Determine which point are collinear (1,9),(6,3) and (10,0)

first find a line that connects (1,9) and (6,3)

(10, 0) cant be collinear coz its on x axis

oh yea good point haha

therefor other two are colinear

I DONT KNOW THE LINE -_-

Nice

Aren;t any two points taken together colinear? The third point might not be. But (1,9) and (10,0) are colinear...as are any two you choose.

idk

well how do you show your work

I'm a little confused by the question. Now if it said, "determine is the points are colinear" it would make sense.

well its determine whether or not they are

whether or not the three points are colinear? In that case, you would find the slope from point A to point B and the slope from point B to point C and see if they are the same.

okay working on this.. First figure out the line that connects first 2 points (1,9) and (6,3): (y - y1) = M (x-x1) 9-3 = M (1-6) 6 = M (-5) M = - 6/5

y-9 = (-6/5) (x-1) y = -6x/5 + 6/5 + 9 y = -6x/5 + 51/5

nice

Now plug in point x = 10 and see if the y value is really 0: y = -6 * 10 / 5 + 51 / 5 y = -60/5 + 51/5 y = (51-60)/5 y = -9/5

see if i try point (10,0), it's not on the line, but the point (10, -9/5) is.

for the second points the slope is (y-y1)=M(x-x1) 0-3=M(10-6) -3=4M M=-3/4 Since the two slopes (bahrom's first slope and this one) aren't the same, they aren't colinear... ANd then you don't even have to come up with an equation for a line.

good point mtbender, sorry I never took this stuff in school haha, so I had to improvise lol

so which one is collinear

No problem...I like the completeness of getting the equation and doing the plugging (means you understand the entire scope). But there a certain joy in elegance. :)

1,9 and 6,3 are collinear

oh ok this is sooooo confusing

Well...so are (1,9) and (10,0); (6,3) and (10,0) Any two of the points taken together are colinear They are three points of a triangle so there is a line that connects any two.

The correct answer is that the three points are not colinear.

ok what about (3,1),(5,5) and (7,9)

Look at the slopes between points A-B and B-C like we did in the above problem. Set it up as \[y-y_{1}=M(x-x_{1})\] solve for M. You should find the slopes to be the same (M=2) in both cases. Since the slopes are the same, the points are all colinear.

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