The length is given by the distance formula
[tex]d = \sqrt{(-4 - -15)^2 + (11 - 9)^2} = \sqrt{11^2 + 2^2}=\sqrt{125}=5 \sqrt{5} [/tex]
So the pedantic answer is none of the above; they want us to approximate.
[tex]d \approx 11.18[/tex]
The midpoint is the average of the coordinates:
[tex]\left( \dfrac{-15 + -4}{2}, \dfrac{9 + 11}{2} \right) = \left( -\dfrac{19}{2}, 10 \right)[/tex]
Choice B
