## Precalculus (6th Edition)

The distance between the second and fourth points is $2\sqrt{10}$ units.
RECALL: The distance $d$ between the points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ Use the formula above to obtain: $d=\sqrt{(1-3)^2+(-3-3)^2} \\d=\sqrt{(-2)^2+(-6)^2} \\d=\sqrt{4+36} \\d=\sqrt{40} \\d=\sqrt{4(10)} \\d=2\sqrt{10}$