How Do You Solve for t When Point E Lies on Line CD?

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Homework Help Overview

The discussion revolves around determining the value of t such that point E, defined by the coordinates (t, 2t-1), lies on the line connecting points C and D. The context involves vector calculations and parametric equations in a coordinate system.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the vectors and lines involved, particularly focusing on the parametric equations of the lines through points C and D. There are attempts to establish the intersection of these lines with the line defined by point E.

Discussion Status

The conversation has progressed with participants exploring various methods to find the intersection of the lines. Some have provided guidance on forming equations, while others have expressed uncertainty about the next steps in the process. There is an acknowledgment of understanding the reasoning behind the calculations, indicating a productive direction in the discussion.

Contextual Notes

Participants are navigating the complexities of vector relationships and parametric equations, with some expressing confusion about how to proceed with the established equations. The original poster has indicated difficulty in solving for t, which remains a focal point of the discussion.

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Homework Statement


A(-4,-4), B(4,-2) and D(-2,2)
a) Determine AB and AD vector
My answer: AB=[8,2] and AD=[2,6]
b) A point C is determined by DC being parallel to AB and the angle ABC = 90. Calculate the coordinates of C.
I found out that C=(46/17, 54/17)
c) Another point E has the coordinates (t, 2t-1) where t = R
1) Find t so that point E is on line through C and D.

I am not able to solve this one

Homework Equations


ED=k*CD
ED=(-2-t, -2t+1)
But I always end up with two unknown variables k and t.

The Attempt at a Solution



ED=[-2-t,-2t+1]
ED must be parallel with CD, therefore is
ED=k*CD=[(-80/17)k,(-20/17)k]
I don't know what to do next
 
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Actually, x=t ; y=2t-1 are the parametric equations of a straight line and they tell you that E is a point on this line. You just have to intersect this straight line, with the straight line the contains the vector DC (Do you know how to obtain the equations of this last straight line? )
 
Hmm, I'm not sure that I fully understand what you mean. I know how to find the parametic to DC, but what should I do with the two parametics?
 
Do you know how to obtain the equations (parametric, or vectorial, or any of the multiple type of equations of a straight line) of the unique straight line that passes through C and D ?
 
CD=[-2-(46/17), 2-(54/17)]=[-(80/17),-(20/17]
The parametric will then be
x=-2-(80/17)t
y=2-(20/17)t
 
Well, given that you already know that \vec{CD} is parallel to vector \vec{AB} and this last one is easier to calculate with, let us use it to form the equations of the unique straight line that passes through C and D. We can use the point D and the vector \vec{AB}, so the parametric equations of that unique straight line are:

x=-2+8s; y=2+2s

Now you only have to obtain the point interesection of those two straight lines (this last one, and the one given by: x=t; y=2t-1 ). Do you know how to do this?
 
1.) -2+8s=t
2.) 2+2s=2t-1

1) s=(t+2)/8

2.) 2+2*((t+2)/8)=2t-1
t=2
 
Good, so you now know that t=2, then, in x=t;y=2t-1 you insert t=2 and obtain E, but the most important thing (far beyond any calculation) is: did you understand the reasoning behind this calculation?
 
Yes, I actually did! To be honest, in the beginning I didn't think I would. After looking through it some times I understand it now. I am sorry, that I am slow learner, and it took some time for me to fully understand what you meant. I really do appriate your help! Thank you so much!
 
  • #10
You're welcome! And you must know that at the beginning we all were kind of "slow" :-)
 
  • #11
Haha, I will remember that :)
 

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