
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
622.2k+ 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
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

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

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

