Answer
The value of $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ with the points $\left( {{x}_{1}},{{y}_{1}} \right)=\left( 7,2 \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)=\left( 1,-1 \right)$ is $3\sqrt{5}$.
Work Step by Step
Assume the points shown below:
$\left( {{x}_{1}},{{y}_{1}} \right)=\left( 7,2 \right)$and$\left( {{x}_{2}},{{y}_{2}} \right)=\left( 1,-1 \right)$
The formula to determine the distance between two points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$.
Put the values $\left( {{x}_{1}},{{y}_{1}} \right)=\left( 7,2 \right),\left( {{x}_{2}},{{y}_{2}} \right)=\left( 1,-1 \right)$ in the above expression,
$\begin{align}
& \sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}=\sqrt{{{\left( 1-7 \right)}^{2}}+{{\left( -1-2 \right)}^{2}}} \\
& =\sqrt{{{\left( -6 \right)}^{2}}+{{\left( -3 \right)}^{2}}} \\
& =\sqrt{36+9}
\end{align}$
And simplify the expression,
$\begin{align}
& \sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}=\sqrt{45} \\
& =\sqrt{9\times 5} \\
& \sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}=3\sqrt{5}
\end{align}$
Thus, the value of $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ with the points $\left( {{x}_{1}},{{y}_{1}} \right)=\left( 7,2 \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)=\left( 1,-1 \right)$ is $3\sqrt{5}$.
