Answer: [tex]\sqrt{13}[/tex] units
Work Shown:
[tex](x_1,y_1) = (-5,-4) \text{ and } (x_2, y_2) = (-2,-6)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(-5-(-2))^2 + (-4-(-6))^2}\\\\d = \sqrt{(-5+2)^2 + (-4+6)^2}\\\\d = \sqrt{(-3)^2 + (2)^2}\\\\d = \sqrt{9 + 4}\\\\d = \sqrt{13}\\\\d \approx 3.6056\\\\[/tex]
I used the distance formula.
A slightly alternate method is to form a right triangle and use the pythagorean theorem. The hypotenuse will have the endpoints (-5,-4) and (-2,-6).
