
The points $\left( { - a, - b} \right),\left( {a,b} \right),\left( {0,0} \right)$ and $\left( {{a^2},ab} \right),a \ne 0,b \ne 0$ are
${\text{A}}{\text{.}}$ Collinear
${\text{B}}{\text{.}}$ Vertices of a parallelogram
${\text{C}}{\text{.}}$ Vertices of rectangle
${\text{D}}{\text{.}}$ Lie on a circle
Answer
514.8k+ views
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.
Recently Updated Pages
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

The correct order of melting point of 14th group elements class 11 chemistry CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE
