The equation of a line

Functions: Lines and linear functions

The equation of a line

Assume that $a$ , $b$ , and $c$ are constant real numbers: parameters.

Line

The solution to the equation $a\cdot x+b\cdot y+c=0$ can be drawn in the plane. They are the points $\rv{x,y}$ satisfying $a\cdot x+b\cdot y+c=0$ . If $a\ne0$ or $b\ne0$ , then these points form a straight line, or simply just a line.

If $b\ne0$ , then the equation can be written as $y=-\frac{a}{b}x -\frac{c}{b}$ . For, these are the solutions if we consider $x$ as a parameter and $y$ as unknown. This indicates that for every value of $x$ there is a point $\rv{x,y}$ with $y$ equal to $-\frac{a}{b}x-\frac{c}{b}$ .

If $a\ne0$ , the line is oblique (by oblique we mean neither horizontal neither vertical).
If $a=0$ , then the value of $y$ is constant, equal to $-\frac{c}{b}$ . In this case the line is horizontal.

In the exceptional case the equation looks like .
- If $a\ne0$ , then the line is vertical.
- If $a=0$ and

Four ways to describe a line

A straight line can be described in different ways.

The solutions $\rv{x,y}$ to an equation $a\cdot x+b\cdot y+c=0$ with unknowns $x$ and $y$ . Here $a$ , $b$ , and $c$ are real numbers such that $a$ and $b$ are not both equal to zero.
The line through two given points in the plane; if $P=\rv{p,q}$ and $Q=\rv{s,t}$ are points in the plane, then the line through $P$ and $Q$ has equation $a\cdot x+b\cdot y+c=0$ with $a=q-t$ , $b=s-p$ , and $c=t\cdot p-q\cdot s$ .
The line through a given point, the base point, and a direction, indicated by the number $-\frac{a}{b}$ , where $a$ and $b$ are as in the equation given above; this number is called the slope of the line.
The line with function representation $y = p\cdot x+q$ if $b\ne0$ and $x=r$ otherwise; here we have $p=-\frac{a}{b}$ (the slope), $q = -\frac{c}{b}$ (the intercept), which is the value of $y$ for $x=0$ and $r = -\frac{c}{a}$ in terms of the above $a$ , $b$ , and $c$ . This can be seen as a special case of the previous description, with base point $\rv{0,q}$ . In the case where $b\ne0$ , the variable $y$ is a function of $x$ , in the other case, $x$ is a constant function of $y$ .

The first description, by means of the equation $ax+by+c=0$ , can be considered to be the definition of a line.

The second description is geometrically inspired, as each pair of points $P=\rv{p,q}$ and $Q=\rv{s,t}$ determines a unique line. Substitution of $\rv{x,y}=\rv{p,q}$ in the equation $(q-t)\cdot x +(s-p)\cdot y +(t\cdot p-q\cdot s) = 0$ shows that $P$ is a solution, and similarly for $Q$ .

The third description is also natural from a geometric viewpoint as a line is uniquely determined by its direction and a point lying on it. If $P=\rv{p,q}$ is the point and $r$ is the slope, then the line is given by the equation $a\cdot x+b\cdot y+c=0$ with $a=-r$ , $b=1$ , and $c=r\cdot p-q$ . In other words, the equation of this line is given by the point-slope formula $y-q=r\cdot(x-p)\tiny.$

The fourth description can be considered as the solution to the equation $a\cdot x+b\cdot y+c=0$ with unknown $y$ .

The line segment between the two points $\rv{3,{{1}\over{3}}}$ and $\rv{9,{{7}\over{3}}}$ is drawn in the figure below.

Give the function rule for this line in the form $y=a\cdot x+b$ .

$y={{1}\over{3}} x-{{2}\over{3}}$

The line is described by the function rule $y=ax+b$ where $a$ is the slope, given as the quotient of the difference of the $y$ -values with the difference of the $x$ -values of two points on the line. Hence $a=\frac{{{7}\over{3}}-{{1}\over{3}}}{9-3}={{1}\over{3}}$ . The value of $b$ follows from $b=y-ax$ , where $\rv{x,y}$ is a random point on the line. Hence $b={{1}\over{3}}-{{1}\over{3}}\cdot3=-{{2}\over{3}}$ .

New example