Answer:
Part a: A square and a rhombus both have sides of equal length.
Let's calculate the length of each side.
[tex]\overline{PL}: \sqrt{(0-5)^2+(3-3)^2}=\sqrt{25+0}=5\\\overline{LA}\sqrt{(5-2)^2+(3-(-1))^2}=\sqrt{3^2+4^2}=5\\\overline{AN}:\sqrt{(2-(-3))^2+(-1-(-1))^2}=\sqrt{25+0}=5\\\overline{NP}:\sqrt{(-3-0)^2+(-1-3)^2}=\sqrt{3^2+4^2}=5[/tex]
All sides measure 5. So it's at least a rhombus (think of a square as a rhombus with extra restrictions).
Part b: A square has extra restriction, more important all angles being 90° and diagonals having the same length. If you discussed about slopes it's easy to see that two sides are horizontal but the other two are NOT vertical. Which might be proof enough.
But the exercise looks about lengths on the plane, so let's measure diagonals.
[tex]\overline{PA}=\sqrt{(0-2)^2+(3-(-1))^2}= \sqrt{2^2+4^2}=\sqrt{20}\\\overline{LN}=\sqrt{(5-(-3))^2+(3-(-1))^2}= \sqrt{8^2+4^2}=\sqrt{80}[/tex]
The two diagonals have different lengths, so the figure cannot be a square.
