#### Chapter 9 Objects in the Two-Dimensional Coordinate System

Section 9.2 Lines in the Plane

# 9.2.2 Coordinate Form of Equations of a Line

Let us first introduce the most general form of a coordinate equation of a line. Using this equation, every line in the plane can be specified as an infinite set of points with respect to a given coordinate system.
##### Info 9.2.2

A line $g$ in $ℝ{}^{2}$ is a set of points

$g=\left\{\left(x;y\right)\in ℝ{}^{2} : px+qy=c\right\} .$

Here, $p$, $q$, $c$ are real numbers that define the line. At least one of the numbers $p$ and $q$ must be non-zero. The linear equation

$px+qy=c$

is called the equation of a line or, more specifically, to distinguish it from other forms of equations of a line, as coordinate form of the equation of a line. A common abbreviation for the explicit set notation given above is to specify only the variable of the line and the equation of the line:

$g: px+qy=c .$

The example below shows a few lines and their set notations or equations.
##### Example 9.2.3

•  $g=\left\{\left(x;y\right)\in ℝ{}^{2} : x-y=0\right\}$
Here, we have $p=1$, $q=-1$, and $c=0$.
•  $h: -x-2=-3y ⇔ h: -x+3y=2$
Here, we have $p=-1$, $q=3$, and $c=2$.
•  $\alpha : 4y=1$
Here, we have $p=0$, $q=4$, and $c=1$.
•  $\beta =\left\{\left(x;y\right)\in ℝ{}^{2} : x-1=0\right\}$
Here, we have $p=1$, $q=0$, and $c=1$.

Now we want to be able to draw a line correctly in a coordinate system in $ℝ{}^{2}$. The line can be uniquely defined by an equation of a line, or by other data. To do this, we have to establish a relation to the graphs of linear affine functions. We also need to know by what kind of data a line in the plane is uniquely defined. This information will be given below.
##### Info 9.2.4

A line $g$ given by an equation of a line in coordinate form

$g: px+qy=c$

can be converted into normal form if $q\ne 0$. In this case, the equation of a line $px+qy=c$ can be solved for $y$, and the normal form of $g$ is then

$g: y=-\frac{p}{q}x+\frac{c}{q} .$

In this form, the line describes a graph of a linear affine function $f$ with the slope $-\frac{p}{q}$ and the $y$-intercept $\frac{c}{q}$:

$f:\mathrm{ }\left\{\begin{array}{ccc}\hfill ℝ& \hfill \to \hfill & ℝ\hfill \\ \hfill x& \hfill ⟼\hfill & y=f\left(x\right)=-\frac{p}{q}x+\frac{c}{q} .\hfill \end{array}$

Since the slope and the $y$-intercept can be read off from the equation of a line in normal form, lines can be drawn in the same way as the graphs of linear affine functions.
##### Example 9.2.5
The line

$g=\left\{\left(x;y\right)\in ℝ{}^{2} : -x-2y=2\right\}$

has the equation $-x-2y=2$ in coordinate form. This equation can be converted into the form $y=-\frac{1}{2}x-1$ by equivalent transformations of linear equations. Thus, the line $g$ has the normal form

$g: y=-\frac{1}{2}x-1 ,$

and it describes the graph of the linear affine function

$f:\mathrm{ }\left\{\begin{array}{ccc}\hfill ℝ& \hfill \to \hfill & ℝ\hfill \\ \hfill x& \hfill ⟼\hfill & y=f\left(x\right)=-\frac{1}{2}x-1\hfill \end{array}$

with the slope $-\frac{1}{2}$ and the $y$-intercept $-1$.
To draw $g$ one has to understand the following: the $y$-intercept $-1$ implies that the point $\left(0;-1\right)$ lies on $g$. Starting from this point, the line $g$ can be drawn correctly by constructing a slope triangle of slope $-\frac{1}{2}$ (by $x=1$ units to the right and by $y=\frac{1}{2}$ units downwards) in the correct direction:

There two special cases: they can be demonstrated by the two lines $\alpha :4y=1$ and $\beta :x-1=0$ in Example 9.2.3 (see figure below).

The line $\alpha$ is parallel to the $x$-axis. Thus, its normal form $\alpha :y=\frac{1}{4}$ describes the graph of a constant function as a special case of the linear affine functions. The line $\beta$ is parallel to the $y$-axis. Its equation of a line cannot be converted into normal form since $q=0$. This is true for all lines that are parallel to the $y$-axis.For such lines a normal form does not exist, since these lines cannot be a graph of a function (as discussed in Section 6.1.4). Lines that are parallel to the $y$-axis have neither have a $y$-intercept (since they do not intersect the $y$-axis) nor a slope. However, for the sake of consistency, they can be assigned a slope of $\infty$.
##### Exercise 9.2.6
Draw the following lines in a coordinate system. Convert the corresponding equation of a line (if required and possible) into normal form first.
• ${g}_{1}: y=-2x+3$
• ${g}_{2}: -2x+y-2=0$
• ${g}_{3}: x+3=0$

##### Exercise 9.2.7
Let a line $h$ be given by the figure below.
Specify the equation of the line $h$ in normal form.
$h: y=$
As well as by an equations, a line in the plane can also be uniquely defined by other data. From these data, the corresponding equation of a line can be derived, and the line can be drawn in a coordinate system.
##### Info 9.2.8

There are two alternative ways to uniquely define a line in the plane:
• "A line is uniquely defined by two points" If two points $P$ and $Q$ in $ℝ{}^{2}$ are given, there exists exactly one line $g$ that passes through the points $P$ and $Q$. The line passing through $P$ and $Q$ is then also denoted by $g={g}_{PQ}={g}_{QP}$ or simply by $g=PQ$.
• "A line is uniquely defined by a point and a slope" If a point $P$ in $ℝ{}^{2}$ and a slope $m$ are given, there exists exactly one line $g$ that passes through $P$ and has the slope $m$.

The two following examples illustrate how the equation of a line can be derived from the data that uniquely define this line, and how this line can be drawn.
##### Example 9.2.9
Let the points $P=\left(-1;-1\right)$ and $Q=\left(2;1\right)$ be given. The line ${g}_{PQ}=PQ$ passing through these two points can be drawn immediately. To determine the line of equation it is useful to construct a slope triangle from the two given points:
From the $x$-coordinates $-1$ and $2$ of $P$ and $Q$ we obtain the width $3$ of the slope triangle. From the corresponding $y$-coordinates $-1$ and $1$ we obtain its height $2$. Thus, the slope in the equation of the line is $m=\frac{2}{3}$. For the normal form of the equation of the line ${g}_{PQ}$ we thus obtain:

${g}_{PQ}: y=mx+b=\frac{2}{3}x+b .$

Now, only the $y$-intercept $b$ has to be determined. We know that the line ${g}_{PQ}$ passes through the two points $P$ and $Q$. Therefore, we can substitute the $x$- and $y$-coordinates of one of these points into the equation of the line, and calculate $b$. Substituting, for example, the coordinates of the point $Q=\left(2;1\right)$ results in

$1=\frac{2}{3}·2+b ⇔ b=1-\frac{4}{3}=-\frac{1}{3} .$

Using the point $P=\left(-1;-1\right)$ would result in the same equation. Thus, the required equation of the line in normal form is

${g}_{PQ}: y=\frac{2}{3}x-\frac{1}{3} .$

##### Example 9.2.10
Let the point $R=\left(2;-1\right)$ and the slope $m=\frac{1}{2}$ be given. Find the line $g$ that passes through the point $R$ and has the slope $m=\frac{1}{2}$. As in Example 9.2.9, the equation of the line $g$ in normal form can be specified immediately while the $y$-intercept is still unknown:

$g: y=mx+b=\frac{1}{2}x+b .$

Moreover, both the coordinates $x$ and $y$ of the point $R$ are given here from which the $y$-intercept can be calculated, as in Example 9.2.9:

$-1=\frac{1}{2}·2+b ⇔ b=-2 .$

Thus, the required equation of the line is

$g: y=mx+b=\frac{1}{2}x-2 .$

Using the point $R=\left(2;-1\right)$ and the slope $m=\frac{1}{2}$, the line $g$ can also be drawn immediately as illustrated by the figure below.

##### Exercise 9.2.11
Every set of data given here defines a unique line. For each one, give the equation of the line and then sketch it.
1. The points $A=\left(1;5\right)$ and $B=\left(3;1\right)$ are on the line.
${g}_{AB}: y=$
2. The points $S=\left(1.5;-0.5\right)$ and $T=\left(\frac{3}{2};2\right)$ are on the line.
${g}_{ST}: x=$
3. The line $g$ passes trough the point $\left(-4;3\right)$ with a slope of $-1$.
$g: y=$
4. The line $h$ passes trough the point $\left(42;2\right)$ with a slope of $0$.
$h: y=$