But how can I calculate the coordinates of a point $Q$ wich lies on $frac13$ of line $PD$ with $P(2,3)$ and $D(4,-8)$?

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To calculate the middle of a line you can just average the points: $x = dfracx_1 + x_22$ and $y = dfracy_1 + y_22$

But how can I calculate the coordinates of a point $Q$ wich lies on $frac13$ of line $PD$ with $P(2,3)$ and $D(4,-8)$?

The next technic does not work is this case:

analytic-geometry

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edited Jul 26 at 14:43

mechanodroid

22.2k52041

asked Jul 26 at 14:31

WinstonCherf

49816

add a comment | 

up vote
1
down vote

favorite

To calculate the middle of a line you can just average the points: $x = dfracx_1 + x_22$ and $y = dfracy_1 + y_22$

But how can I calculate the coordinates of a point $Q$ wich lies on $frac13$ of line $PD$ with $P(2,3)$ and $D(4,-8)$?

The next technic does not work is this case:

analytic-geometry

share|cite|improve this question

edited Jul 26 at 14:43

mechanodroid

22.2k52041

asked Jul 26 at 14:31

WinstonCherf

49816

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

To calculate the middle of a line you can just average the points: $x = dfracx_1 + x_22$ and $y = dfracy_1 + y_22$

But how can I calculate the coordinates of a point $Q$ wich lies on $frac13$ of line $PD$ with $P(2,3)$ and $D(4,-8)$?

The next technic does not work is this case:

analytic-geometry

share|cite|improve this question

edited Jul 26 at 14:43

mechanodroid

22.2k52041

asked Jul 26 at 14:31

WinstonCherf

49816

To calculate the middle of a line you can just average the points: $x = dfracx_1 + x_22$ and $y = dfracy_1 + y_22$

But how can I calculate the coordinates of a point $Q$ wich lies on $frac13$ of line $PD$ with $P(2,3)$ and $D(4,-8)$?

The next technic does not work is this case:

analytic-geometry

share|cite|improve this question

edited Jul 26 at 14:43

mechanodroid

22.2k52041

asked Jul 26 at 14:31

WinstonCherf

49816

share|cite|improve this question

share|cite|improve this question

edited Jul 26 at 14:43

mechanodroid

22.2k52041

edited Jul 26 at 14:43

mechanodroid

22.2k52041

edited Jul 26 at 14:43

mechanodroid

22.2k52041

22.2k52041

asked Jul 26 at 14:31

WinstonCherf

49816

asked Jul 26 at 14:31

WinstonCherf

49816

asked Jul 26 at 14:31

WinstonCherf

49816

49816

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
2
down vote



accepted

The section formula gives a formula for a point $Q$ between $P = (x_1, y_1)$ and $D = (x_2, y_2)$ such that $fracPQQD = fracmn$ as:

$$Q = left(fracmx_2 + nx_1m+n, fracmy_2 + ny_1m+nright)$$

so in our case we have $fracPQQD = frac12$ so

$$Q = left(frac1cdot 4 + 2cdot 21+2, frac1cdot (-8) + 2cdot 31+2right) = left(frac83, -frac23right)$$

share|cite|improve this answer

edited Jul 26 at 15:13

answered Jul 26 at 14:42

mechanodroid

22.2k52041

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The answer should be (8/3,-2/3)..
  – WinstonCherf
  Jul 26 at 15:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @WinstonCherf True, if $PQ$ is $frac13$ of $PD$, then $fracPQQD = frac12$. Fixed now.
  – mechanodroid
  Jul 26 at 15:14
  

  
  
  
  

add a comment | 

up vote
1
down vote

It's simply, in terms of vectors:
$$Q=P+frac13overrightarrowPD$$
or in terms of barycentres:
$$Q=frac23P+frac13D.$$

share|cite|improve this answer

answered Jul 26 at 14:52

Bernard

110k635102

add a comment | 

up vote
0
down vote

$$x=x_1+lambda(x_2-x_1)，~y=y_1+lambda(y_2-y_1).$$

share|cite|improve this answer

answered Jul 26 at 14:42

mengdie1982

2,825216

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

The section formula gives a formula for a point $Q$ between $P = (x_1, y_1)$ and $D = (x_2, y_2)$ such that $fracPQQD = fracmn$ as:

$$Q = left(fracmx_2 + nx_1m+n, fracmy_2 + ny_1m+nright)$$

so in our case we have $fracPQQD = frac12$ so

$$Q = left(frac1cdot 4 + 2cdot 21+2, frac1cdot (-8) + 2cdot 31+2right) = left(frac83, -frac23right)$$

share|cite|improve this answer

edited Jul 26 at 15:13

answered Jul 26 at 14:42

mechanodroid

22.2k52041

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The answer should be (8/3,-2/3)..
  – WinstonCherf
  Jul 26 at 15:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @WinstonCherf True, if $PQ$ is $frac13$ of $PD$, then $fracPQQD = frac12$. Fixed now.
  – mechanodroid
  Jul 26 at 15:14
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

The section formula gives a formula for a point $Q$ between $P = (x_1, y_1)$ and $D = (x_2, y_2)$ such that $fracPQQD = fracmn$ as:

$$Q = left(fracmx_2 + nx_1m+n, fracmy_2 + ny_1m+nright)$$

so in our case we have $fracPQQD = frac12$ so

$$Q = left(frac1cdot 4 + 2cdot 21+2, frac1cdot (-8) + 2cdot 31+2right) = left(frac83, -frac23right)$$

share|cite|improve this answer

edited Jul 26 at 15:13

answered Jul 26 at 14:42

mechanodroid

22.2k52041

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The answer should be (8/3,-2/3)..
  – WinstonCherf
  Jul 26 at 15:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @WinstonCherf True, if $PQ$ is $frac13$ of $PD$, then $fracPQQD = frac12$. Fixed now.
  – mechanodroid
  Jul 26 at 15:14
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

The section formula gives a formula for a point $Q$ between $P = (x_1, y_1)$ and $D = (x_2, y_2)$ such that $fracPQQD = fracmn$ as:

$$Q = left(fracmx_2 + nx_1m+n, fracmy_2 + ny_1m+nright)$$

so in our case we have $fracPQQD = frac12$ so

$$Q = left(frac1cdot 4 + 2cdot 21+2, frac1cdot (-8) + 2cdot 31+2right) = left(frac83, -frac23right)$$

share|cite|improve this answer

edited Jul 26 at 15:13

answered Jul 26 at 14:42

mechanodroid

22.2k52041

The section formula gives a formula for a point $Q$ between $P = (x_1, y_1)$ and $D = (x_2, y_2)$ such that $fracPQQD = fracmn$ as:

$$Q = left(fracmx_2 + nx_1m+n, fracmy_2 + ny_1m+nright)$$

so in our case we have $fracPQQD = frac12$ so

$$Q = left(frac1cdot 4 + 2cdot 21+2, frac1cdot (-8) + 2cdot 31+2right) = left(frac83, -frac23right)$$

share|cite|improve this answer

edited Jul 26 at 15:13

answered Jul 26 at 14:42

mechanodroid

22.2k52041

share|cite|improve this answer

share|cite|improve this answer

edited Jul 26 at 15:13

edited Jul 26 at 15:13

edited Jul 26 at 15:13

answered Jul 26 at 14:42

mechanodroid

22.2k52041

answered Jul 26 at 14:42

mechanodroid

22.2k52041

answered Jul 26 at 14:42

mechanodroid

22.2k52041

22.2k52041

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The answer should be (8/3,-2/3)..
  – WinstonCherf
  Jul 26 at 15:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @WinstonCherf True, if $PQ$ is $frac13$ of $PD$, then $fracPQQD = frac12$. Fixed now.
  – mechanodroid
  Jul 26 at 15:14
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The answer should be (8/3,-2/3)..
  – WinstonCherf
  Jul 26 at 15:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @WinstonCherf True, if $PQ$ is $frac13$ of $PD$, then $fracPQQD = frac12$. Fixed now.
  – mechanodroid
  Jul 26 at 15:14
  

  
  
  
  

The answer should be (8/3,-2/3)..
– WinstonCherf
Jul 26 at 15:01

The answer should be (8/3,-2/3)..
– WinstonCherf
Jul 26 at 15:01

1

1

@WinstonCherf True, if $PQ$ is $frac13$ of $PD$, then $fracPQQD = frac12$. Fixed now.
– mechanodroid
Jul 26 at 15:14

@WinstonCherf True, if $PQ$ is $frac13$ of $PD$, then $fracPQQD = frac12$. Fixed now.
– mechanodroid
Jul 26 at 15:14

add a comment | 

up vote
1
down vote

It's simply, in terms of vectors:
$$Q=P+frac13overrightarrowPD$$
or in terms of barycentres:
$$Q=frac23P+frac13D.$$

share|cite|improve this answer

answered Jul 26 at 14:52

Bernard

110k635102

add a comment | 

up vote
1
down vote

It's simply, in terms of vectors:
$$Q=P+frac13overrightarrowPD$$
or in terms of barycentres:
$$Q=frac23P+frac13D.$$

share|cite|improve this answer

answered Jul 26 at 14:52

Bernard

110k635102

add a comment | 

up vote
1
down vote

up vote
1
down vote

It's simply, in terms of vectors:
$$Q=P+frac13overrightarrowPD$$
or in terms of barycentres:
$$Q=frac23P+frac13D.$$

share|cite|improve this answer

answered Jul 26 at 14:52

Bernard

110k635102

It's simply, in terms of vectors:
$$Q=P+frac13overrightarrowPD$$
or in terms of barycentres:
$$Q=frac23P+frac13D.$$

share|cite|improve this answer

answered Jul 26 at 14:52

Bernard

110k635102

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 14:52

Bernard

110k635102

answered Jul 26 at 14:52

Bernard

110k635102

answered Jul 26 at 14:52

Bernard

110k635102

110k635102

add a comment | 

add a comment | 

up vote
0
down vote

$$x=x_1+lambda(x_2-x_1)，~y=y_1+lambda(y_2-y_1).$$

share|cite|improve this answer

answered Jul 26 at 14:42

mengdie1982

2,825216

add a comment | 

up vote
0
down vote

$$x=x_1+lambda(x_2-x_1)，~y=y_1+lambda(y_2-y_1).$$

share|cite|improve this answer

answered Jul 26 at 14:42

mengdie1982

2,825216

add a comment | 

up vote
0
down vote

up vote
0
down vote

$$x=x_1+lambda(x_2-x_1)，~y=y_1+lambda(y_2-y_1).$$

share|cite|improve this answer

answered Jul 26 at 14:42

mengdie1982

2,825216

$$x=x_1+lambda(x_2-x_1)，~y=y_1+lambda(y_2-y_1).$$

share|cite|improve this answer

answered Jul 26 at 14:42

mengdie1982

2,825216

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 14:42

mengdie1982

2,825216

answered Jul 26 at 14:42

mengdie1982

2,825216

answered Jul 26 at 14:42

mengdie1982

2,825216

2,825216

add a comment | 

add a comment | 

 

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