Chapter 9 Objects in the Two-Dimensional Coordinate System
Section 9.2 Lines in the Plane9.2.2 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 $\mathbb{R}{}^{2}$ is a set of points
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
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:
The example below shows a few lines and their set notations or equations.
Example 9.2.3
- $g=\{(x;y)\in \mathbb{R}{}^{2}\hspace{0.5em}:\hspace{0.5em}x-y=0\}$
Here, we have $p=1$, $q=-1$, and $c=0$.
- $h:\hspace{0.5em}-x-2=-3y\hspace{0.5em}\hspace{0.5em}\iff \hspace{0.5em}\hspace{0.5em}h:\hspace{0.5em}-x+3y=2$
Here, we have $p=-1$, $q=3$, and $c=2$.
- $\alpha :\hspace{0.5em}4y=1$
Here, we have $p=0$, $q=4$, and $c=1$.
- $\beta =\{(x;y)\in \mathbb{R}{}^{2}\hspace{0.5em}:\hspace{0.5em}x-1=0\}$
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 $\mathbb{R}{}^{2}$. 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
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
In this form, the line describes a graph of a linear affine function $f$ with the slope $-\frac{p}{q}$ and the $y$-intercept $\frac{c}{q}$:
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
$g=\{(x;y)\in \mathbb{R}{}^{2}\hspace{0.5em}:\hspace{0.5em}-x-2y=2\}$
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
$g:\hspace{0.5em}y=-\frac{1}{2}x-1\hspace{0.5em},$
and it describes the graph of the linear affine function
$f:\mathrm{\hspace{0.5em}\hspace{0.5em}}\{\begin{array}{ccc}\hfill \mathbb{R}& \hfill \to \hfill & \mathbb{R}\hfill \\ \hfill x& \hfill \u27fc\hfill & y=f(x)=-\frac{1}{2}x-1\hfill \end{array}$
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 $(0;-1)$ 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:
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
and it describes the graph of the linear affine function
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 $(0;-1)$ 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.
- ${g}_{1}:\hspace{0.5em}y=-2x+3$
- ${g}_{2}:\hspace{0.5em}-2x+y-2=0$
- ${g}_{3}:\hspace{0.5em}x+3=0$
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 $\mathbb{R}{}^{2}$ 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 $\mathbb{R}{}^{2}$ 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 $P=(-1;-1)$ and $Q=(2;1)$ 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:
${g}_{PQ}:\hspace{0.5em}y=mx+b=\frac{2}{3}x+b\hspace{0.5em}.$
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 $Q=(2;1)$ results in
$1=\frac{2}{3}\xb72+b\hspace{0.5em}\hspace{0.5em}\iff \hspace{0.5em}\hspace{0.5em}b=1-\frac{4}{3}=-\frac{1}{3}\hspace{0.5em}.$
Using the point $P=(-1;-1)$ would result in the same equation. Thus, the required equation of the line in normal form is
${g}_{PQ}:\hspace{0.5em}y=\frac{2}{3}x-\frac{1}{3}\hspace{0.5em}.$
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:
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 $Q=(2;1)$ results in
Using the point $P=(-1;-1)$ would result in the same equation. Thus, the required equation of the line in normal form is
Example 9.2.10
Let the point $R=(2;-1)$ 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:
$g:\hspace{0.5em}y=mx+b=\frac{1}{2}x+b\hspace{0.5em}.$
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:
$-1=\frac{1}{2}\xb72+b\hspace{0.5em}\hspace{0.5em}\iff \hspace{0.5em}\hspace{0.5em}b=-2\hspace{0.5em}.$
Thus, the required equation of the line is
$g:\hspace{0.5em}y=mx+b=\frac{1}{2}x-2\hspace{0.5em}.$
Using the point $R=(2;-1)$ and the slope $m=\frac{1}{2}$, the line $g$ can also be drawn immediately as illustrated by the figure below.
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:
Thus, the required equation of the line is
Using the point $R=(2;-1)$ 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.
- The points $A=(1;5)$ and $B=(3;1)$ are on the line.
${g}_{AB}:\hspace{0.5em}y=$
- The points $S=(1.5;-0.5)$ and $T=(\frac{3}{2};2)$ are on the line.
${g}_{ST}:\hspace{0.5em}x=$
- The line $g$ passes trough the point $(-4;3)$ with a slope of $-1$.
$g:\hspace{0.5em}y=$
- The line $h$ passes trough the point $(42;2)$ with a slope of $0$.
$h:\hspace{0.5em}y=$