Prerequisites - P.6 - The Rectangular Coordinate System and Graphs - Exercises - Page 57: 20

$\sqrt{29} \approx 5.39$

Work Step by Step

$(x_{1},y_{1}) = (1,3)$ and $(x_{2},y_{2}) = (3,-2)$ Then apply the distance formula ,ie, $d = \sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ $d = \sqrt {(3-1)^{2}+(-2-3)^{2}}$ $d = \sqrt {(2)^{2}+(-5)^{2}}$ $d = \sqrt {4+25} = \sqrt{29}$

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