Slope Criterion for Perpendicular Lines

Task

Suppose $\ell$ and $k$ are lines with slopes $m$ and $-\frac{1}{m}$ respectively where $m$ is a non-zero real number. The goal of this task is to show that $\ell$ and $k$ are perpendicular. Below is a sample picture of $\ell$ and $k$ along with several marked points.

In this picture, $\overleftrightarrow{PQ}$ is a horizontal line and $\overleftrightarrow{OR}$ is a vertical line.

1. Suppose $O$ is chosen as the origin and that $P = (m,-1)$. Find the coordinates of $Q$.
2. Show that $\triangle QRO$ is similar to $\triangle ORP$ with a scale factor of $m$.
3. Deduce that $\ell$ is perpendicular to $k$.

IM Commentary

The goal of this task is to use similar triangles to establish the slope criterion for perpendicular lines. Students need to be familiar with scaling and with the side-side-side congruence criterion. Other arguments using rigid motions with coordinates and trigonometric functions are presented in http://www.illustrativemathematics.org/illustrations/1871. The teacher may also wish to pose the slope criterion as an open question and allow students to choose which approach to take. Because this criterion is so often used, it is important for students to see several arguments establishing the result.

The picture and similarity established in this task form the foundation for a proof of the Pythagorean Theorem as done here: http://www.illustrativemathematics.org/illustrations/1568. These two tasks could be done together to reinforce the importance of similarity for proving theorems about triangles.

The argument used in this task can also be adapted to show that if $\ell$ and $m$ are perpendicular lines (which are not horizontal and vertical) then the slopes are inverse reciprocals of one another, using the similarity of $\triangle QRO$ and $\triangle ORP$.

Solution

1. Since $\overleftrightarrow{PQ}$ is a horizontal line, the $y$-coordinate of $Q$ is the same as the $y$-coordinate of $P$, namely $-1$. For the $x$-coordinate of $Q$ we know that the slope of $\overleftrightarrow{OQ} = \ell$ is $m$. Letting $x_Q$ denote the $x$-coordinate of $Q$ this means that $$\frac{-1 - 0}{x_Q - 0} = m.$$ Solving for $x_Q$ gives $x_Q = -\frac{1}{m}$.
2. For $\triangle QRO$ side $\overline{QR}$ is horizontal and we find, using part (a), $|QR| = \frac{1}{m}$. Side $\overline{RO}$ is vertical and we have $|RO| = 1$. For $\triangle ORP$ side $\overline{OR}$ is vertical and $|OR| = 1$ and side $\overline{RP}$ is horizontal with $|RP| = m$. So there is a common scale factor of $m$ going from $\overline{QR}$ to $\overline{OR}$ and from $\overline{RO}$ to $\overline{RP}$. Because $\triangle QRO$ and $\triangle ORP$ are both right triangles, there is also a scale factor of $m$ going from $\overline{QO}$ to $\overline{OP}$ (as can be shown, for example, with the Pythagorean theorem). So there is a scale factor of $m$ from each side of $\triangle QRO$ to each corresponding side of $\triangle{ORP}$.
3. By part (b), $\triangle QRO$ is similar to $\triangle{ORP}$. This means that $m(\angle QRO) = m(\angle ORP)$. Since these angles are supplementary, they are both right angles. This means that $m(\angle RPO) + m(\angle POR) = 90$. Since $\triangle QRO$ is similar to $\triangle ORP$ we have $m(\angle RPO) = m(\angle ROQ)$. Thus $$m(\angle ROQ) + m(\angle POR) = 90$$ and this means that $\ell$ and $m$ are perpendicular.