
For example: Add \(2\) and \(7\) on the number line.
\(2+7=\Box\)
First, draw a number line from \(0\) to \(10\).
Start from zero and move \(2\) points forward to show \('2'\).
Now, start from \(2\) and move \(7\) points forward to add \('7'\) to \('2'\).
We reach the number \(9\).
Thus, \(2+7=9\).
(1) For negative integers
\(\longleftarrow\;'–'\)
(2) For adding negative integers
\(\longleftarrow\;'+(-)'\)
For example: Add \(-2\) and \(-7\) on the number line.
\(-2+(-7)=\Box\)
First, draw a number line from \(-10\) to \(0.\)
Start from zero and move \(2\) points backward to show '\(-2\)'.
Now, start from \(-2\) and move \(7\) points backward to add \(-7\) to \(-2\).
We reach the number \(-9\).
Thus, \(-2+(-7)=-9\)
Adding negative and positive integers means we are adding losses and gains.
We can also add negative and positive integers on the number line.
\('+'\longrightarrow\)
(1) For negative integers
\(\longleftarrow\,'-'\)
(2) For adding negative integers
\(\longleftarrow\,'+(-)'\)
For example: Add \(-4\) and \(6\) on the number line.
\(-4+6=\Box\)
To add \(-4\) and \(6\) on the number line, first draw a number line from \(-4\) to \(6.\)
Start from \(0\) and move \(4\) points backward to represent \(-4\).
Now, start from \(-4\) and move \(6\) points forward to add \(6\) to \(-4\).
We reach the number \(2.\)
Thus, \(-4+6=2\)
A \(-4+4=0\)
B \(8+(-4)=4\)
C \(4+(-8)=-4\)
D \(-4+8=4\)