I believe the author means you have to evaluate the expression to find k coordinates of a point, not k points. Since you have k-dimensional space you need to know all k's coordinates to define any point in that space.
For your example this would look like:
Here is the explanation of how to come up with such graphical representation.
- First you need to evaluate the boundary. To do that you should prented your expression is an equation. This turns your example into
a-2b = 5. This can be converted to
a = 5 + 2b. Now you can draw a line.
- The line drawn is your boundary (any point on it is on) and the points which make your original expression be
in points since they belong to a domain. The points which make your expression turn into true are all the points up-left of your line.
b = -3 and
a = 6. Then
a - 2b = 6 - 2*(-3) = 6 + 6 = 12.
12 > 5 ->
true. Point in the domain.
Now lets take
b = 0 and
a = 0. Then
a - 2b = 0 - 2*(0) = 0.
0 > 5 ->
false. The point is outside the domain.