Question

Find the distance between the given numbers

$\mathbf{\left(}\mathbf{a}\mathbf{\right)}\frac{7}{15}\mathbf{and}\mathbf{-}\frac{1}{21}$

localid="1651072093557" $\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{-}\mathbf{38}\mathbf{}\mathbf{and}\mathbf{-}\mathbf{57}$

localid="1651072099360" $\mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{and}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{8}$

Short answer

1. The distance of$(\frac{7}{15},-\frac{1}{21})is0.52$
2. The distance of$(-38,-57)is19$
3. The distance of$(-\text{2.6,}-1.8)is0.8$

Step-by-step solution

. Apply the concept of absolute value

If a is a real number, then the absolute value of a is given by:

$\left|a\right|=\{\begin{array}{l}a\text{;if}a\ge 0\\ -a\text{;if}a<0\end{array}$

. Apply the concept of distance

If a and b are real numbers, then the distance between the points a and b on the real line is$d(a,b)=|b-a|$

. Analyze the given numbers

Given numbers are$\frac{7}{15}\text{and}-\frac{1}{21}$

. Find the distance

Distance between$\frac{7}{15}\text{and}-\frac{1}{21}$is given by:

$\begin{array}{l}d(\frac{7}{15},-\frac{1}{21})=|-\frac{1}{21}-\frac{7}{15}|\\ \Rightarrow d(\frac{7}{15},-\frac{1}{21})=|-0.048-0.467|\\ \Rightarrow d(\frac{7}{15},-\frac{1}{21})=|-0.048-0.467|\\ \Rightarrow d(\frac{7}{15},-\frac{1}{21})=|-0.515|\\ \Rightarrow d(\frac{7}{15},-\frac{1}{21})=-(-0.515)\mathrm{..................}[-0.515<0]\\ \Rightarrow d(\frac{7}{15},-\frac{1}{21})=0.515\approx 0.52\end{array}$

. Apply the concept of absolute value

If a is a real number, then the absolute value of a is given by:

$\left|a\right|=\{\begin{array}{l}a\text{;if}a\ge 0\\ -a\text{;if}a<0\end{array}$

. Apply the concept of distance

If a and b are real numbers, then the distance between the points a and b on the real line is$d(a,b)=|b-a|$

. Analyze the given numbers

Given numbers are$\text{-38and-57}$

. Find the distance

Distance between$\text{-38and-57}$ is given by:

$\begin{array}{l}d(-38,-57)=\left|-57-(-38)\right|\\ \Rightarrow d(-38,-57)=|-57+38|\\ \Rightarrow d(-38,-57)=|-19|\\ \Rightarrow d(-38,-57)=-(-19)\mathrm{..................}[-19<0]\\ \Rightarrow d(-38,-57)=19\end{array}$

. Apply the concept of absolute value

If a is a real number, then the absolute value of a is given by:

$\left|a\right|=\{\begin{array}{l}a\text{;if}a\ge 0\\ -a\text{;if}a<0\end{array}$

. Apply the concept of distance

If a and b are real numbers, then the distance between the points a and b on the real line is$d(a,b)=|b-a|$

. Analyze the given numbers

Given numbers are$-\text{2.6and}-1.8$

. Find the distance

Distance between $-\text{2.6and}-1.8$is given by:

$\begin{array}{l}d(-\text{2.6,}-1.8)=\left|-1.8-(-\text{2.6)}\right|\\ \Rightarrow d(-\text{2.6,}-1.8)=|-1.8+\text{2.6}|\\ \Rightarrow d(-\text{2.6,}-1.8)=\left|0.8\right|\\ \Rightarrow d(-\text{2.6,}-1.8)=\mathrm{0.8..................}[0.8>0]\end{array}$