Mixing
How do I find the middle(1/2), 1/3, 1/4, etc, of a line?
$begingroup$
Similar to this question: How to calculate the middle of a line? where it's explained how to find the middle of a line (x,y), so that's half the line 1/2, but I also need to find one third of the line, one fourth, and so on.
I tried dividing by 4 instead of 2 to get 1/4 but it didn't seem to work:
1/2 (works):
$x = dfrac{x_1 + x_2}{2}$ $y = dfrac{y_1 + y_2}{2}$
1/4 (doesn't work):
$x = dfrac{x_1 + x_2}{4}$ $y = dfrac{y_1 + y_2}{4}$
geometry coordinate-systems
edited Apr 13 '17 at 12:21
Community♦
1
asked Nov 12 '13 at 7:08
01AutoMonkey01AutoMonkey
165138
$endgroup$
add a comment |
$begingroup$
Similar to this question: How to calculate the middle of a line? where it's explained how to find the middle of a line (x,y), so that's half the line 1/2, but I also need to find one third of the line, one fourth, and so on.
I tried dividing by 4 instead of 2 to get 1/4 but it didn't seem to work:
1/2 (works):
$x = dfrac{x_1 + x_2}{2}$ $y = dfrac{y_1 + y_2}{2}$
1/4 (doesn't work):
$x = dfrac{x_1 + x_2}{4}$ $y = dfrac{y_1 + y_2}{4}$
geometry coordinate-systems
edited Apr 13 '17 at 12:21
Community♦
1
asked Nov 12 '13 at 7:08
01AutoMonkey01AutoMonkey
165138
$endgroup$
add a comment |
$begingroup$
Similar to this question: How to calculate the middle of a line? where it's explained how to find the middle of a line (x,y), so that's half the line 1/2, but I also need to find one third of the line, one fourth, and so on.
I tried dividing by 4 instead of 2 to get 1/4 but it didn't seem to work:
1/2 (works):
$x = dfrac{x_1 + x_2}{2}$ $y = dfrac{y_1 + y_2}{2}$
1/4 (doesn't work):
$x = dfrac{x_1 + x_2}{4}$ $y = dfrac{y_1 + y_2}{4}$
geometry coordinate-systems
edited Apr 13 '17 at 12:21
Community♦
1
asked Nov 12 '13 at 7:08
01AutoMonkey01AutoMonkey
165138
$endgroup$
Similar to this question: How to calculate the middle of a line? where it's explained how to find the middle of a line (x,y), so that's half the line 1/2, but I also need to find one third of the line, one fourth, and so on.
I tried dividing by 4 instead of 2 to get 1/4 but it didn't seem to work:
1/2 (works):
$x = dfrac{x_1 + x_2}{2}$ $y = dfrac{y_1 + y_2}{2}$
1/4 (doesn't work):
$x = dfrac{x_1 + x_2}{4}$ $y = dfrac{y_1 + y_2}{4}$
geometry coordinate-systems
geometry coordinate-systems
edited Apr 13 '17 at 12:21
Community♦
1
asked Nov 12 '13 at 7:08
01AutoMonkey01AutoMonkey
165138
edited Apr 13 '17 at 12:21
Community♦
1
asked Nov 12 '13 at 7:08
01AutoMonkey01AutoMonkey
165138
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
edited Apr 13 '17 at 12:21
Community♦
1
1
asked Nov 12 '13 at 7:08
01AutoMonkey01AutoMonkey
165138
asked Nov 12 '13 at 7:08
01AutoMonkey01AutoMonkey
165138
asked Nov 12 '13 at 7:08
01AutoMonkey01AutoMonkey
165138
165138
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Think about the formula for the midpoint of a line as
$$x = frac{1}{2}x_{1} + frac{1}{2}x_{2}.$$
We find the midpoint by taking equal contributions from either end of the line.
If we want to find a point which is one third of the way along the line we should use the following
$$x = frac{2}{3}x_{1} + frac{1}{3}x_{2}$$
where we take uneven contributions from the two endpoints.
The point one quarter of the way along the line can be found using
$$x = frac{3}{4}x_{1} + frac{1}{4}x_{2}.$$
The y-coordinates are found in a similar manner.
answered Nov 12 '13 at 7:15
in_mathematica_we_trustin_mathematica_we_trust
2,530819
$endgroup$
add a comment |
$begingroup$
Divide your segment into even parts.
From the last point you can find that
$$
x_2 = x_1 + nDelta implies Delta = frac {x_2 - x_1}n
$$
You need to find second point in that division, so
$$
x' = x_1 + Delta = x_1 + frac {x_2-x_1}n = frac {n-1}n x_1 + frac 1n x_2
$$
So, let's say you want to divide your segment into 2 even parts, then $n = 2$ and
$$
x' = frac {x_1}2 + frac {x_2}2
$$
If $n = 3$ (3 even parts)
$$
x' = frac 23 x_1 + frac 13 x_2
$$
$n = 4$ (4 even parts)
$$
x' = frac 34 x_1 + frac 14 x_2
$$
edited Jan 17 at 23:26
Community♦
1
answered Nov 12 '13 at 7:29
KasterKaster
9,06221729
$endgroup$
add a comment |
$begingroup$
Treat $x_1$ as the offset, and $x_2-x_1$ as the distance that needs sectioning: $$x=x_1+frac{x_2-x_1}nquad,quad y=y_1+frac{y_2-y_1}nquad,quad z=z_1+frac{z_2-z_1}n$$ For $x_1=y_1=z_1=0$, the result becomes self-evident.
answered Nov 12 '13 at 7:33
LucianLucian
41.3k159130
$endgroup$
add a comment |
$begingroup$
Let $P$ and $Q$ be two distinct points and define the (ruler) function
$f : mathbb R to overleftrightarrow{PQ}$, from the set of real numbers into the line $overleftrightarrow{PQ}$, by $f(t) = (1-t)P + tQ$. This function lays a ruler over the line $overleftrightarrow{PQ}$ such that the ordinate of $P$ is $0$ ($f(0)= P$) and the ordinate of $Q$ is $1$ $(f(1) = Q)$. So, for example, the point $dfrac34$ of the way from $P$ to $Q$ is $fleft( dfrac 34 right) = dfrac 14 P + dfrac 34 Q$.
answered Jun 24 '17 at 2:26
steven gregorysteven gregory
18.2k32258
$endgroup$
add a comment |
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4 Answers
4
active
oldest
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4 Answers
4
active
oldest
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active
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votes
active
oldest
votes
$begingroup$
Think about the formula for the midpoint of a line as
$$x = frac{1}{2}x_{1} + frac{1}{2}x_{2}.$$
We find the midpoint by taking equal contributions from either end of the line.
If we want to find a point which is one third of the way along the line we should use the following
$$x = frac{2}{3}x_{1} + frac{1}{3}x_{2}$$
where we take uneven contributions from the two endpoints.
The point one quarter of the way along the line can be found using
$$x = frac{3}{4}x_{1} + frac{1}{4}x_{2}.$$
The y-coordinates are found in a similar manner.
answered Nov 12 '13 at 7:15
in_mathematica_we_trustin_mathematica_we_trust
2,530819
$endgroup$
add a comment |
$begingroup$
Think about the formula for the midpoint of a line as
$$x = frac{1}{2}x_{1} + frac{1}{2}x_{2}.$$
We find the midpoint by taking equal contributions from either end of the line.
If we want to find a point which is one third of the way along the line we should use the following
$$x = frac{2}{3}x_{1} + frac{1}{3}x_{2}$$
where we take uneven contributions from the two endpoints.
The point one quarter of the way along the line can be found using
$$x = frac{3}{4}x_{1} + frac{1}{4}x_{2}.$$
The y-coordinates are found in a similar manner.
answered Nov 12 '13 at 7:15
in_mathematica_we_trustin_mathematica_we_trust
2,530819
$endgroup$
add a comment |
$begingroup$
Think about the formula for the midpoint of a line as
$$x = frac{1}{2}x_{1} + frac{1}{2}x_{2}.$$
We find the midpoint by taking equal contributions from either end of the line.
If we want to find a point which is one third of the way along the line we should use the following
$$x = frac{2}{3}x_{1} + frac{1}{3}x_{2}$$
where we take uneven contributions from the two endpoints.
The point one quarter of the way along the line can be found using
$$x = frac{3}{4}x_{1} + frac{1}{4}x_{2}.$$
The y-coordinates are found in a similar manner.
answered Nov 12 '13 at 7:15
in_mathematica_we_trustin_mathematica_we_trust
2,530819
$endgroup$
Think about the formula for the midpoint of a line as
$$x = frac{1}{2}x_{1} + frac{1}{2}x_{2}.$$
We find the midpoint by taking equal contributions from either end of the line.
If we want to find a point which is one third of the way along the line we should use the following
$$x = frac{2}{3}x_{1} + frac{1}{3}x_{2}$$
where we take uneven contributions from the two endpoints.
The point one quarter of the way along the line can be found using
$$x = frac{3}{4}x_{1} + frac{1}{4}x_{2}.$$
The y-coordinates are found in a similar manner.
answered Nov 12 '13 at 7:15
in_mathematica_we_trustin_mathematica_we_trust
2,530819
answered Nov 12 '13 at 7:15
in_mathematica_we_trustin_mathematica_we_trust
2,530819
answered Nov 12 '13 at 7:15
in_mathematica_we_trustin_mathematica_we_trust
2,530819
answered Nov 12 '13 at 7:15
in_mathematica_we_trustin_mathematica_we_trust
2,530819
2,530819
add a comment |
add a comment |
$begingroup$
Divide your segment into even parts.
From the last point you can find that
$$
x_2 = x_1 + nDelta implies Delta = frac {x_2 - x_1}n
$$
You need to find second point in that division, so
$$
x' = x_1 + Delta = x_1 + frac {x_2-x_1}n = frac {n-1}n x_1 + frac 1n x_2
$$
So, let's say you want to divide your segment into 2 even parts, then $n = 2$ and
$$
x' = frac {x_1}2 + frac {x_2}2
$$
If $n = 3$ (3 even parts)
$$
x' = frac 23 x_1 + frac 13 x_2
$$
$n = 4$ (4 even parts)
$$
x' = frac 34 x_1 + frac 14 x_2
$$
edited Jan 17 at 23:26
Community♦
1
answered Nov 12 '13 at 7:29
KasterKaster
9,06221729
$endgroup$
add a comment |
$begingroup$
Divide your segment into even parts.
From the last point you can find that
$$
x_2 = x_1 + nDelta implies Delta = frac {x_2 - x_1}n
$$
You need to find second point in that division, so
$$
x' = x_1 + Delta = x_1 + frac {x_2-x_1}n = frac {n-1}n x_1 + frac 1n x_2
$$
So, let's say you want to divide your segment into 2 even parts, then $n = 2$ and
$$
x' = frac {x_1}2 + frac {x_2}2
$$
If $n = 3$ (3 even parts)
$$
x' = frac 23 x_1 + frac 13 x_2
$$
$n = 4$ (4 even parts)
$$
x' = frac 34 x_1 + frac 14 x_2
$$
edited Jan 17 at 23:26
Community♦
1
answered Nov 12 '13 at 7:29
KasterKaster
9,06221729
$endgroup$
add a comment |
$begingroup$
Divide your segment into even parts.
From the last point you can find that
$$
x_2 = x_1 + nDelta implies Delta = frac {x_2 - x_1}n
$$
You need to find second point in that division, so
$$
x' = x_1 + Delta = x_1 + frac {x_2-x_1}n = frac {n-1}n x_1 + frac 1n x_2
$$
So, let's say you want to divide your segment into 2 even parts, then $n = 2$ and
$$
x' = frac {x_1}2 + frac {x_2}2
$$
If $n = 3$ (3 even parts)
$$
x' = frac 23 x_1 + frac 13 x_2
$$
$n = 4$ (4 even parts)
$$
x' = frac 34 x_1 + frac 14 x_2
$$
edited Jan 17 at 23:26
Community♦
1
answered Nov 12 '13 at 7:29
KasterKaster
9,06221729
$endgroup$
Divide your segment into even parts.
From the last point you can find that
$$
x_2 = x_1 + nDelta implies Delta = frac {x_2 - x_1}n
$$
You need to find second point in that division, so
$$
x' = x_1 + Delta = x_1 + frac {x_2-x_1}n = frac {n-1}n x_1 + frac 1n x_2
$$
So, let's say you want to divide your segment into 2 even parts, then $n = 2$ and
$$
x' = frac {x_1}2 + frac {x_2}2
$$
If $n = 3$ (3 even parts)
$$
x' = frac 23 x_1 + frac 13 x_2
$$
$n = 4$ (4 even parts)
$$
x' = frac 34 x_1 + frac 14 x_2
$$
edited Jan 17 at 23:26
Community♦
1
answered Nov 12 '13 at 7:29
KasterKaster
9,06221729
edited Jan 17 at 23:26
Community♦
1
edited Jan 17 at 23:26
Community♦
1
edited Jan 17 at 23:26
Community♦
1
1
answered Nov 12 '13 at 7:29
KasterKaster
9,06221729
answered Nov 12 '13 at 7:29
KasterKaster
9,06221729
answered Nov 12 '13 at 7:29
KasterKaster
9,06221729
9,06221729
add a comment |
add a comment |
$begingroup$
Treat $x_1$ as the offset, and $x_2-x_1$ as the distance that needs sectioning: $$x=x_1+frac{x_2-x_1}nquad,quad y=y_1+frac{y_2-y_1}nquad,quad z=z_1+frac{z_2-z_1}n$$ For $x_1=y_1=z_1=0$, the result becomes self-evident.
answered Nov 12 '13 at 7:33
LucianLucian
41.3k159130
$endgroup$
add a comment |
$begingroup$
Treat $x_1$ as the offset, and $x_2-x_1$ as the distance that needs sectioning: $$x=x_1+frac{x_2-x_1}nquad,quad y=y_1+frac{y_2-y_1}nquad,quad z=z_1+frac{z_2-z_1}n$$ For $x_1=y_1=z_1=0$, the result becomes self-evident.
answered Nov 12 '13 at 7:33
LucianLucian
41.3k159130
$endgroup$
add a comment |
$begingroup$
Treat $x_1$ as the offset, and $x_2-x_1$ as the distance that needs sectioning: $$x=x_1+frac{x_2-x_1}nquad,quad y=y_1+frac{y_2-y_1}nquad,quad z=z_1+frac{z_2-z_1}n$$ For $x_1=y_1=z_1=0$, the result becomes self-evident.
answered Nov 12 '13 at 7:33
LucianLucian
41.3k159130
$endgroup$
Treat $x_1$ as the offset, and $x_2-x_1$ as the distance that needs sectioning: $$x=x_1+frac{x_2-x_1}nquad,quad y=y_1+frac{y_2-y_1}nquad,quad z=z_1+frac{z_2-z_1}n$$ For $x_1=y_1=z_1=0$, the result becomes self-evident.
answered Nov 12 '13 at 7:33
LucianLucian
41.3k159130
edited Nov 12 '13 at 8:18
answered Nov 12 '13 at 7:33
LucianLucian
41.3k159130
answered Nov 12 '13 at 7:33
LucianLucian
41.3k159130
answered Nov 12 '13 at 7:33
LucianLucian
41.3k159130
41.3k159130
add a comment |
add a comment |
$begingroup$
Let $P$ and $Q$ be two distinct points and define the (ruler) function
$f : mathbb R to overleftrightarrow{PQ}$, from the set of real numbers into the line $overleftrightarrow{PQ}$, by $f(t) = (1-t)P + tQ$. This function lays a ruler over the line $overleftrightarrow{PQ}$ such that the ordinate of $P$ is $0$ ($f(0)= P$) and the ordinate of $Q$ is $1$ $(f(1) = Q)$. So, for example, the point $dfrac34$ of the way from $P$ to $Q$ is $fleft( dfrac 34 right) = dfrac 14 P + dfrac 34 Q$.
answered Jun 24 '17 at 2:26
steven gregorysteven gregory
18.2k32258
$endgroup$
add a comment |
$begingroup$
Let $P$ and $Q$ be two distinct points and define the (ruler) function
$f : mathbb R to overleftrightarrow{PQ}$, from the set of real numbers into the line $overleftrightarrow{PQ}$, by $f(t) = (1-t)P + tQ$. This function lays a ruler over the line $overleftrightarrow{PQ}$ such that the ordinate of $P$ is $0$ ($f(0)= P$) and the ordinate of $Q$ is $1$ $(f(1) = Q)$. So, for example, the point $dfrac34$ of the way from $P$ to $Q$ is $fleft( dfrac 34 right) = dfrac 14 P + dfrac 34 Q$.
answered Jun 24 '17 at 2:26
steven gregorysteven gregory
18.2k32258
$endgroup$
add a comment |
$begingroup$
Let $P$ and $Q$ be two distinct points and define the (ruler) function
$f : mathbb R to overleftrightarrow{PQ}$, from the set of real numbers into the line $overleftrightarrow{PQ}$, by $f(t) = (1-t)P + tQ$. This function lays a ruler over the line $overleftrightarrow{PQ}$ such that the ordinate of $P$ is $0$ ($f(0)= P$) and the ordinate of $Q$ is $1$ $(f(1) = Q)$. So, for example, the point $dfrac34$ of the way from $P$ to $Q$ is $fleft( dfrac 34 right) = dfrac 14 P + dfrac 34 Q$.
answered Jun 24 '17 at 2:26
steven gregorysteven gregory
18.2k32258
$endgroup$
Let $P$ and $Q$ be two distinct points and define the (ruler) function
$f : mathbb R to overleftrightarrow{PQ}$, from the set of real numbers into the line $overleftrightarrow{PQ}$, by $f(t) = (1-t)P + tQ$. This function lays a ruler over the line $overleftrightarrow{PQ}$ such that the ordinate of $P$ is $0$ ($f(0)= P$) and the ordinate of $Q$ is $1$ $(f(1) = Q)$. So, for example, the point $dfrac34$ of the way from $P$ to $Q$ is $fleft( dfrac 34 right) = dfrac 14 P + dfrac 34 Q$.
answered Jun 24 '17 at 2:26
steven gregorysteven gregory
18.2k32258
answered Jun 24 '17 at 2:26
steven gregorysteven gregory
18.2k32258
answered Jun 24 '17 at 2:26
steven gregorysteven gregory
18.2k32258
answered Jun 24 '17 at 2:26
steven gregorysteven gregory
18.2k32258
18.2k32258
add a comment |
add a comment |
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