Posted by sally on Sunday, January 27, 2013 at 5:20pm.
sorry B=(-2,11)
I agree with you that this is a rather ridiculous question
There are 21 possible pairs of points, each pair could give us the equation of a line containing such a pair.
If the objective is to use the method you suggest, it could just as easily be effictive with , let's say, 4 points.
I would have allowed my students to make a sketch, thus eliminating some of the more obvious non-collinear points.
You don't actually have to find the equation of the line. If you find the slopes of pairs that have a common point, and those two slopes are equal, then the 3 points are collinear.
e.g. If slope PQ = slope PR , the P, Q, and R are collinear, notice P was the "link"
but, if slope PQ = slope RS, all we know is that PQ || RS
As to your point #2, I think you must be making some kind of error,
Using a point such as (0,3) in y=mx +b will let you find b in the same way as using any other point
e.g. suppose we know y = 4x + b and we are told that (0,3) is on it
3 = 4(0) + b
b = 3
notice that in y = mx + b, the b value is the y-inercept
and of course (0,3) is the point of the y-intercept, so obviously if in a point, the x = 0, the y must be the value of b in the equation.
I played with this for a while but could see no practical way to do it without sketching the points.
I dis put the points in order of increasing x from -6 to +6 which makes it easier to visualize
Thank you very much Reiny and also Damon. This is the first time I felt embarrassed helping my son. Sign of things to come of course. As Reiny suggested, we ended up plotting and then selecting less than half a dozen candidates. I also cheated and made a spreadsheet just to speed things up. The spreadsheet quickly showed the Trio is: C, H AND F.
Thank you again