Linear Equations and Half-Planes
Let us understand in more detail how linear inequations correspond to half-planes. First, consider an arbitrary linear equation, which we can always write as follows:
\[y=mx+b\;\;\;\;\;\;\;\;\;(how?)\]
Suppose that the graph of this equation is the following line:
Now, let us take a point A in the upper half-plane, and the point B which lies vertically below it on the line, as shown:
Note that since A and B are vertically aligned, they will have the same x coordinates. Let’s assume the coordinates of these two points as follows:
\[A \equiv \left({h,\;{k_1}} \right),\,\,\,B \equiv \left( {h,\;{k_2}} \right)\]
We note that the coordinates of the point B must satisfy the equation of the line, since it lies on the line:
\[{k_2} = mh + b\]
We further note that the y coordinate of A must be greater than the y coordinate of B, since A lies above B. Thus, we can say that
\[{k_1} > {k_2}\]
This further means that
\[{k_1}> mh + b\]
We therefore note that every point which lies in the upper half-plane will satisfy the linear inequation
\[y > mx + b\]
Similarly, every point lying in the lower half-plane will satisfy the linear inequation
\[y < mx + b\]
And of course, any point lying exactly on the line will satisfy the linear equation
\[y = mx + b\]
We note the following two special cases:
(1) Horizontal Line
If a line is horizontal, we have seen that it will have an equation of the form \(y= k,\) where k is a constant. In this case, the upper half-plane will correspond to the inequation \(y > k,\) while the lower half-plane will correspond to the inequation \(y < k\). This is shown in the following figure:
(2) Vertical Line
If a line is vertical, we have seen that it will have an equation of the form \(x= k\) where k is a constant. In this case, the right half-plane will correspond the inequation \(x > k\), while the left half-plane will correspond to the inequation \(x < k\). This is shown in the following figure:
Example 1: Plot the graph of the line \(x - 2y = 1\), and find out the inequations corresponding to the upper and lower half-planes.
Solution: First, we find out any two points on the line and draw the graph:
\[\left.\begin{array}{l}x = 1 & \Rightarrow \;\;\;y = 0\\x = - 1 & \Rightarrow \;\;\;y = - 1\end{array}\right\}\,\,\,\,\,\begin{array}{*{20}{c}}{A\left( {1,\;0} \right)}\\{B\left( {- 1,\; - 1} \right)}\end{array}\]
The graph is plotted below:
To determine the inequation corresponding to the upper half-plane, we write the equation of the line as follows:
\[y = \frac{1}{2}x -\frac{1}{2}\]
Now, any point lying in the upperhalf-plane will satisfy
\[\begin{align}&y> \frac{1}{2}x - \frac{1}{2}\\& \Rightarrow \;\;\;2y > x - 1\\& \Rightarrow\;\;\;x - 2y < 1\end{align}\]
On the other hand, any point lying in the lower half-plane will satisfy
\[\begin{align}&y< \frac{1}{2}x - \frac{1}{2}\\& \Rightarrow \;\;\;2y < x - 1\\& \Rightarrow\;\;\;x - 2y > 1\end{align}\]
The following diagram summarizes these facts:
Example 2: Plot the half-plane corresponding to the following linear inequation:
\[3x - 4y < 5\]
Solution: We first write this inequation as follows:
\[\begin{align}&4y> 3x - 5\\& \Rightarrow \;\;\;y > \frac{3}{4}x - \frac{5}{4} &...\;(*)\end{align}\]
Now, we convert this inequation to an equation:
\[y = \frac{3}{4}x -\frac{5}{4}\]
Next, we plot the line corresponding to this equation. For that, we find out two points on the line:
\[\left.\begin{align}&x = 0\;\; \Rightarrow \;\;y = - \frac{5}{4}\\&x = 3\;\;\Rightarrow \;\;y = 1\end{align} \right\}\begin{array}{*{20}{c}}{A\left( {0,\;- \frac{5}{4}} \right)}\\\!\!\!\!\!\!\!\!\!{B\left( {3,\;1} \right)}\end{array}\]
The plot of the line is shown below:
Finally, the starred inequation will correspond to the upper half-plane, as shown below:
