The points $A(-a,-b)$, B $(0,0), C(a, b)$ and $D\left(a^{2}, a b\right)$ are
The points $A(-a,-b)$, B $(0,0), C(a, b)$ and $D\left(a^{2}, a b\right)$ are
collinear
vertices of a parallelogram
vertices of a square
vertices of a rectangle
Solution
Distance between the points $A(-a,-b)$ and $B(0,0)=\sqrt{(0+a)^{2}+(0+b)^{2}}=\sqrt{a^{2}+b^{2}}$
Distance between the points $B(0,0)$ and $C(a, b), \sqrt{(a-0)^{2}+(b-0)^{2}}=\sqrt{a^{2}+b^{2}}$
Distance between the points $C(a, b)$ and $D\left(a^{2}, a b\right)$
$\begin{array}{l}
=\sqrt{\left(a^{2}-a\right)^{2}+(a b-b)^{2}}=\sqrt{[a(a-1)]^{2}+[b(a-1)]^{2}} \\
=\sqrt{a^{2}(a-1)^{2}+b^{2}(a-1)^{2}}=\sqrt{\left(a^{2}+b^{2}\right)(a-1)^{2}}=(a-1) \sqrt{a^{2}+b^{2}}
\end{array}$
Similarly, distance between the points $A(-a,-b)$ and $D\left(a^{2}, a b\right)$
$\begin{aligned}
=\sqrt{\left(a^{2}+a\right)+(a b+b)^{2}} &=(a+1) \sqrt{a^{2}+b^{2}} \\
A B+B C+C D &=\sqrt{a^{2}+b^{2}}+\sqrt{a^{2}+b^{2}}+(a-1) \sqrt{a^{2}+b^{2}} \\
&=(a+1) \sqrt{a^{2}+b^{2}}=A D
\end{aligned}$
Hence the point are collinear.