The how to represent an inequality in a graph question

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  • #1
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if x+y ≥ 2 it contains all the point in the line x+y =2 and the half plane above it. but ,graph if x-y ≥ 2 then if consider a line x-y= 2 the inequality represents the line and the half plane below it . i don't understand why it represents the half line below it why not above ?
 

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  • #2
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That is because of the way we set up our axes on a conventional graph - with the y-axis rotated 90 degrees anticlockwise from the x-axis. This makes it natural for us to characterise lines via equations of the form [itex]y = m x + c[/itex]. So naturally, if we have an inequality of form [itex]y \geq m x + c[/itex], the region is "above" the line, while for [itex]y \leq m x + c[/itex], the region is "below" the line, since our axes are oriented such that the y-axis increases along the vertical direction.
 
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You changed the sign of y which is a horizontal reflection along the x-axis so above becomes below and vice versa.
 
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if x+y ≥ 2 it contains all the point in the line x+y =2 and the half plane above it. but ,graph if x-y ≥ 2 then if consider a line x-y= 2 the inequality represents the line and the half plane below it . i don't understand why it represents the half line below it why not above ?
Clearly the graph of ##x - y \ge 2## includes the line x - y = 2. To determine which half-plane makes up the rest of the graph, pick a point that isn't on the line, and see if it makes the inequality a true statement. For example, does the point (0, 0) satisfy the inequality? Does the point (2, -2) satisfy the inequality?
 
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