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Parallel Lines in the Coordinate Plane

Alignments to Content Standards: G-GPE.B.5


  1. Suppose $\ell$ and $m$ are parallel lines in the plane. What can you deduce about the slopes of $\ell$ and $m$? Justify your answer.

  2. Suppose $\ell$ and $m$ are distinct lines in the plane and slope($\ell$) = slope($m$). Are $\ell$ and $m$ parallel? Justify your answer.

IM Commentary

The goal of this task is to prove the slope criterion for parallel lines. There are several subtleties to address, including:

  • separating out the case of vertical lines which do not have well-defined slope,
  • establishing both implications (parallel lines have equal slope and lines of equal slope are parallel),
  • relating the algebra of the linear equations to the geometry of lines.

In the solution, an algebraic description is provided as well as an indication, touching on the third bullet above, about how this algebra can be visualized. One of the advantages to using algebraic equations to describe lines is that they allow for a nice characterization of when lines are perpendicular or parallel in terms of these equations. At the same time, it is important to have a geometric intuition of parallel lines as a line and one of its vertical or horizontal displacements.

A second geometric solution uses properties of similar triangles.  This task provides an opportunity to examine many important aspects of the high school curriculum unified by a common mathematical theme.


Solution: 1 Algebra and geometry

  1. In order for slope($\ell$) and slope($m$) to be well-defined, neither of the lines can be vertical. Assuming that $\ell$ and $m$ are not vertical, we can define $\ell$ by a linear function $y = m_1x + b_1$ and $m$ by a linear function $y = m_2x + b_2$. Since $\ell$ and $m$ are distinct and parallel they share no common points. So $y = m_1x + b_1$ and $y = m_2x + b_2$ have no common solutions. Setting them equal and manipulating we find $$ (m_1-m_2)x = b_2 - b_1. $$ This equation can be solved (giving the $x$-coordinate of a point in common to $\ell$ and $m$) unless $m_1 = m_2$, that is unless the slopes of $\ell$ and $m$ are the same. Thus for any pair of non-vertical parallel lines $\ell$ and $m$, we have slope($\ell$) = slope($m$).

    From a geometric perspective, we can explain why two lines with different slope intersect and hence why two parallel lines must have the same slope.  One way to see this is to apply a rotation of the plane so that one of our lines maps to the $x$-axis and the other line maps to a line with non-zero slope. All lines with non-zero slope (and all vertical lines) cross the $x$-axis exactly once.  We conclude that lines with different slope must meet and so if $\ell$ and $m$ are parallel then they must have the same slope or both be vertical.

  2. As in part (a) we can define $\ell$ by a linear function $y = m_1x + b_1$ and $m$ by a linear function $y = m_2x + b_2$. By hypothesis $m_1 = m_2$. This means that $b_1 \neq b_2$ since $\ell$ and $m$ are distinct. The lines are parallel because $y = m_1x + b_1$ and $y = m_2x + b_2$ have no common solution: as above, setting the two equations equal gives $(m_1 - m_2)x = b_2 - b_1$. This is equivalent to $0x = b_2 - b_1$ which is not possible because $b_2 \neq b_1$.

    From a geometric perspective, lines with the same slope can be viewed as vertical translations of one another. For example, suppose we want to look at all lines with slope 2. We have the line defined by the equation $y = 2x$ which passes through the origin. For each real number $b$ we can displace this line vertically by $b$, producing the line defined by the equation $y = 2x + b$. A line and one of its vertical displacements do not meet and so they are parallel.

Solution: 2 Similar Triangles

  1. One possibility is that $\ell$ and $m$ have no well-defined slope and are parallel vertical lines. If $\ell$ and $m$ are parallel horizontal lines then slope($\ell$) and slope($m$) are both 0. We now focus on the case where $\ell$ and $m$ are parallel lines which are neither vertical nor horizontal.

    Since $\ell$ and $m$ are neither horizontal nor vertical, they both intersect the $x$-axis and they also both intersect the $y$-axis. These intersection points are labeled in the picture below:

    We know that $m(\angle OAB) = m(\angle OCD)$ because $\ell$ and $m$ are parallel and the $y$-axis is a transverse to these parallel lines. Triangles $ABO$ and $CDO$ share right angle $O$ and we can conclude that they are similar. Since corresponding sides of similar triangles are proportional we have $$\frac{|AO|}{|BO|} = \frac{|CO|}{|DO|}.$$ This means that slope($\ell$) = slope($m$).

  2. If slope($\ell$) = slope($m$) = 0 then both $\ell$ and $m$ are horizontal and are therefore parallel. So we may assume that slope($\ell$) = slope($m$) and that this common value is not 0. This means that $\ell$ and $m$ intersect both the $x$-axis and the $y$-axis as in the picture above for part (a). The reasoning for part (a) can now be run through in reverse. Since slope($\ell$) = slope($m$) this means that $$\frac{|AO|}{|BO|} = \frac{|CO|}{|DO|}.$$ Since $\triangle AOB$ and $\triangle COD$ are both right triangles they are similar. Thus $m(\angle OAB) = m(\angle OCD)$ since they are corresponding angles in similar triangles. This means that $\ell$ and $m$ are parallel.