Different descriptions of a line

Geometry: Lines

Different descriptions of a line

We have already seen that the equation of a line is uniquely determined by two distinct points on the line. We have also seen that the graph of a linear function is a straight line and we described two different ways of writing the equation of a line. We will recall these different descriptions and add a third equation for a line.

Comparison of a line

We can write the equation of a line in three ways:

We can write the equation of a line as \[\blue p \cdot x + \green q \cdot y+ r=0\] with #p# and #q# not both equal to #0#.
We can write the equation of a line as \[\begin{array}{c}y=a \cdot x +b\\ \text{ for a non-vertical line or }\\ x=b\\ \text{ for a vertical line}\end{array}\]
If the line intersects the axes in #\rv{\orange k,0}# and #\rv{0, \purple l}# with #\orange k \ne 0# and #\purple l\ne 0#, then the equation of the line can be written as \[\frac{x}{\orange k}+\frac{y}{\purple l}=1\]

example

The following equations determine the same line: \[\frac{x}{\orange2}+\frac{y}{\purple5}=1\] \[\blue {5} \cdot x+\green2 \cdot y-{10}=0\] \[y=-\frac{5}{2} \cdot x +5\]

geogebra picture

Trace back

From any of the three ways of writing down a line we can rewrite it to any of the two other forms by means of basic algebra.

In the example we can multiply both sides by #12# we go from the third form to the first form. By subtracting #4x# on both sides and dividing by #3# we go from the first form to the second form.

example

\[\begin{array}{rcl}\frac{x}{\orange3}+\frac{y}{\purple4}&=&1\\ \blue4x+\green3y&=&12 \\ 3y&=&12-4x \\ y&=&4-\frac{4}{3}x \end{array}\]

slope

We recall that in the linear formula #y=ax+b# the number #a# is called the slope of the line. The slope says something about the direction of the line. More precisely, a slope of #a# means that the line moves #a# steps vertically whenever we go #1# step to the right.

A vertical line of the form #x=b# does not have a (well-defined) slope.

Non-vertical lines that are not of the form #y=ax+b# can always be rewritten to the form #y=ax+b#. In this way we can determine the slope.

Rewrite the equation \[{{6}\over{7}}\cdot x-{{1}\over{2}}\cdot y={{5}\over{6}}\] as #y=a\cdot x+b# with the correct values of #a# and #b#. If this is not possible, then write #x=b# for the correct value of #b#.

#y={{12}\over{7}}\cdot x-{{5}\over{3}}#

Because the coefficient of #y# in the given equation is not equal to zero, it's posible to rewrite the equation as #y=a\cdot x+b#. We get to this form using reduction:
\[\begin{array}{rcl}
{{6}\over{7}}\cdot x-{{1}\over{2}}\cdot y&=&{{5}\over{6}}\\&&\phantom{xxx}\blue{\text{the given equation}}\\
-{{1}\over{2}}\cdot y&=&-{{6}\over{7}}\cdot x+{{5}\over{6}}\\&&\phantom{xxx}\blue{{{6}\over{7}}\cdot x\text{ subtracted}\text{ on both sides}}\\
y&=&{{12}\over{7}}\cdot x-{{5}\over{3}}\\&&\phantom{xxx}\blue{\text{left and right hand side divided by } -{{1}\over{2}} \text{, the coeffient of } y}
\end{array}\]

New example