Parameterizing a line through P(3,-5) with direction vector (2,8)

[vanitymdl]

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.

 

[tiny-tim]

welcome to pf!

hi vanitymdl! welcome to pf!

let's start with the first part

can you do …

Parameterize a line, L, such that is crosses through the point P=(3, -5) and direction v=(2, 8).

 

[vanitymdl]

okay all I know is where the point is located, so with the direction is it going towards that point?

 

[DryRun]

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.

 

[vanitymdl]

Oh so that line is going to stop at (2,8)?

 

[tiny-tim]

no, it means that the direction is parallel to the line going through (0,0) and (2,8)

(its "clock direction" is (2,8))

 

[vanitymdl]

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

 

[tiny-tim]

yup!

are you ok with the second part now? ​

 

[vanitymdl]

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?

 

[vanitymdl]

If my t's for the x and y give me different value does that mean that they don't belong to the line?

 

[DryRun]

For the same point, t is a constant, so it should stay the same for both x and y.

 

[tiny-tim]

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!

 

[vanitymdl]

Thank you SO MUCH :)