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**Hint:**The slope $ m $ of the line joining the two points $ \left( {{x}_{1}},{{y}_{1}} \right) $ and $ \left( {{x}_{2}},{{y}_{2}} \right) $ is $ m=\dfrac{{{y}_{1}}-{{y}_{2}}}{{{x}_{1}}-{{x}_{2}}} $ .

The slopes of parallel lines are the same.

The product of the slopes of a pair of perpendicular lines is $ -1 $ .

If the lines have the same slope and also have a common point, then they are the same line.

Both the pairs of opposite sides of a parallelogram are parallel to each other.

All pairs of adjacent sides of the rectangle are perpendicular to each other.

**Complete step-by-step answer:**

Let's say that the points are $ A(-a,-b) $ , $ B(0,0) $ , $ C(a,b) $ and $ D({{a}^{2}},ab) $ .

Using the slope formula, the slope of the line AB will be:

\[ \Rightarrow \]$ {{m}_{AB}}=\dfrac{-b-0}{-a-0}=\dfrac{-b}{-a}=\dfrac{b}{a} $

Also, the slopes of the lines BC, CD and AD are:

\[ \Rightarrow \]$ {{m}_{BC}}=\dfrac{0-b}{0-a}=\dfrac{-b}{-a}=\dfrac{b}{a} $

\[ \Rightarrow \]$ {{m}_{CD}}=\dfrac{b-ab}{a-{{a}^{2}}}=\dfrac{b(1-a)}{a(1-a)}=\dfrac{b}{a} $

\[ \Rightarrow \]$ {{m}_{AD}}=\dfrac{-b-ab}{-a-{{a}^{2}}}=\dfrac{b(-1-a)}{a(-1-a)}=\dfrac{b}{a} $

Since, the slopes of all the lines are the same and they also have some common points between them, we can say that the points are actually collinear.

**The correct answer is A. Collinear.**

**Note:**There are many ways to show that three points A, B and C are collinear (in a straight line):

Section formula $ \dfrac {Ratio} {Slope\ method}$.

Distance method: $ AB+BC=AC $ .

Area method: $ \text{Area of }\Delta ABC=0 $ .

The angle $ \theta $ between a pair of lines $ y=mx+c $ and $ y=nx+d $ is given by $ \tan \theta =\dfrac{|m-n|}{1+mn} $ .

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