Lines in the plane

Systems of linear equations and matrices: Systems of Linear Equations

Lines in the plane

One of the many ways of describing a straight line in the flat plane is a linear equation. We discuss how to use linear equations for determining an equation for the line through two points looks and for determining the points of intersection of two lines.

An equation for a line in the flat plane

Let \(a\), \(b\), and \(c\) be fixed real numbers. The solution of the equation \(a\cdot x+b\cdot y+c=0\) can be drawn in the plane. If at least one of #a#, #b# is distinct from #0#, then the solution points form a straight line, for short, a line.

If \(b\ne0\), then we can write the equation as \(y=-\frac{a}{b}x-\frac{c}{b}\). After all, these are the solutions if we view \(x\) as a parameter and \(y\) as an unknown. It indicates that, for each value of \(x\), there is a point \(\rv{x,y}\) with \(y\) equal to \(-\frac{a}{b}x-\frac{c}{b}\).
- If \(a \ne 0\), then we have an inclined line.
- If \(a=0\), then the value of \(y\) is always equal to \(-\frac{c}{b}\), and, we have a horizontal line.
In the exceptional case \(b=0\), the equation looks like \(a\cdot x+c=0\).
- If \(a \ne 0\), then we have a vertical line.
- If \(a=0\) and
  - \(c\ne 0\), then there are no solutions;
  - \(c=0\), then every value \(x\) and every value for \(y\) is a solution.

Not unique

We speak of an equation of a line because more equations are possible: for each number \(k\ne 0\) the equations \(a\cdot x+b\cdot y+c=0\) and \((k\cdot a)\cdot x+(k\cdot b)\cdot y+(k\cdot c)=0\) describe same straight line. In other words, the triplets \(\rv{a,b,c}\) and \(\rv{k\cdot a,k\cdot b,k\cdot c} \) describe the same straight line.

Parameter

In case of \(y=-\frac{a}{b}x-\frac{c}{b}\), we talk about #x# as a parameter. This corresponds to the parametric representation \(\rv{x,y}=\rv{x,-\frac{a}{b}x-\frac{c}{b}}\) discussed in an earlier chapter.

Not only the lines themselves are solutions of linear equations, also determining intersections of lines and finding a line through two points can be reduced to solving linear equations:

Operations with lines in the plane

The detection of points of intersection of two lines is equivalent to finding the solutions of the system of equations for the corresponding lines.
An equation for the line through two distinct points #\rv{p,q}# and #\rv{r,s}# can be found by finding a solution distinct from #\rv{0,0,0}#of the system of equations \[\eqs{a\cdot p+b\cdot q+c&=&0\\ a\cdot r+b\cdot s+c&=&0}\] in the unknowns #a#, #b#, #c#.

Intersection

Points of intersections of lines are points that lie on both lines. The intersection points are thus solutions of each equation for one of the two lines. Conversely, any solution of the system of two equations, consisting of one equation for each line, is a point of intersection.

Line through 2 points

The line through two distinct points #\rv{p,q}# and #\rv{r,s}# has an equation of the form \[a\cdot x+b\cdot y+c=0\] for certain numbers #a#, #b#, #c# with at least one of #a#, #b# distinct from #0#. The two equations of the theorem express the fact that #\rv{p,q}# and #\rv{r,s}# must be solutions of the equation.

Note that #\rv{a,b,c}=\rv{0,0,0}# indeed forces that at least one of #a#, #b# is distinct from #0#, because, if #a=b=0#, then the given equations imply that #c=0#.

The line through the two points #\rv{p,q}# and #\rv{r,s}# can be given in a single formula:

\[(q-s)\cdot x+(r-p)\cdot y+p\cdot s-q\cdot r= 0\]

This is easy to check by filling in the two points. Or, by solving the equations for #a#, #b#, #c# with #p#, #q#, #r# and #s# as parameters; this gives #\rv{a,b,c}=\lambda\cdot \rv{q-s,r-p,p\cdot s-q\cdot r}# for #\lambda\ne0#.

Below are examples of these two operations.

Determine the intersection points of the two lines given by the equations \(2x+3y-1=0\) and \(x+y-1=0 \).

In other words: what is the solution of the following system of equations?
\[\eqs{2x+3y-1&=&0\\ x+y-1&=&0 }\]

The lines intersect at the point #\rv{2,-1}#

By subtracting twice the first equation from the second, the equation \(y+1=0\) arises. It implies #y=-1#. Substituting this value for #y# in the first equation gives #2x-4=0# , from which it follows that #x=2#. Thus, the solution of the system is #x=2\land y= -1#. This means that #\rv{x,y} = \rv{2,-1}# is the point of intersection of the two lines.

In the figure below, the first line is drawn in red and the second is drawn in blue.

New example