Answer:
It's a rectangle, but not a rhombus (or square).
Step-by-step explanation:
Let's see the vectors of each next vertex:
[tex]Q - P = (4, 5) - (-5, 2) = (9, 3)\\R - Q = (6, -1) - (4, 5) = (2, -6)\\S - R = (-3, -4) - (6, -1) = (-9, -3)\\P - S = (-5, 2) - (-3, -4) = (-2, 6)\\[/tex]
Firstly, we can notice that it's a parallelogram - because the Q-P side is parallel to the S - R side (if the x:y ratios are the same, the sides are parallel).
A rhombus needs the sides to be of the same length. But the length of a (x, y) vector is [tex]\sqrt{x^2 + y^2}[/tex] and [tex]\sqrt{9^2 + 3^2} > \sqrt{2^2 + (-6)^2}[/tex], we don't even have to compute it exactly. If it's not a rhombus, it's also not a square (every square is a rhombus).
The last thing left is to know if it's a rectangle - for it to be a rectangle, we need to check if the vectors are perpendicular.
We can compute the dot product of the vectors - perpendicular vectors always have a dot product equal to zero. The dot product of two vectors [tex](x_1, y_1)[/tex] and [tex](x_2, y_2)[/tex] is equal to [tex]x_1x_2 + y_1y_2[/tex].
[tex](9, 3) \cdot (2, -6) = 9\cdot 2 + 3\cdot(-6) = 18 - 18 = 0[/tex]
So the sides are perpendicular, and as such - the figure is a rectangle.
