You have a sign wrong in first line of your attempt x'=2x-y gives 2x=x'+y
The question statement says find f(L), which I have highlighted in bold. However, the book's answer seems to continue on and calcualte for y from the f(L) equation, which I proceed to work out, although my final answer is slightly off. I'm slightly confused as to whether or not the highlihted value for f(L) is in fact f(L)! Can anyone help?
Many thanks.
Q. L is the line x - y + 1 = 0. f is the transformation f: (x, y) ---> (x', y') where: x' = 2x - y & y' = y. Find f(L).
Attempt: If y = y', then x' = 2x - y => 2x = x' - y => 2x = x' - y' => x =
If L = x - y + 1 = 0, then f(L) = - y' + 1 = 0 => x' - y' - 2y' + 2 = 0 => x' - 3y' + 2 = 0
Now apply x/ y values to f(L) = x' - 3y' + 2 = 0 => 2x - y - 3y + 2 = 0 => 2x - 4y + 2 = 0 => x - 2y + 1 = 0 => 2y = x + 1 => y =
Ans: (From text book): y = x + 2
I don't know what you mean by this. From the equation of the line, x- y+ 1= 0, y= x+ 1 so the set is {(x, x+1)}. Applying L to that we have x'= 2x- y= 2x- (x+ 1)= x- 1 and y'= y= x+ 1 so that (x, x+1) is mapped to (x- 1, x+ 1). If x'= x- 1 then x= x'+ 1 and x+ 1= x'+ 2, so we can also write this set as {(x', x'+ 2)}.