Precalculus (6th Edition) Blitzer

The distance between the pair of points $\left( -2,3 \right)$ and $\left( 3,-9 \right)$ is $13$ units.
Consider the provided pair of points: $\left( -2,3 \right)$ and $\left( 3,-9 \right)$ The distance $d$, between the points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is given by: $d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ Substitute $3$ for ${{x}_{2}}$, $-2$ for ${{x}_{1}}$, $-9$ for ${{y}_{2}}$ and $3$ for ${{y}_{1}}$ in the above formula and simplify: \begin{align} & d=\sqrt{{{\left( 3-\left( -2 \right) \right)}^{2}}+{{\left( -9-3 \right)}^{2}}} \\ & =\sqrt{{{\left( 3+2 \right)}^{2}}+{{\left( -12 \right)}^{2}}} \\ & =\sqrt{25+144} \\ & =\sqrt{169} \end{align} On further simplification, $d=13$ Therefore, the distance between the pair of points $\left( -2,3 \right)$ and $\left( 3,-9 \right)$ is $13$ units.