Question about transformations and algebra

[Raanikeri]

Can anyone help me understand how to derive the answer for part (b) please?

I can now get the answer to (a) and (c) but can't get the answer to (b).

For (b), I plotted the points out with values of a from 1 to 6. I got the coordinates (-1,1), (1,2), (3,3), (5,4), (7,5), (9,6). I still don't get how to get the answer of (2a-3, a) from here. I can understand that the y coordinate always remains same as value of a. But I don't get how to obtain the x coordinate as an algebraic expression of 2a-3 from the values I just plotted out. Can anyone help?

I also don't know how to find the answer to part (d)... any suggestion how to go about doing that?

 

[lex]

Draw a picture for part (b) (assuming that a is greater than 3) - draw the line x=a and plot the point (3,a). Draw the reflected point and work out the distance over to the line. You should then be able to work out the co-ordinates of the reflected point.
(If you draw the picture again with x<=3, you find you get the same answer)!

For (d) do the same having first noted that a reflection in the line y=x just reverses the x and y co-ordinates. I.e. do what you did for part (b), but use the point (a,3) and the reflection line y=b+1 (assume b+1 >3, first)

 

[Raanikeri]

Draw a picture for part (b) (assuming that a is greater than 3) - draw the line x=a and plot the point (3,a). Draw the reflected point and work out the distance over to the line. You should then be able to work out the co-ordinates of the reflected point.
(If you draw the picture again with x<=3, you find you get the same answer)!

For (d) do the same having first noted that a reflection in the line y=x just reverses the x and y co-ordinates. I.e. do what you did for part (b), but use the point (a,3) and the reflection line y=b+1 (assume b+1 >3, first)

I did plot out the points for values of a from 1 to 6. I said previously that this gave me the coordinates (-1,1), (1,2), (3,3), (5,4), (7,5), (9,6). Isn't this what you said to do?

My question is, how do I get the answer of a general formula of "2a-3" for these x coordinates of -1, 1, 3, 5, 7, and 9, from the points I plotted out, or for any value of x plotted out on a graph reflecting the point (3,a) about the line x=a? That's the real problem I have.

By the way the question never said the value of a must be greater than 3, so why did you say you are assuming a is greater than 3? How did you know that from the question? What did I miss ??

 

[Raanikeri]

Actually ... I just figured out how to find answer to (b) by finding the nth term for my x coordinate points that I have plotted already... ugh wish I knew earlier... damn I had just woken up from a nap about half an hour earlier and I dreamt it but didn't believe it would work till I tried it out now ?

Now I just need help for (d)... I'll be trying it out as well.. but damn. Nothing annoys me more than not being able to solve a question ?

The problem with (d) is that they have introduced a 2nd variable "b" into the linear equation y=b+1! How do you plot the value of "b" on the graph? It seems wrong to assume that "b" is the same as "x" and plotting the values of "b" on the "x" axis just seems wrong to me by the way. Especially when they have already mentioned another linear equation in the same question, y=x, which actually has an actual "x" value, so surely the "b" value in y=b+1 cannot be treated as an "x" value?? ??‍

Unless we plot a series of lines y=0, y=1, y=2, etc based on b values of say, -1 to 2, and see how the point reflects on those lines..?

 

[lex]

Sorry, I should have referred to your working. That was a very clever solution to part (b)!
The following is what I had been suggesting:

 

[Raanikeri]

Sorry, I should have referred to your working. That was a very clever solution to part (b)!
The following is what I had been suggesting:
View attachment 30998View attachment 30999

Thanks lex! I really loved seeing your solution. It's fantastic to know how others have different ways of solving this!

I could also get the answer but by comparing values of y and a and b. I enclose my working. Basically I assigned values of a (and thus b, since a=b due to the reflection in y=x) to be same as the nth term of a sequence, so 1, 2, 3, 4... then found the corresponding final answer's y coordinate. Wrote the y coordinate answers as a sequence and finding nth term again, I could derive the formula for the y coordinate as either 2b-1 or 2a-1, even though the original answer for this question says only 2b-1. I suppose that is so because the question stated to express answer in terms of both a and b, so the 2b-1 is the one to pick.

Your solution is a good one for explaining to someone who doesn't yet know how to find the nth term ? Graphs aren't really my strong point... I just couldn't "see" the solution just looking at the graph, but once you explained it, I could. Thank you for helping me out with this.

 

[lex]

I'm glad you found the solution interesting.
You certainly are a determined 'problem-solver' and show good 'lateral thinking'.

Just like when drawing pictures, you need to make sure you are aware of any 'restrictions' you may have imposed on the problem.
E.g. your restriction a=b is not necessary for this problem. Reflection in the line y=x does not mean that a=b. (For any point on the line y=x, the y co-ordinate = x co-ordinate, but simply because we are reflecting in the line y=x, does not place any restrictions on a and b).
In other words, you have been working with a restricted version of the problem.
You worked on:
(3,a) is reflected in the line y=x, followed by y=b+1, where a=b
e.g. * (3,5) reflected in the line y=x, followed by y=6
But you have not considered a question like:
** (3,5) reflected in the line y=x, followed by y=7

As you rightly point out, your method gives several possible answers:
(1) (a,2b-1)
(2) (b,2b-1)
(3) (a,2a-1)
(4) (b,2a-1)
All 4 of these work for your restricted problem (where a=b), e.g. * above.
However your method does not show which of the 4 might be true for the more general problem where a may not be equal to b, e.g. ** above.
It turns out that (1) works and the others do not.

My method has the benefit of producing a convincing argument that (1) (a,2b-1) is the correct solution.

In fact even for part (b): (3,a) reflected in the line x=a,
using the pattern in the first 6 points is not enough to show that (2a-3,a) is the correct answer, since e.g.
[imath](\tfrac{1}{80} (a^6-21a^5+175a^4-735a^3+1624a^2-1604a+480), \tfrac{1}{48} (a^6-21a^5+175a^4-735a^3+1624a^2-1716a+720))[/imath]
satisfies your 6 points, when a=1, a=2..., a=6
(and does not work for a=7)
No matter how many points you test, I can find rules which work for them, but not for the next point!

The advantage of my pictorial method, is that it gives a convincing argument why (2a-3,a) is the correct answer.