Question

The points $( - a, - b),(0,0),(a,b),({a^2},ab)$ are
A) Collinear
B) Vertices of rectangle
C) Vertices of parallelogram
D) None of these

Answer 1

Hint:
We will check by option in this type of question. For four points to be collinear we will first find the equation of line passing through two points by the Two – point form method then we will check that if the remaining third and fourth point will satisfy the equation or not. For four points to be vertices of the rectangle first we will find all the distances through four points there will be six such distances by using Distance formula, as we know in rectangle the opposite sides are equal and the diagonals are also equal so we will check it. For four points to be vertices of parallelogram first we will find the distances through by using distance formula, as we know in parallelogram opposite sides are equal and midpoints of diagonals are equal.

Complete step by step solution:
Let the points be $A( - a, - b), B(0,0), C(a,b), D({a^2},ab)$
For the points to be collinear first we will find the equation of line passing through two points by the Two – point form method given by the formula
 $y - {y_1} = \dfrac{{({y_2} - {y_1})}}{{({x_2} - {x_1})}}(x - {x_1})$
 Where $({x_1},{x_2}),\left( {{y_1},{y_2}} \right)$ here are the two points
So, equation of line is given by $y - ( - b) = \dfrac{{(0 - ( - b))}}{{(0 - ( - a))}}(x - ( - a))$
Where $({x_1},{x_2}),\left( {{y_1},{y_2}} \right)$ Corresponds to $A( - a, - b),B(0,0)$
So $y + b = \dfrac{b}{a}(x + a)$
On multiplying the equation by a we get,
 $ \Rightarrow ay + ab = bx + ab$
On simplification we get,
 $ \Rightarrow ay - bx = 0$ (i)
Now we will check that if the remaining third and fourth point will satisfy the equation or not
Putting the points $C(a,b),D({a^2},ab)$ in the equation $ay - bx = 0$
For point C(a,b)
 $ \Rightarrow ab - ba = 0$
So. we get,
 $ \Rightarrow 0 = 0$
It satisfy the equation
For point $D({a^2},ab)$
 $ \Rightarrow aab - b{a^2} = 0$
We get,
 $ \Rightarrow 0 = 0$
It satisfies the equation
Since both the points satisfy the equation (i)
Therefore, the points $A( - a, - b),B(0,0),C(a,b),D({a^2},ab)$ are collinear

So, option A is correct.

Note:
For four points to be vertices of rectangle first finding all distances through four points
 $A( - a, - b),B(0,0),C(a,b),D({a^2},ab)$ By using the distance formula $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $
Where $({x_1},{x_2}),\left( {{y_1},{y_2}} \right)$ the two are points here and ‘d’ is the distance between them
  $AB = \sqrt {{{(0 - ( - a))}^2} + {{(0 - ( - b))}^2}} = \sqrt {{{(a)}^2} + {{(b)}^2}} $
 $BC = \sqrt {{{(a - 0)}^2} + {{(b - 0)}^2}} = \sqrt {{{(a)}^2} + {{(b)}^2}} $
 $CD = \sqrt {{{({a^2} - a)}^2} + {{(ab - b)}^2}} = \sqrt {({a^4} + {a^2} - 2{a^3} + {{(ab)}^2} + {b^2} - 2a{b^2}} $
 $DA = \sqrt {{{( - a - {a^2})}^2} + {{( - b - ab)}^2}} = \sqrt {({a^4} + {a^2} + 2{a^3} + {{(ab)}^2} + {b^2} + 2a{b^2}} $
So $AB = BC = \sqrt {{{(a)}^2} + {{(b)}^2}} $ but $CD \ne DA$
Therefore, the given points are not vertices of rectangle
Now for points to be vertices of parallelogram
 $AB = BC = \sqrt {{{(a)}^2} + {{(b)}^2}} $ But $CD \ne DA$
Since the opposite sides are not equal therefore the given points are not vertices of parallelogram
So, there is no need to check for midpoint of diagonals.