It is an interesting problem.
Method 1: (simplest, can be done with pencil and paper)
Graph the points, and do a visual inspection.
See graph attached.
Method 2: by observations, without need to take square-roots (for distances)
We note that:
1. the x-coordinates of the first and third points are the same (-3), meaning that they lie on the same vertical line.
2. the y-coordinates of the second and fourth points are the same (1), that means that they lie on the same horizontal line. So the diagonals are at right angles to each other, intersecting at P(-3,1).
These two observations tell us that the figure is either a square, rhombus, or a kite, or an arbitrary quadrilateral that has orthogonal diagonals.
3. AC=4-(-2)=6, BD=-1-(-5)=4, diagonals are not equal, so cannot be a square.
4. PA=(4-1)=3, PB=(-3-(-1))=2, diagonals bisect each other, so it is a rhombus (not a kite).
