My son has been given the collinear problem below that has be stumped two folds.
1- By trying to solve the problem using (y2-y1)/(x2-x1)=slope -- and then, find the (b) in y=mx+b for two points takes way way too long to be a 7th grade probem. My son has already been working on it for over two hours. There must be a shortcut…
2- Why is it that; y=mx+b for points with a zero never give the righ b? e.g. (0,3)
Problem:
The coordinates of the points are:
A= (0,3) | B=(2.11) | C=(3,-3) | D=(-1,9) | E=(-3,10) |F=(6,10) | G=(-5,19) | H=(-6,18)
Three, and only three of these points are collinear. The student’s task is to determine “Analytically” and without plotting, which three is the collinear trio.
5 answers
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.
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
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
Thank you again
now you guys are 20 years old
1 answer