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Raehya

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SHOW THAT THE FOLLOWING SETS OF POINTS ARE ON THE SAME LINE.

b.(a,b+c), (b,c+a), (c,a+b)


Please can someone answer this? I know how to solve it but only if its numbers. My teacher forgot to taught us how to solve it when it comes to variables/letters!!!
 
Raehya said:
SHOW THAT THE FOLLOWING SETS OF POINTS ARE ON THE SAME LINE.

b.(a,b+c), (b,c+a), (c,a+b)


Please can someone answer this? I know how to solve it but only if its numbers. My teacher forgot to taught us how to solve it when it comes to variables/letters!!!
Click to expand...
Suppose a=2, b=3 and c=4

How would you show that THE FOLLOWING SETS OF POINTS ARE ON THE SAME LINE.

b.(a,b+c), (b,c+a), (c,a+b)

Please show your work in detail.
 
Raehya said:
SHOW THAT THE FOLLOWING SETS OF POINTS ARE ON THE SAME LINE.

b.(a,b+c), (b,c+a), (c,a+b)


Please can someone answer this? I know how to solve it but only if its numbers. My teacher forgot to taught us how to solve it when it comes to variables/letters!!!I
Click to expand...
It's something that you are going to have to get used to. But the proof works exactly the same way as it does for numbers, the only difference is that you can say something like 2 + 3 = 5 and you can't do that (generally) with a + b. But you can leave it as 2 + 3...no one says you have to put that as 5. Give Subhotosh Khan's idea a go. Then do it again using a, b, and c.

-Dan
 
Worked solution removed by moderator.
 
Last edited by a moderator:
Shiloh, you shouldn't be posting a solution before four days, or a complete solution before seven days.
Did you see the second post where the OP was asked to show the work in detail?
 
Raehya said:
SHOW THAT THE FOLLOWING SETS OF POINTS ARE ON THE SAME LINE.
(a,b+c), (b,c+a), (c,a+b)
Click to expand...
May I suggest that you name these three points: [imath]A:(a,b+c),~B:(b,c+a)~\&~C:(c,a+b)[/imath].
I see no need for numbers, Moreover, we assume that these three are points in [imath]\mathbb{R}^2[/imath]
Now think about this carefully: if three points pairwise determine the same slope then they are colinear.
[imath][/imath][imath][/imath]
 
First, you posted this in calculus. The only hard thing about using calculus is the algebra involved. If you are still having difficulty representing unknown numbers by letters, drop calculus immediately and take a review course in algebra.

Second, two points determine a line, right? How do you calculate the equation of a line from the coordinates of a pair of points? (You probably learned this in the first few weeks of algebra, but you were asked to get answers expressible in numerals. But the principal is the same.)

[math]\dfrac{y - (b + c)}{x - a} = \dfrac{(b + c) - b}{a - (c + a)} \implies y = \text {WHAT?}[/math]
Tell us what algebraic answer you get after doing the algebra to solve for y.

Now if the third point is co-linear with the other two, that means that it is on the same line and so satisfies the equation of that line.

Show whether the third point does or does not satisfy the equation.

(This is a far less efficient method than pka’s, but it goes back to stuff you did in the first semester of first year algebra.)
 
JeffM said:
(This is a far less efficient method than pka’s, but it goes back to stuff you did in the first semester of first year algebra.)
Click to expand...
It also comes mightily close to simply giving the answer.
 
pka said:
It also comes mightily close to simply giving the answer.
Click to expand...
Looking at the contribution by OP - that could be "yet so far away". S/he will be "forced" to do some work to get full credit for the assignment. And in the process (completing close answer to creditable answer), the OP cannot avoid learning "something".
 
Raehya,

Please respond so that we know how you are going with this problem.

If you were given three numerical points, would you find the gradients between each pair and see if they are all equal?

You can do exactly the same thing with these three points, together with a bit of algebra to tidy the gradients up. What do you notice?
 
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