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Let $P_{1},P_{2},\ldots,P_{n}$ be $n$ points in the $xy-$plane such that no three of them are collinear. For every pair of points $P_{i}$ and $P_{j}$, let $L_{ij}$ be the line passing through them. Let $L_{ab}$ be the line with the steepest gradient amongst all $\frac{n(n -1)}{2}$ lines.

Which one of the following properties should necessarily be satisfied ?

  1. $P_{a}$ and $P_{b}$ are adjacent to each other with respect to their $x$-coordinate
  2. Either $P_{a}$ or $P_{b}$ has the largest or the smallest $y$-coordinate among all the points
  3. The difference between $x$-coordinates $P_{a}$ and $P_{b}$ is minimum
  4. None of the above

9 Answers

Best answer
38 38 votes

A is correct.

If we arrange all points in ascending order of their $x-$ coordinates, then the steepest gradient will be between two adjacent points.

Ref: https://stackoverflow.com/questions/8222108/max-slope-from-set-of-points

Counter example for (C) is as follows:

Consider $3$ points 

  1. $(x_1, y_1) = (1,10)$
  2. $(x_2, y_2) = (2,2)$
  3. $(x_3, y_3) = (4,22)$

$\text{Grad}_{12} = \frac{y_2-y_1}{x_2-x_1}=\frac{2-10}{2-1} = -8/1  = -8$
$\text{Grad}_{32} = \frac{y_2-y_1}{x_2-x_1}=\frac{22-2}{4-2} =  20/2 = 10$
$\text{Grad}_{31} = \frac{y_2-y_1}{x_2-x_1}=\frac{22-10}{4-1} = 12/3 = 4$

Here, $\text{Grad}_{32}$ is steepest, but $x_3-x_2 = 4-2 = 2$ is not minimum.

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13 13 votes

Now understand the question. It is of the form, 

IF Lab is the line with the steepest gradient then (%%%%%%%)
Where %%%%%%% is one of the options in the question.

 

This is what is meant by necessary. It means that for the statements of the form p->q, q has to be true if we have p as true. 

 

Now the question asks which of the options has to be true (definitely) if the premise (i.e. Lab us the line with the steepest gradient) is true.

 

OPTION A:

 

Here adjacent points wrt to x co-ordinates mean, that there exists no point having an x co-ordinate b/w these two points. 


Now based on this, let us see, option A. Suppose for contradiction, assume that option A is not true. Then, for two adjacent points with respect to x co-ordinates, shall not have the steepest slope, rather two nonadjacent points wrt to x co-ordinate will have the steepest slope.

 

 

The (Xp,Yp) is the intermediate point adjacent to (Xa,Xb). We supposed that Lab (the purple line) where Pa and Pb are non-adjacent wrt x co-ordinates is having the steepest gradient, but due to point (Xp.Yp), we got a steeper gradient, indicated by the orange line. So we reach a contradiction.  So OPTION A is correct.

 

OPTION B: it is not a necessary condition for the y co-ordinates to be maximum and or minimum, because, the gradient is given by the ratio delta y/ delta x. So it shall also depend on the difference between the x coordinates.

 

OPTION C:

 

The difference b/w Xa and Xb need not be minimum, necessarily. It shall depend on the y co-ordinates as well. For counter example:

 

1 1 vote
Consider 2 point a,c  and line of infinite length passing through them point b can be above line ac or below as 3 point can't be collinear, let us assume that b is below line ac now we find slop ab= Delta p and bc = delta q either of them will be greater ( assume delta p> delta q)

Similary if point b would be above line then result would be exactly opposite (delta q> p)

This is only possible when  are sorted as per their x coordinate

consider 3 points  
A=(x1, y1) = (1,10)
B=(x2, y2) = (2,2)
C=(x3, y3) = (4,22)

Cleary we can never see ab= -8 bc= 10 and ac= 4 so bc it's our gradient.

Thus answer is A
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correct answer is A
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