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

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To find the value of t for point E to lie on line CD, the coordinates of E are expressed as (t, 2t-1). The discussion involves determining the parametric equations for lines CD and DC, with the equations derived from points C and D. The intersection of these lines is calculated, leading to the conclusion that t equals 2. The participant expresses gratitude for the assistance received in understanding the problem-solving process.
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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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