The how to represent an inequality in a graph question

[#neutrino]

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 ?

 

[Fightfish]

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 $y = m x + c$. So naturally, if we have an inequality of form $y \geq m x + c$, the region is "above" the line, while for $y \leq m x + c$, the region is "below" the line, since our axes are oriented such that the y-axis increases along the vertical direction.

 

[fresh_42]

You changed the sign of y which is a horizontal reflection along the x-axis so above becomes below and vice versa.

 

[Mark44]

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?