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What is the number of integer coordinates on a line segment?

8

3

$begingroup$

What is the number of integer coordinates(end points included) on a line segment with integer end coordinates $(x_1,y_1)$ and $(x_2,y_2)$?
Integer coordinate means that both abscissa and ordinate are integers.

coordinate-systems

share|cite|improve this question

edited May 20 '16 at 0:50

M47145

3,29831132

asked Sep 3 '14 at 17:08

PleaseHelpPleaseHelp

481526

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @vadim123 why do you need the coordinates to be positive ? Also kindly explain how.
  $endgroup$
  – PleaseHelp
  Sep 3 '14 at 17:27
  

  
  
  
  

add a comment | 

8

3

$begingroup$

What is the number of integer coordinates(end points included) on a line segment with integer end coordinates $(x_1,y_1)$ and $(x_2,y_2)$?
Integer coordinate means that both abscissa and ordinate are integers.

coordinate-systems

share|cite|improve this question

edited May 20 '16 at 0:50

M47145

3,29831132

asked Sep 3 '14 at 17:08

PleaseHelpPleaseHelp

481526

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @vadim123 why do you need the coordinates to be positive ? Also kindly explain how.
  $endgroup$
  – PleaseHelp
  Sep 3 '14 at 17:27
  

  
  
  
  

add a comment | 

$begingroup$

What is the number of integer coordinates(end points included) on a line segment with integer end coordinates $(x_1,y_1)$ and $(x_2,y_2)$?
Integer coordinate means that both abscissa and ordinate are integers.

coordinate-systems

share|cite|improve this question

edited May 20 '16 at 0:50

M47145

3,29831132

asked Sep 3 '14 at 17:08

PleaseHelpPleaseHelp

481526

$endgroup$

share|cite|improve this question

edited May 20 '16 at 0:50

M47145

3,29831132

asked Sep 3 '14 at 17:08

PleaseHelpPleaseHelp

481526

share|cite|improve this question

edited May 20 '16 at 0:50

M47145

3,29831132

asked Sep 3 '14 at 17:08

PleaseHelpPleaseHelp

481526

edited May 20 '16 at 0:50

M47145

3,29831132

edited May 20 '16 at 0:50

M47145

3,29831132

asked Sep 3 '14 at 17:08

PleaseHelpPleaseHelp

481526

asked Sep 3 '14 at 17:08

PleaseHelpPleaseHelp

481526

2 Answers
2

active

oldest

votes

9

$begingroup$

Write the slope $fracy_2 - y_1x_2 - x_1$ in lowest terms as $fracab$. Then you can obtain every point with integer coordinates by starting at $(x_1, y_1)$ and repeatedly going left $b$ units and up $a$ units. So how many points does that give you?

share|cite|improve this answer

answered Sep 3 '14 at 17:13

Daniel McLauryDaniel McLaury

16.1k33081

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Great answer. I was thinking around the same area. No. of points could be [$(x_2-x_1)/b$ or $(y_2-y_1)/a$] +$1$ and $(x_2-x_1)= GCD*b$ and $(y_2-y_1)= GCD*a$
  $endgroup$
  – PleaseHelp
  Sep 3 '14 at 17:44
  
  

  
  
  
  

add a comment | 

10

$begingroup$

It is simply $GCD(x_1-x_2,y_2-y_1)$ . I can explain if anyone wants to, just because this post is so old , I dont think anyone will bother to see it

EDIT:
Let the equation of line be :

$fracy - y_1x - x_1$ = $fracy_2 - y_1x_2 - x_1$

Now just switching the positions of these expressions we get:

$fracy- y_1y_2 - y_1$ = $fracx- x_1x_2 - x_1$

Now ,
$y - y_1$ should be equal to $k times frac y_2 - y_1GCD(y_2-y_1,x_2-x_1)$ .

Why ? $x$ should be integer! If we write $x$ in terms of everything else We will see

$fracy - y_1y_2 - y_1$ should be Integer.

Now I say that $k_1=0$ Gives me $y1$ , to find the final $k_n$ , I put $y= y_2$ into this equation ($x_2,y_2$ is the last integral point) .

That gives me kn=$GCD( y_2 - y_1, x_2 - x_1)$ . Now , every k between 0 and this $k_n$ are integral points!

If you include the initial point too ans is $GCD( y_2 - y_1, x_2 - x_1) +1$ .

If you exclude initial and final points ans is $GCD( y_2 - y_1, x_2 - x_1) -1$.

share|cite|improve this answer

edited Sep 13 '18 at 6:16

Mooncrater

296214

answered Jun 27 '15 at 15:50

Shubham SharmaShubham Sharma

258313

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Show us if you can!
  $endgroup$
  – Maestro13
  Jun 27 '15 at 15:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Maestro13 Do comment if you find something wrong. I thought of the proof today only . So looking forward to some sort of bug
  $endgroup$
  – Shubham Sharma
  Jun 27 '15 at 18:25
  

  
  
  
  

add a comment | 

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9

$begingroup$

Write the slope $fracy_2 - y_1x_2 - x_1$ in lowest terms as $fracab$. Then you can obtain every point with integer coordinates by starting at $(x_1, y_1)$ and repeatedly going left $b$ units and up $a$ units. So how many points does that give you?

share|cite|improve this answer

answered Sep 3 '14 at 17:13

Daniel McLauryDaniel McLaury

16.1k33081

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Great answer. I was thinking around the same area. No. of points could be [$(x_2-x_1)/b$ or $(y_2-y_1)/a$] +$1$ and $(x_2-x_1)= GCD*b$ and $(y_2-y_1)= GCD*a$
  $endgroup$
  – PleaseHelp
  Sep 3 '14 at 17:44
  
  

  
  
  
  

add a comment | 

9

$begingroup$

Write the slope $fracy_2 - y_1x_2 - x_1$ in lowest terms as $fracab$. Then you can obtain every point with integer coordinates by starting at $(x_1, y_1)$ and repeatedly going left $b$ units and up $a$ units. So how many points does that give you?

share|cite|improve this answer

answered Sep 3 '14 at 17:13

Daniel McLauryDaniel McLaury

16.1k33081

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Great answer. I was thinking around the same area. No. of points could be [$(x_2-x_1)/b$ or $(y_2-y_1)/a$] +$1$ and $(x_2-x_1)= GCD*b$ and $(y_2-y_1)= GCD*a$
  $endgroup$
  – PleaseHelp
  Sep 3 '14 at 17:44
  
  

  
  
  
  

add a comment | 

$begingroup$

Write the slope $fracy_2 - y_1x_2 - x_1$ in lowest terms as $fracab$. Then you can obtain every point with integer coordinates by starting at $(x_1, y_1)$ and repeatedly going left $b$ units and up $a$ units. So how many points does that give you?

share|cite|improve this answer

answered Sep 3 '14 at 17:13

Daniel McLauryDaniel McLaury

16.1k33081

$endgroup$

share|cite|improve this answer

answered Sep 3 '14 at 17:13

Daniel McLauryDaniel McLaury

16.1k33081

answered Sep 3 '14 at 17:13

Daniel McLauryDaniel McLaury

16.1k33081

answered Sep 3 '14 at 17:13

Daniel McLauryDaniel McLaury

16.1k33081

10

$begingroup$

It is simply $GCD(x_1-x_2,y_2-y_1)$ . I can explain if anyone wants to, just because this post is so old , I dont think anyone will bother to see it

EDIT:
Let the equation of line be :

$fracy - y_1x - x_1$ = $fracy_2 - y_1x_2 - x_1$

Now just switching the positions of these expressions we get:

$fracy- y_1y_2 - y_1$ = $fracx- x_1x_2 - x_1$

Now ,
$y - y_1$ should be equal to $k times frac y_2 - y_1GCD(y_2-y_1,x_2-x_1)$ .

Why ? $x$ should be integer! If we write $x$ in terms of everything else We will see

$fracy - y_1y_2 - y_1$ should be Integer.

Now I say that $k_1=0$ Gives me $y1$ , to find the final $k_n$ , I put $y= y_2$ into this equation ($x_2,y_2$ is the last integral point) .

That gives me kn=$GCD( y_2 - y_1, x_2 - x_1)$ . Now , every k between 0 and this $k_n$ are integral points!

If you include the initial point too ans is $GCD( y_2 - y_1, x_2 - x_1) +1$ .

If you exclude initial and final points ans is $GCD( y_2 - y_1, x_2 - x_1) -1$.

share|cite|improve this answer

edited Sep 13 '18 at 6:16

Mooncrater

296214

answered Jun 27 '15 at 15:50

Shubham SharmaShubham Sharma

258313

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Show us if you can!
  $endgroup$
  – Maestro13
  Jun 27 '15 at 15:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Maestro13 Do comment if you find something wrong. I thought of the proof today only . So looking forward to some sort of bug
  $endgroup$
  – Shubham Sharma
  Jun 27 '15 at 18:25
  

  
  
  
  

add a comment | 

10

$begingroup$

It is simply $GCD(x_1-x_2,y_2-y_1)$ . I can explain if anyone wants to, just because this post is so old , I dont think anyone will bother to see it

EDIT:
Let the equation of line be :

$fracy - y_1x - x_1$ = $fracy_2 - y_1x_2 - x_1$

Now just switching the positions of these expressions we get:

$fracy- y_1y_2 - y_1$ = $fracx- x_1x_2 - x_1$

Now ,
$y - y_1$ should be equal to $k times frac y_2 - y_1GCD(y_2-y_1,x_2-x_1)$ .

Why ? $x$ should be integer! If we write $x$ in terms of everything else We will see

$fracy - y_1y_2 - y_1$ should be Integer.

Now I say that $k_1=0$ Gives me $y1$ , to find the final $k_n$ , I put $y= y_2$ into this equation ($x_2,y_2$ is the last integral point) .

That gives me kn=$GCD( y_2 - y_1, x_2 - x_1)$ . Now , every k between 0 and this $k_n$ are integral points!

If you include the initial point too ans is $GCD( y_2 - y_1, x_2 - x_1) +1$ .

If you exclude initial and final points ans is $GCD( y_2 - y_1, x_2 - x_1) -1$.

share|cite|improve this answer

edited Sep 13 '18 at 6:16

Mooncrater

296214

answered Jun 27 '15 at 15:50

Shubham SharmaShubham Sharma

258313

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Show us if you can!
  $endgroup$
  – Maestro13
  Jun 27 '15 at 15:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Maestro13 Do comment if you find something wrong. I thought of the proof today only . So looking forward to some sort of bug
  $endgroup$
  – Shubham Sharma
  Jun 27 '15 at 18:25
  

  
  
  
  

add a comment | 

$begingroup$

It is simply $GCD(x_1-x_2,y_2-y_1)$ . I can explain if anyone wants to, just because this post is so old , I dont think anyone will bother to see it

EDIT:
Let the equation of line be :

$fracy - y_1x - x_1$ = $fracy_2 - y_1x_2 - x_1$

Now just switching the positions of these expressions we get:

$fracy- y_1y_2 - y_1$ = $fracx- x_1x_2 - x_1$

Now ,
$y - y_1$ should be equal to $k times frac y_2 - y_1GCD(y_2-y_1,x_2-x_1)$ .

Why ? $x$ should be integer! If we write $x$ in terms of everything else We will see

$fracy - y_1y_2 - y_1$ should be Integer.

Now I say that $k_1=0$ Gives me $y1$ , to find the final $k_n$ , I put $y= y_2$ into this equation ($x_2,y_2$ is the last integral point) .

That gives me kn=$GCD( y_2 - y_1, x_2 - x_1)$ . Now , every k between 0 and this $k_n$ are integral points!

If you include the initial point too ans is $GCD( y_2 - y_1, x_2 - x_1) +1$ .

If you exclude initial and final points ans is $GCD( y_2 - y_1, x_2 - x_1) -1$.

share|cite|improve this answer

edited Sep 13 '18 at 6:16

Mooncrater

296214

answered Jun 27 '15 at 15:50

Shubham SharmaShubham Sharma

258313

$endgroup$

share|cite|improve this answer

edited Sep 13 '18 at 6:16

Mooncrater

296214

answered Jun 27 '15 at 15:50

Shubham SharmaShubham Sharma

258313

edited Sep 13 '18 at 6:16

Mooncrater

296214

edited Sep 13 '18 at 6:16

Mooncrater

296214

answered Jun 27 '15 at 15:50

Shubham SharmaShubham Sharma

258313

answered Jun 27 '15 at 15:50

Shubham SharmaShubham Sharma

258313