line it represents and conversely. Consider case (ii), let us first assume that b > . Consider a point P ( α , β ) on the line ax + by = c, b > , so that a α + b β = c. Take an arbitrary point Q ( α , γ ) in the half plane II (Fig .
). Now, from Fig . , we interpret, γ > β (Why?) or b γ > b β or a α + b γ > a α + b β (Why?) or a α + b γ > c i.e., Q( α, γ ) satisfies the inequality ax + by > c . Thus, all the points lying in the half plane II above the line ax + by = c satisfies the inequality ax + by > c .
Conversely, let ( α , β ) be a point on line ax + by = c and an arbitrary point Q( α , γ ) satisfying ax + by > c so that a α + b γ > c ⇒ a α + b γ > a α + b β (Why?) ⇒ γ > β (as b > ) This means that the point ( α , γ ) lies in the half plane II. Thus, any point in the half plane II satisfies ax + by > c , and conversely any point satisfying the inequality ax + by > c lies in half plane II. In case b < , we can similarly prove that any point satisfying ax + by > c lies in the half plane I, and conversely. Hence, we deduce that all points satisfying ax + by > c lies in one of the half planes II or I according as b > or b < , and conversely.
Thus, graph of the inequality ax + by > c will be one of the half plane (called solution region ) and represented by shading in the corresponding half plane. Note The region containing all the solutions of an