Say I have a line segment with two points (2,1) and (5,10). I need to find all points on the line starting at (2,1) and moving towards (5,10) that have a distance of 3 from each other. Is there a formula I can use? I know the slope and the y intercept will help me find points on the line but randomly plugging in numbers to see if it is at a certain distance does not seem right. Any suggestions?
Points on a line at a specified distance interval?
- Thread starter Big Bossy
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First off, I'd check to see that there are ANY.
\(\displaystyle \sqrt{(5-2)^{2}+(10-1)^{2}}\;=\;\sqrt{9+81}\;=\;\sqrt{90}\;=\;3\cdot\sqrt{10}\ge 3\)
Okay, I'm convinced.
Try this:
Start at (2,1).
Describe all the points on the line segment from (2,1) to (5,10): (2,1) + t*((5-2),(10-1)) = (2,1) + t(3,9) for \(\displaystyle 0 \le t \le 1\)
Okay, all the points look lile this: (2,1) + t(3,9) or (2+3t,1+9t). Since t' was ust made up, it makes no difference if we use something else.
All those same points can be described as (2+3s,1+9s) for \(\displaystyle 0 \le s \le 1\)
Can yuo express the s-points vs. the t-points in an expression that suggests they are exactly 3 apart?
\(\displaystyle \sqrt{(5-2)^{2}+(10-1)^{2}}\;=\;\sqrt{9+81}\;=\;\sqrt{90}\;=\;3\cdot\sqrt{10}\ge 3\)
Okay, I'm convinced.
Try this:
Start at (2,1).
Describe all the points on the line segment from (2,1) to (5,10): (2,1) + t*((5-2),(10-1)) = (2,1) + t(3,9) for \(\displaystyle 0 \le t \le 1\)
Okay, all the points look lile this: (2,1) + t(3,9) or (2+3t,1+9t). Since t' was ust made up, it makes no difference if we use something else.
All those same points can be described as (2+3s,1+9s) for \(\displaystyle 0 \le s \le 1\)
Can yuo express the s-points vs. the t-points in an expression that suggests they are exactly 3 apart?
HINT:
Plot the line as A(2,1), B(5,10); complete right triangle ABC; BC = hypotenuse = 3SQRT(10) ; OK?
In corner A, draw smaller right triangle ADE: AD = 3 is hypotenuse; you now have 2 similar triangles.
EDIT: AB is hypotenuse, NOT BC; D is on AB, of course...
Plot the line as A(2,1), B(5,10); complete right triangle ABC; BC = hypotenuse = 3SQRT(10) ; OK?
In corner A, draw smaller right triangle ADE: AD = 3 is hypotenuse; you now have 2 similar triangles.
EDIT: AB is hypotenuse, NOT BC; D is on AB, of course...
D
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Denis said:
Denis,
You are confusing me!!
The points (that are 3 apart ) must be on the line joining (2,1) and (5,10).
Where is point C coming from to make the right-triangle?
D
Deleted member 4993
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Big Bossy said:
You can easily see that there are are infinite number of points that will meet those criteria. From the calculation that TK showed you, you can define the domain of x where there will be another point at a distance 3 (along the given line).
Any point (x[sub:w6yphmi1]1[/sub:w6yphmi1],y[sub:w6yphmi1]1[/sub:w6yphmi1])on the given line must satisfy the equation:
y[sub:w6yphmi1]1[/sub:w6yphmi1] = 3x[sub:w6yphmi1]1[/sub:w6yphmi1] - 5
Suppose there is another point on the line at a distance 3 from the first point whose cordinates are (x[sub:w6yphmi1]2[/sub:w6yphmi1], y[sub:w6yphmi1]2[/sub:w6yphmi1]) where again y[sub:w6yphmi1]2[/sub:w6yphmi1]= 3x[sub:w6yphmi1]2[/sub:w6yphmi1]-5
Then we can write:
\(\displaystyle (x_2 \ - \ x_1)^2 \ + \ (y_2 \ - \ y_1)^2 \ = \ 9\)
Now you will get a relation between x[sub:w6yphmi1]1[/sub:w6yphmi1] and x[sub:w6yphmi1]2[/sub:w6yphmi1]
Hello, Big Bossy!
\(\displaystyle \text{The slope of line }AB\text{ is: }\:m \:=\:\frac{10-1}{5-2} \:=\:3\)
\(\displaystyle \text{The line through }A(2,1)\text{ with slope 3 is: }\:y - 1 \:=\:3(x-2) \quad\Rightarrow\quad y \:=\:3x - 5\)
\(\displaystyle \text{Let the two points be: }\
(p,3p-5)\text{ and }Q(q,3q-5),\,\text{ where }p < q.\)
\(\displaystyle \text{Then we have: }\Q = 3 \quad\Rightarrow\quad \sqrt{(q-p)^2 + ([3q-5] - [3p-5])^2} \:=\:3\)
. . \(\displaystyle \sqrt{(q-p)^2 + (3q-3p)^2} \:=\:3 \quad\Rightarrow\quad \sqrt{(q-p)^2 + 9(q-p)^2} \;=\;3\)
. . \(\displaystyle \sqrt{10(q-p)^2} \:=\:3 \quad\Rightarrow\quad \sqrt{10}(q-p) \:=\:3 \quad\Rightarrow\quad q-p \:=\:\frac{3}{\sqrt{10}}\)
. . \(\displaystyle q \;=\;p + \frac{3}{\sqrt{10}}\)
\(\displaystyle \text{The points are: }\(p,\,3p-5),\;\; Q\left(p + \tfrac{3}{\sqrt{10}},\,3p +\tfrac{9}{\sqrt{10}} - 5\right)\)
. . \(\displaystyle \text{for }p \in \left[2,\:5-\tfrac{3}{\sqrt{10}}\right]\;\;\)
\(\displaystyle \text{The slope of line }AB\text{ is: }\:m \:=\:\frac{10-1}{5-2} \:=\:3\)
\(\displaystyle \text{The line through }A(2,1)\text{ with slope 3 is: }\:y - 1 \:=\:3(x-2) \quad\Rightarrow\quad y \:=\:3x - 5\)
\(\displaystyle \text{Let the two points be: }\
(p,3p-5)\text{ and }Q(q,3q-5),\,\text{ where }p < q.\)\(\displaystyle \text{Then we have: }\
Q = 3 \quad\Rightarrow\quad \sqrt{(q-p)^2 + ([3q-5] - [3p-5])^2} \:=\:3\). . \(\displaystyle \sqrt{(q-p)^2 + (3q-3p)^2} \:=\:3 \quad\Rightarrow\quad \sqrt{(q-p)^2 + 9(q-p)^2} \;=\;3\)
. . \(\displaystyle \sqrt{10(q-p)^2} \:=\:3 \quad\Rightarrow\quad \sqrt{10}(q-p) \:=\:3 \quad\Rightarrow\quad q-p \:=\:\frac{3}{\sqrt{10}}\)
. . \(\displaystyle q \;=\;p + \frac{3}{\sqrt{10}}\)
\(\displaystyle \text{The points are: }\
(p,\,3p-5),\;\; Q\left(p + \tfrac{3}{\sqrt{10}},\,3p +\tfrac{9}{\sqrt{10}} - 5\right)\). . \(\displaystyle \text{for }p \in \left[2,\:5-\tfrac{3}{\sqrt{10}}\right]\;\;\)
Wow, you all are very smart. It would have been hard to figure this out without all the responses. Thanks.
So based on the repsonses I can find all x values that are at a given distance from a starting point using
D represents the desired distance to plot the next point on the line.
X1 represents the starting x point and of course m=slope and b=y intercept
\(\displaystyle x=X\)[sub:t1ath38c]1[/sub:t1ath38c]\(\displaystyle + D \div (\sqrt{m^2 + 1})\)
I then plug x into the line equation to find y.
\(\displaystyle y=mx+b\)
Now, depending on the slope, I believe when it is negative, my results sometimes go in the opposite direction. So I have to determine the direction and reverse it. I do not have a clean way to do that yet.
So based on the repsonses I can find all x values that are at a given distance from a starting point using
D represents the desired distance to plot the next point on the line.
X1 represents the starting x point and of course m=slope and b=y intercept
\(\displaystyle x=X\)[sub:t1ath38c]1[/sub:t1ath38c]\(\displaystyle + D \div (\sqrt{m^2 + 1})\)
I then plug x into the line equation to find y.
\(\displaystyle y=mx+b\)
Now, depending on the slope, I believe when it is negative, my results sometimes go in the opposite direction. So I have to determine the direction and reverse it. I do not have a clean way to do that yet.