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Suppose $P_1=(x_1,y_1)$, $P_2=(x_2,y_2)$ are two points.

Also suppose that we have a rectangle which we just know the value of its sides $a$ and $b$.

I am looking for some kind of formulation which can show whether $P_1$ and $P_2$ are inside of rectangle or not.

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    What exactly do you mean by formulation? Are you looking for an algorithm?2012-03-21
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    I am looking for a mathematical relation between a rectangle properties and points positions.2012-03-21
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    Suppose $a$ is $1$, $b$ is $100$, $P_1 = (0, 0)$, and $P_2 = (10, 10)$. What are you asking in this case? Are you asking whether or not there exists a $1$ by $100$ rectangle containing both $P_1$ and $P_2$?2012-03-21
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    @sepideh: Many people interested in game programming think of a rectangle as having sides parallel to the edges of the screen. Is this what you mean by rectangle, or can a rectangle have arbitrary orientation?2012-03-21
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    @TannerL.Swett: yes, exactly.2012-03-21
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    @AndréNicolas: my work is related to the network not game programming. however, I suppose that I have a big rectangle area which is divided equally to the small rectangles.(grid shape)2012-03-21
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    @AndréNicolas: now I need to decide if two points belong to the same rectangle.2012-03-21
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    If your rectangle has base $a$ and height $b$, but you are allowed to move it freely sideways or up/down, the condition is $|x_1-x_2|\le a$ **and** $|y_1-y_2|\le b$.2012-03-21
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    @AndréNicolas: Thanks. However, I don't think that is practical for my real problem.2012-03-21
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    @sepideh: If we don't know where the rectangle is (you only gave us the side lengths) how can we know whether the points are inside it?2012-03-21
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    If the two points are in the same rectangle (wherever that rectangle may be) then $$\lfloor\frac{x_1}{a}\rfloor = \lfloor\frac{x_2}{a}\rfloor \text{ and } \lfloor\frac{y_1}{b}\rfloor = \lfloor\frac{y_2}{b}\rfloor. $$2012-03-21

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Without knowing the position of the rectangle, it is impossible to tell whether any particular point lies inside of it. The most one can say (as the commenters pointed out), is that the conditions $$|x_1-x_2|\le a \quad \text{and } \ |y_1-y_2|\le b$$ are necessary for the two points to belong to the same (axes-aligned) rectangle $a\times b$. It is also sufficient if you are allowed to move the rectangle.