Isosceles acute, answer a.
Here's the logic:
AC^2=(1-(-1))^2+(-1-3)^2=4+16=20.
BC^2=(-1-3)^2+(1-3)^2=16+4=20. Therefore AC=BC, isosceles.
Angle C=tan^-1((3-(-1))/(3-1))-tan^-1((1-(-1))/(3-(-1))=tan^-1(2)-tan^-1(1/2)=36.87° approx, which is acute.
(The angle C is measured in relation to the triangle. Because AC and BC have negative slope the angle between them at C is also negative, but for the purpose of establishing angle C in relation to the triangle rather than to the reference frame (x-y coordinate system), only the magnitudes of the relevant angles have been considered.)