# More on 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: