# collinear points

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• May 3rd 2009, 03:04 AM
collinear points
The number of real numbers a such that (a,1),(1,a) & (a-1,a-1) are three distinct points is

a)0
b)1
c)at least 2 but finitely many
d)infinitely many

i tried putting a=-1,0,1/2,1,2 which lead me to the answer a) which isn't correct.........
• May 3rd 2009, 03:40 AM
NonCommAlg
Quote:

The number of real numbers a such that (a,1),(1,a) & (a-1,a-1) are three distinct points is

a)0
b)1
c)at least 2 but finitely many
d)infinitely many

i tried putting a=-1,0,1/2,1,2 which lead me to the answer a) which isn't correct.........

distinct or collinear? (why am i asking this? because the title of your post is "collinear points"!!)
• May 3rd 2009, 06:14 AM
both
they should be distinct as well as coillinear!
• May 3rd 2009, 06:15 AM
i saw that now. it was basically a typo. i meant "...distinct collinear points..."
(Itwasntme)
• May 3rd 2009, 06:29 AM
Jester
Quote:

The number of real numbers a such that (a,1),(1,a) & (a-1,a-1) are three distinct points is

a)0
b)1
c)at least 2 but finitely many
d)infinitely many

i tried putting a=-1,0,1/2,1,2 which lead me to the answer a) which isn't correct.........

Clearly $a \ne 1$. Create 2 vectors connecting your points and the require that they be parallel. This will give you your $a$ value.
• May 3rd 2009, 06:35 AM
Quote:

Originally Posted by danny arrigo
Clearly $a \ne 1$. Create 2 vectors connecting your points and the require that they be parallel. This will give you your $a$ value.

hello danny,

we basically need the number of values of a & not the value of a.
• May 3rd 2009, 06:51 AM
Jester
Quote:

hello danny,

we basically need the number of values of a & not the value of a.

But in finding the value(s) you can then answer your question. It is probably the most direct way.
• May 3rd 2009, 06:52 AM
Quote:

Originally Posted by danny arrigo
But in finding the value(s) you can then answer your question. It is probably the most direct way.

well, i said that because the answer comes out to be d. (infinitely many)
how will you lead to that with that method ? (Nerd)
• May 3rd 2009, 07:08 AM
Jester
Quote:

well, i said that because the answer comes out to be d. (infinitely many)
how will you lead to that with that method ? (Nerd)

How did you arrive at that answer?
• May 3rd 2009, 07:11 AM
Quote:

Originally Posted by danny arrigo
How did you arrive at that answer?

its given in the text. i did not. (Itwasntme)
• May 3rd 2009, 07:21 AM
Jester
Quote:

its given in the text. i did not. (Itwasntme)

I guess it comes back to NonCommAlg's question - distinct or collinear?
• May 3rd 2009, 07:25 AM
HallsofIvy
If (1, a), (a, 1), and (a-1, a-1) are co-linear, then we must have (a-1)/(1- a)= -1 (the slope of the line from (1,a) to (a,1)) equal to (a-1-a/ a-1-1)= -1/a (the slope of the line from (a, 1) to (a-1, a-1)). -1= -1/a is satisfied only by a= 1. But if a= 1, (1, a)= (1, 1)= (a, 1) so the three points are not distinct.

There is NO value of a that will make these points distinct and collinear.
• May 3rd 2009, 07:26 AM
distinct AND collinear
• May 3rd 2009, 07:46 AM
The points are distinct and collinear for $a=3:$ $\;\;(3,1),~(1,3),~(2,2)$.