I have a question about parallel movement! No spams. Because of (x, y) → (x+a, y+b) the function f(x, y)=0 becomes f(x-a, y-b)=0. During the proof, new variables x' and y' are introduced. And point (x', y') = (x+a, y+b). Then, it is solved for x and y but not for x' and y'. I'm confused about this one, why would we not use x' and y' when the new graph becomes f(x', y') = 0???
Answers
Answer:
Because of the introduction of a and b.
Step-by-step explanation:
Because of the parallel movement (x, y) → (x+a, y+b), the variables x and y are no longer related. To be specific, the variables are different in both equations.
Imagine the graph of y=x² and y=1. If y in both graphs is related, it will lead to a contradiction. The x, y in the graph is just for showing each equation.
We introduce x' and y' because we cannot draw a graph without the introduction.
x' = x + a, y' = y + b
∴x = x' - a, y = y' - b
If we put the value back in, we get a new graph. This is because, of course, the variables x, y are not related. Because a, b are the parallel movement of (x, y), we must substitute the value of x, y. a, b are only related to the vertice of (x, y).
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