## Precalculus (6th Edition) Blitzer

The distance between the pair of points $\left( -4,3 \right)$ and $\left( -2,5 \right)$ is $2.83$ units.
Consider the provided pair of points: $\left( -4,3 \right)$ and $\left( -2,5 \right)$ 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 $-2$ for ${{x}_{2}}$, $-4$ for ${{x}_{1}}$, $5$ for ${{y}_{2}}$ and $3$ for ${{y}_{1}}$ in the above formula and simplify: \begin{align} & d=\sqrt{{{\left( -2-\left( -4 \right) \right)}^{2}}+{{\left( 5-3 \right)}^{2}}} \\ & =\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 2 \right)}^{2}}} \\ & =\sqrt{4+4} \\ & =\sqrt{8} \end{align} On further simplification, \begin{align} & d=2\sqrt{2} \\ & \approx 2.83 \end{align} Therefore, the distance between the pair of points $\left( -4,3 \right)$ and $\left( -2,5 \right)$ is $2.83$ units.