- #1

odolwa99

- 85

- 0

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