How do you know if points are collinear

[Blair1994]

e.g (2,1) , (1, -3) and (3, 5)

[m@tty]

"collinear" means lying on the same line. So construct a line that passes through two of the points and see if the third satisfies the relationship.

[xZero]

make a line from (1,-3) to (3,5)

u = (1,-3) +t((3,5)-(1,-3))
u = (1,-3) +t(2,8)

x(t)=1+2t
y(t)=-3+8t

if (2,1) lines on this line, then the points are collinear, so sub the points in and see if t is the same for both parametric equations

[Blair1994]

with a question like that where it asks about the relationship between points, is the answer usually collinear? or should i test other stuff?

[m@tty]

Ah, you can also have co-planar points if they give you ordered triples [ (x,y,z) ] but I can't remember if that was in spesh or not..

But you need to approach each question independent of previous ones you've done.. after you deduce what it is asking for, formulate a method to tackle it. Don't use stock standard approaches for similar questions.

make a line from (1,-3) to (3,5)

u = (1,-3) +t((3,5)-(1,-3))
u = (1,-3) +t(2,8)

x(t)=1+2t
y(t)=-3+8

if (2,1) lines on this line, then the points are collinear, so sub the points in and see if t is the same for both parametric equations

I would just use normal cartesian equations - easier. (I still find it simpler, even though the working here is longer haha)

so, using (2,1) and (1,-3):



then sub one of the points into


sub (2,1)


Try the third point(3,5):







And so the points are not collinear.

[Blair1994]

the other point is 3,5

[xZero]

Haha I always prefer parametric equations, also you subbed the points wrong. It should be (3,5) and they are collinear

[m@tty]

the other point is 3,5

Haha I always prefer parametric equations, also you subbed the points wrong. It should be (3,5) and they are collinear

Indeed xD I'm at work, let's say it was rushed..

And you prefer parametric? Fair enough

btw, you're missing a t in your parametric equation for y...

[vea]

There is also the vector method:

If any two vectors from one point to another are parallel, then the three points are collinear.
e.g (2,1) , (1, -3) and (3, 5)

(2,1)-(1,-3)=(1,4)
(2,1)-(3,5)=(-1,-4)

Now (2,1)-(3,5)=-[(2,1)-(1,-3)]
Therefore the points (2,1), (1,-3) and (3,5) are collinear.

I don't think I've done it formally here but to do it properly, you would assign the points above vectors and show it using vectors.