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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
in Linear Algebra by Boss (16.3k points)
edited by | 1.1k views

5 Answers

+11 votes
Best answer

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.

by Boss (15.7k points)
selected by
0
So then what should be the answer according to you if it x2-x1 should not be minimum .
+1
I think , it shud be A .
0
Amsar Sokt, how is it A either?  If A is the answer, G12 (in your example) should be the steepest. What about D? :p
0
A) is correct. steepest gradient is always between two adjacent (x-axis) points.
0

@reena_kandari @Himanshu1

The difference between x-coordinates Pa and Pb is minimum

I think they are saying difference between x cordinates of Point A and Point B only which is must to get steepest gradient for line passing from A and B

Not the difference between x cordinates of any two points..

 

0

@Himanshu1 are these 3 points of your example situated in same straight line?

0
Then what should be the correct answer A or C?
0
Options A and C seem very similar. Can anyone please point out the difference between them?

Isn't the slope of 2 points lying on the x-axis adjacent to each other = 0? Does it serve as a counter example for Option A?
+1
Option (A) means -

"Two points are adjacent" means, i.e. when you plot the points on the graph, there's no third point between the two points which constitute to a line with steepest gradient.

Now, it has one more implicit assumption that, "when two points have SAME $x$ co-ordinates, then they are adjacent".

Option (C) means -

Steepest line always has minimum difference between its two $x$ co-ordinates, which is false since gradient depends on BOTH $x$ and $y$ co-ordinates, and NOT only on the $x$ co-ordinates.
+6 votes
Answer: C

Gradient $= \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

For gradient to be maximum $(x_{2}-x_{1})$ should be minimum.
by Boss (33.8k points)
edited by
+1 vote
by Boss (13.3k points)
+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
by Boss (18.3k points)
edited by
0
Can you give example where b is voilated?
0 votes
correct answer is A
by (15 points)

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