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${\text{A}}{\text{.}}$ Collinear

${\text{B}}{\text{.}}$ Vertices of a parallelogram

${\text{C}}{\text{.}}$ Vertices of rectangle

${\text{D}}{\text{.}}$ Lie on a circle

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Hint- Here, we will find the slopes of the lines joining these points in order to find the relation between the points.

Let the given points be \[{\text{A}}\left( { - a, - b} \right),{\text{B}}\left( {a,b} \right),{\text{C}}\left( {0,0} \right)\] and \[{\text{D}}\left( {{a^2},ab} \right)\]

Since, we know that the slope of the joining any two points \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] is given by \[m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\].

Here, let us find out the slopes of all the lines i.e., AB, BC, CD, AD obtained by joining any two adjacent points.

Slope of line AB, \[{m_{AB}} = \dfrac{{b - \left( { - b} \right)}}{{a - \left( { - a} \right)}} = \dfrac{{b + b}}{{a + a}} = \dfrac{{2b}}{{2a}} = \dfrac{b}{a}\]

Slope of line BC, \[{m_{BC}} = \dfrac{{0 - b}}{{0 - a}} = \dfrac{{ - b}}{{ - a}} = \dfrac{b}{a}\]

Slope of line CD, \[{m_{CD}} = \dfrac{{ab - 0}}{{{a^2} - 0}} = \dfrac{{ab}}{{{a^2}}} = \dfrac{b}{a}\]

Slope of line AD, \[{m_{AD}} = \dfrac{{ab - \left( { - b} \right)}}{{{a^2} - \left( { - a} \right)}} = \dfrac{{ab + b}}{{{a^2} + a}} = \dfrac{{b\left( {a + 1} \right)}}{{a\left( {a + 1} \right)}} = \dfrac{b}{a}\]

Clearly, slopes of all the lines AB, BC, CD and AD are equal i.e., \[{m_{AB}} = {m_{BC}} = {m_{CD}} = {m_{AD}}\]

Therefore, we can say that all the given points \[{\text{A}}\left( { - a, - b} \right),{\text{B}}\left( {a,b} \right),{\text{C}}\left( {0,0} \right)\] and \[{\text{D}}\left( {{a^2},ab} \right)\] are collinear since the lines joining these points are having equal slopes.

Therefore, option A is correct.

Note- These type of problems can be solved by comparing the values of the slopes obtained by joining any two adjacent points. If all the values of these slopes are equal then, these points are collinear and if not then these are non-collinear.

Let the given points be \[{\text{A}}\left( { - a, - b} \right),{\text{B}}\left( {a,b} \right),{\text{C}}\left( {0,0} \right)\] and \[{\text{D}}\left( {{a^2},ab} \right)\]

Since, we know that the slope of the joining any two points \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] is given by \[m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\].

Here, let us find out the slopes of all the lines i.e., AB, BC, CD, AD obtained by joining any two adjacent points.

Slope of line AB, \[{m_{AB}} = \dfrac{{b - \left( { - b} \right)}}{{a - \left( { - a} \right)}} = \dfrac{{b + b}}{{a + a}} = \dfrac{{2b}}{{2a}} = \dfrac{b}{a}\]

Slope of line BC, \[{m_{BC}} = \dfrac{{0 - b}}{{0 - a}} = \dfrac{{ - b}}{{ - a}} = \dfrac{b}{a}\]

Slope of line CD, \[{m_{CD}} = \dfrac{{ab - 0}}{{{a^2} - 0}} = \dfrac{{ab}}{{{a^2}}} = \dfrac{b}{a}\]

Slope of line AD, \[{m_{AD}} = \dfrac{{ab - \left( { - b} \right)}}{{{a^2} - \left( { - a} \right)}} = \dfrac{{ab + b}}{{{a^2} + a}} = \dfrac{{b\left( {a + 1} \right)}}{{a\left( {a + 1} \right)}} = \dfrac{b}{a}\]

Clearly, slopes of all the lines AB, BC, CD and AD are equal i.e., \[{m_{AB}} = {m_{BC}} = {m_{CD}} = {m_{AD}}\]

Therefore, we can say that all the given points \[{\text{A}}\left( { - a, - b} \right),{\text{B}}\left( {a,b} \right),{\text{C}}\left( {0,0} \right)\] and \[{\text{D}}\left( {{a^2},ab} \right)\] are collinear since the lines joining these points are having equal slopes.

Therefore, option A is correct.

Note- These type of problems can be solved by comparing the values of the slopes obtained by joining any two adjacent points. If all the values of these slopes are equal then, these points are collinear and if not then these are non-collinear.

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