1. ## Partitions

Let R2 denote the usual (x, y)-plane. We can define a partition of R2 by using the
lines with slope 2. Describe the corresponding equivalence relation ∼ by giving the
conditions on the coordinates so that (x1, y1) ∼ (x2, y2).
I have we must have (y2-y1)/(x2-x1)=2
Then y2-y1=2x2-2x1
y2-2x2=y1-2x1
So x1=2x2?

2. Originally Posted by kathrynmath
Let R2 denote the usual (x, y)-plane. We can define a partition of R2 by using the
lines with slope 2. Describe the corresponding equivalence relation ∼ by giving the
conditions on the coordinates so that (x1, y1) ∼ (x2, y2).
I have we must have (y2-y1)/(x2-x1)=2
Then y2-y1=2x2-2x1
y2-2x2=y1-2x1
So x1=2x2?
$\displaystyle \dfrac{y_2-y_1}{x_2-x_1}=2$ means $\displaystyle y_2-y_1 = 2x_2-2x_1$.
Nothing more can be done. You can rearrange it, but not combine terms.

3. Okay, your equivalence classes are the individual lines. Two points are equivalent, $\displaystyle (x_1, y_1)~(x_2, y_2)$ if and only if they lie on the same line. That is, as you say, $\displaystyle \frac{y_2- y_1}{x_2- x_1}= 2$ which is the same as saying that $\displaystyle y_2- y_1= 2x_2- 2x_1$ and the same as $\displaystyle y_2- 2x_2= y_1- 2x_1$. But I cannot see how you got "$\displaystyle x_1= 2x_2$" from that! Any one of those first three equations will do. The simplest, in my opinion, is the first: $\displaystyle (x_1, y_1)~ (x_2, y_2)$ if and only if $\displaystyle \frac{y_2- y_1}{x_2- x_1}= 2$.

4. So that's all I would have to say for describing the equivalence relating by saying they are only related if (y2-y1)/(x2-x1)=2?

5. Yes.