Parameterizing a Line: Cross Through P(3,-5) with Direction V(2,8)

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In summary, the line L passes through the point P and has direction v, so it goes away from the point P, say, to another point Q on its path.
  • #1
vanitymdl
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Homework Statement



Parameterize a line, L, such that is crosses through the point P=(3, -5) and direction v=(2, 8). Now, using this parametrization determine the following points belong to L: P1=(73, -180) and P2=(5, -14)

Homework Equations





The Attempt at a Solution


I really need help with this type of questions in major need of help and explanations.
 
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  • #2
welcome to pf!

hi vanitymdl! welcome to pf! :smile:

let's start with the first part

can you do …
vanitymdl said:
Parameterize a line, L, such that is crosses through the point P=(3, -5) and direction v=(2, 8).
 
  • #3
okay all I know is where the point is located, so with the direction is it going towards that point?
 
  • #4
vanitymdl said:
okay all I know is where the point is located, so with the direction is it going towards that point?

The line L passes through the point P and has direction v, so it goes away from the point P, say, to another point Q on its path.
 
  • #5
Oh so that line is going to stop at (2,8)?
 
  • #6
no, it means that the direction is parallel to the line going through (0,0) and (2,8)

(its "clock direction" is (2,8))
 
  • #7
okay I think I have an idea now
since P is (3,-5) and v (2,8)

then (x,y) = (3,-5) + t(2,8)
which is (3+2t, -5+8t)

x = 3+2t
y = -5+8t
 
  • #8
yup! :biggrin:

are you ok with the second part now? :smile:
 
  • #9
Ah I'm excited I figured that out. Okay just to clarify the second part, I get my x and y then equal it to the point that I'm trying to figure out if its in the line?
 
  • #10
If my t's for the x and y give me different value does that mean that they don't belong to the line?
 
  • #11
For the same point, t is a constant, so it should stay the same for both x and y.
 
  • #12
vanitymdl said:
If my t's for the x and y give me different value does that mean that they don't belong to the line?

again … yup! :biggrin:
 
  • #13
Thank you SO MUCH :)
 

1. What does it mean to parameterize a line?

Parameterizing a line means to represent a line in terms of one or more variables, usually denoted as t or s. This allows us to describe all the points on the line using a single equation, rather than listing out each individual point.

2. How do you parameterize a line?

To parameterize a line, we first need to determine the direction of the line. This can be done by finding the slope of the line using the two given points. Next, we choose a point on the line and assign it as the starting point. Then, we use the slope to determine the change in x and y values for each unit of t or s. Finally, we can write out the equation of the line using the starting point and the change in values for t or s.

3. Why do we parameterize lines?

Parameterizing lines allows us to easily describe and manipulate lines in a more concise and general way. It also allows us to easily find specific points on the line by plugging in different values for t or s. Additionally, it is useful in applications such as computer graphics and physics.

4. Can any line be parameterized?

Yes, any line can be parameterized as long as we know at least one point on the line and its direction. The equation for the parameterized line may vary depending on the chosen point and direction, but it will still represent the same line.

5. How is parameterizing a line different from writing it in slope-intercept form?

Parameterizing a line is different from writing it in slope-intercept form because it uses a different set of variables (t or s) and allows for more general representations of lines. Slope-intercept form is specific to lines with a constant slope and y-intercept, while parameterizing a line can represent lines with varying slopes and intercepts.

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