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odolwa99
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Having a bit of trouble with this one. 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) and investigate if f(L) is parallel to L.
If y = y', then x' = 2x - y => 2x = x' - y => 2x = x' - y' => x = [itex]\frac{x' - y'}{2}[/itex]
If L = x - y + 1 = 0, then f(L) = [itex]\frac{x' - y'}{2}[/itex] - 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 = [itex]\frac{x - 1}{2}[/itex]
Ans: (From textbook): y = x + 2
Many thanks.
Homework Statement
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) and investigate if f(L) is parallel to L.
Homework Equations
The Attempt at a Solution
If y = y', then x' = 2x - y => 2x = x' - y => 2x = x' - y' => x = [itex]\frac{x' - y'}{2}[/itex]
If L = x - y + 1 = 0, then f(L) = [itex]\frac{x' - y'}{2}[/itex] - 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 = [itex]\frac{x - 1}{2}[/itex]
Ans: (From textbook): y = x + 2