the points (-a , -b) (0,0) (a,b) (a²,ab) are :

collinear

vertices of a parallelogrm

vertices of a rectangle

none of these

By visualising the first three points on the Cartesian plane .... they form a line. So first 3 points are collinear.

lets find out about 4th point. If last 3 points are collinear then all points are in same line
using formula of area of triangle

 $$\begin{gathered} points are(0,0), (a,b), (a^2,ab) \\ = \frac{1}{2}\left(x_1\left(y_2 -y_3\right) + x_2\left(y_3 -y_1\right) + x_3\left(y_1 -y_2\right)\right) \\ =\frac{1}{2}\left(0 + a\left(ab - 0\right) + a^2\left(0 -b\right)\right) \\ =\frac{1}{2}\left( a^2 b+ a^2\left(-b\right)\right) \\ = 0 \\ it means last 3 points are also collinear \\ it means all 4 points are collinear \end{gathered}$$