Chapter 9 Objects in the Two-Dimensional Coordinate System

Section 9.4 Regions in the plane

9.4.2 Regions bounded by Lines and Circles

The Info Box below lists the regions that can occur if the equals sign in the equation of a line is replaced by an inequality sign.

Info 9.4.1

Let a line

g

in the plane (with slope

m

and

y

-intercept

b

) be given by

g : y = m x + b

in normal form with respect to a fixed coordinate system. Substituting an inequality sign for the equals sign results in the following sets that describe regions in the plane:

$B_{1} : = {(x; y) \in ℝ^{2} : y > m x + b} =$ "region above the line excluding the points on the line itself"
$B_{2} : = {(x; y) \in ℝ^{2} : y \geq m x + b} =$ "region above the line including the points on the line itself"
$B_{3} : = {(x; y) \in ℝ^{2} : y < m x + b} =$ "region below the line excluding the points on the line itself"
$B_{4} : = {(x; y) \in ℝ^{2} : y \leq m x + b} =$ "region below the line including the points on the line itself"

For equations of a line that cannot be transformed into normal form, the line of thought on the resulting regions is analogous. The example below shows two simple cases.

Example 9.4.2

Let two lines be given by the equations

g : y = - x + 1

h : x = - 1 .

Find and sketch the following sets:

A = "region above g excluding the points on the line g itself",

B = "region to the right of line h including the points on the line h itself",

and

A \cap B

.
From the Info Box above, we have

A = {(x; y) \in ℝ^{2} : y < - x + 1}

and

B = {(x; y) \in ℝ^{2} : x \geq - 1}

since the points to the right of line

h

have

x

-coordinates that are greater than

- 1

. Thus, the intersection

A \cap B

is the set of points with coordinates that satisfy both the conditions:

A \cap B = {(x; y) \in ℝ^{2} : y < - x + 1 \land x \geq - 1} .

This is illustrated by the figures below.

Here, the intersection point

(- 1; 2)

of the two lines is not an element of

A \cap B

since points on the line

g

are generally excluded from the region

A \cap B

. In the figure above this is indicated by a small empty circle at this point.

The example above illustrates the following: regions given by coordinate inequalities derived from equations of lines are easy to specify. It becomes more difficult if intersections of such regions are considered. The following, more difficult example shows that even absolute values can be involved.

Example 9.4.3

Describe the set defined by

M = {(x; y) : | x - y | < 1}

in words and sketch it.
For absolute values (see Section 2.2) a case analysis is required as usual:

$x - y \geq 0 \Leftrightarrow x \geq y$
In this case the inequality $| x - y | < 1$ can be solved for

$| x - y | < 1 \Leftrightarrow x - y < 1 \Leftrightarrow y > x - 1 .$
Thus, the set $M$ contains all points $(x; y)$ with coordinates that satisfy the inequalities $x \geq y$ and $y > x - 1$ , i.e. those points that lie above the line $y = x - 1$ but below the angle bisector $y = x$ .
$x - y < 0 \Leftrightarrow x < y$
In this case the inequality $| x - y | < 1$ can be solved for

$| x - y | < 1 \Leftrightarrow - (x - y) < 1 \Leftrightarrow y < x + 1 .$
Thus, the set $M$ contains all points $(x; y)$ with coordinates that satisfy the inequalities $x < y$ and $y < x + 1$ , i.e. those points that lie below the line $y = x + 1$ but above the angle bisector $y = x$ .

From these two cases, we obtain the following description of the set

M

M = "all points between the lines y = x - 1 and y = x + 1 that do not lie on those lines"

The figure below shows the corresponding sketch.

Exercise 9.4.4

Describe and sketch the following sets:

$A = {(x; y) \in ℝ^{2} : y \geq 1} \cap {(x; y) \in ℝ^{2} : x \geq 1}$
$B = {(x; y) \in ℝ^{2} : | 2 x - y | \geq 1}$
$C = {(x; y) \in ℝ^{2} : | y | > x + 1}$

All points with coordinates that satisfy the inequality $y \geq 1$ lie on or above the line $y = 1$ , and all points with coordinates that satisfy the inequality $x \geq 1$ lie on or to the right of the line $x = 1$ . The set $A$ contains the points that satisfy both conditions (intersection of two sets). Thus, we have:
Case analysis:
1. $2 x - y \geq 0 \Leftrightarrow y \leq 2 x$ :
  
  $| 2 x - y | \geq 1 \Leftrightarrow 2 x - y \geq 1 \Leftrightarrow y \leq 2 x - 1$
  This case includes all points that lie on or below the line $y = 2 x - 1$ .
2. $2 x - y < 0 \Leftrightarrow y > 2 x$ :
  
  $| 2 x - y | \geq 1 \Leftrightarrow - (2 x - y) \geq 1 \Leftrightarrow y \geq 2 x + 1$
  This case includes all points that lie on or above the line $y = 2 x + 1$ .
All in all the following regions are included:
Case analysis:
1. $y \geq 0$ :
  
  $| y | > x + 1 \Leftrightarrow y > x + 1$
  This case includes all points with a non-negative $y$ -coordinate that lie above the line $y = x + 1$ .
2. $y < 0$ :
  
  $| y | > x + 1 \Leftrightarrow - y > x + 1 \Leftrightarrow y < - x - 1$
  This case includes all points with a negative $y$ -coordinate that lie below the line $y = - x - 1$ .
All in all the following region results:

The point (-1;0) does not belong to the set C.

The Info Box below lists the regions in the plane that can be bounded by a circle.

Info 9.4.5

Let a circle

K

in the plane (with the centre

M = (x_{0}; y_{0})

and the radius

r

) be given by the equation

K : (x - x_{0})^{2} + (y - y_{0})^{2} = r^{2}

in normal form with respect to a fixed coordinate system. Then replacing the equals sign with an inequality results in the following sets that describe regions an a plane:

$B_{1} : = {(x; y) \in ℝ^{2} : (x - x_{0})^{2} + (y - y_{0})^{2} < r^{2}} =$ "region within the circle excluding the points on the circle itself"
$B_{2} : = {(x; y) \in ℝ^{2} : (x - x_{0})^{2} + (y - y_{0})^{2} \leq r^{2}} =$ "region within the circle including the points on the circle itself"
$B_{3} : = {(x; y) \in ℝ^{2} : (x - x_{0})^{2} + (y - y_{0})^{2} > r^{2}} =$ "region outside the circle excluding the points on the circle itself"
$B_{4} : = {(x; y) \in ℝ^{2} : (x - x_{0})^{2} + (y - y_{0})^{2} \geq r^{2}} =$ "region outside the circle including the points on the circle itself"

The example below shows a few special cases of regions that are bounded by circles as well as a few more complex cases that arise by combining several regions bounded by circles or lines.

Example 9.4.6

Let two circles

K_{1}

and

K_{2}

be given by the equations

K_{1} : x^{2} + y^{2} = 4

and

K_{2} : (x - 2)^{2} + y^{2} = 1,

and let the line

g

be given by the equation

g : y = - x + 1 .

The set $A_{1} : = {(x; y) \in ℝ^{2} : (x - 2)^{2} + y^{2} \leq 1}$ consists of all points within and on the circle $K_{2}$ :
The set $A_{2} : = {(x; y) \in ℝ^{2} : (x - 2)^{2} + y^{2} < 1} \cap {(x; y) \in ℝ^{2} : x^{2} + y^{2} < 4}$ consists of all points that lie both within the circle $K_{1}$ and within the circle $K_{2}$ , i.e. within the intersection of the two discs, excluding the points on the circles themselves:
The set $A_{3} : = {(x; y) \in ℝ^{2} : (x - 2)^{2} + y^{2} < 1 \land y \geq - x + 1}$ consists of all points that lie both within the circle $K_{2}$ - excluding the points on the circle - and above the line $g$ :

The intersection points between the circle and the line do not belong to the set $A_{3}$ .

Exercise 9.4.7

Sketch the given sets:

$A = {(x; y) \in ℝ^{2} : (x - 1)^{2} + (y - 1)^{2} \geq 1} \cap {(x; y) \in ℝ^{2} : y > - \frac{1}{2} x + 1}$
$B = {(x; y) \in ℝ^{2} : | x | < 1} \cup {(x; y) \in ℝ^{2} : (x + 3)^{2} + (y - 1)^{2} \leq 4}$
$C = {(x; y) \in ℝ^{2} : x^{2} + y^{2} < 4 \land x^{2} + (y + 1)^{2} \geq 1}$

The set $A$ consists of all points that lie both outside or on the circle at $(1; 1)$ with radius $1$ and above the line $y = - \frac{1}{2} x + 1$ :

The intersection points of the circle and the line do not belong to $A$ .
The set $B$ consists of the points in the plane with coordinates that satisfy the inequality $| x | < 1$ , i.e. that satisfy $- 1 < x < 1$ . Additionally, the (union!) set $B$ contains the points within and on the circle $(x + 3)^{2} + (y - 1)^{2} = 4$ :

The point $(- 1; 1)$ belongs to the set $B$ .
The set $C$ consists of all points that lie both within the circle $x^{2} + y^{2} = 4$ and outside the circle $x^{2} + (y + 1)^{2} = 1$ :

The point $(0; - 2)$ does not belong to $C$ .

Onlinebrückenkurs Mathematik

1. Elementary Arithmetic

2. Equations in one Variable

3. Inequalities in one Variable

4. System of Linear Equations

5. Geometry

6. Elementary Functions

7. Differential Calculus

8. Integral Calculus

9. Objects in the Two-Dimensional Coordinate System

10. Basic Concepts of Descriptive Vector Geometry

11. Language of Descriptive Statistics