Question

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

Answer 1

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} $ .