Find a Point for x=5: Help Appreciated | Direction Vector = (0,1)

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In summary, the conversation discusses finding a point on the line x=5 and determining the direction vector for the line x+0y-5=0. The participants also discuss how setting x to 5 restricts the value of y, and how any point on the line can be represented as (5,a), where a is any real number. The conversation concludes with the understanding that both (5,0) and (5,6) are valid points on the line x=5.
  • #1
thomasrules
243
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for each of the following, find a point on each line:

x=5

:D LOl ok this is what i thought

x+0y-5=0

so direction vector=(0,1)

If I set x to 5 then y should = 0 shouldn't it?
 
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  • #2
thomasrules said:
for each of the following, find a point on each line:

x=5

:D LOl ok this is what i thought

x+0y-5=0

so direction vector=(0,1)

If I set x to 5 then y should = 0 shouldn't it?

First think about what would happen if you didn't set x equal to 5, say you set x equal to 6 what happens?

You don't set x equal to 5, x is 5 period end of story, that is what you're given all points on that line are of the form (5,a) where a is any real number. You just need one point, I think you can figure it out from here.
 
  • #3
yea I just saw at the back of the book it said (5,6) and I had no idea where that came from
 
  • #4
READ the problem! It said "find a point on each line". How many points are there on a line? Since the equation is x= 5, obviously the first coordinate must be 5. What about the second coordinate? What does that equation tell you about y? How does that equation restrict y?

Your answer, (5, 0), is a perfectly good answer. So is (5, 6). So is
(5, 323121232344).
 

FAQ: Find a Point for x=5: Help Appreciated | Direction Vector = (0,1)

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