Reflections and Foot of a Point

So, we did rotations with matrices. Now what about reflections? The basic reflections—of the identity matrix, say—aren’t worth mentioning at the moment. The more puzzling reflections—those about a line that is not horizontal or vertical—are worth looking at.

The more complicated way, though, to do this we’ll save for another time. The simpler way involves something called the foot of the point. Back when we were working out the distance of a point to a line, naturally we were thinking about the perpendicular distance of that point from the line. And where that perpendicular distance to the point intersects the line is called the foot of the point.

This point is also the perpendicular bisector of \(\mathtt{\overline{rr’}}\), or the line segment connecting the point \(\mathtt{r}\) with its reflection across the line. So, if we can get the foot of the point we are reflecting, we can get the reflected point.

Determining the Foot of the Point

Let’s start with a different diagram. The line shown here can be represented by the following vector equation: \[\mathtt{p +\, α}\begin{bmatrix}\mathtt{2}\\\mathtt{1}\end{bmatrix}\] What is the ordered pair for point r’, the reflection of point r across the line?

Let’s start by finding the location of q, the foot of the point. Since we know p (it’s [0, 4]), and we know that the line is described by the vector \(\mathtt{(2α, α)}\), what we need to know is the scalar that scales us from p to q. We’ll call that scalar t.

To get at the scalar t, we can equate two cosine equations. The equation on the left shows the cosine of β that we learned when we looked at the dot product. And the equation on the right shows the cosine of β as the simple adjacent over hypotenuse ratio: \[\mathtt{\text{cos(β)} = \frac{(q\,-\,p) \cdot (r\,-\,p)}{|q\,-\,p||r\,-\,p|} \quad\quad\quad\text{cos(β)} = \frac{|t(q\,-\,p)|}{|r\,-\,p|}}\]

When we set the two right-hand expressions equal to each other and solve for t, we get the scalar t. (The difference in points, q – p, is just the vector [2, 1] and r – p is just the vector [5, –1].) \[\mathtt{t = \frac{(q\,-\,p) \cdot (r\,-\,p)}{|q\,-\,p|^2}} \,\,\longrightarrow\,\, \mathtt{t =} \frac{\begin{bmatrix}\mathtt{2}\\\mathtt{1}\end{bmatrix} \cdot \begin{bmatrix}\mathtt{\,\,\,\,5}\\\mathtt{-1}\end{bmatrix}}{5} \,\,\longrightarrow\,\,\mathtt{t = 1.8}\]

Using the equation for the line at the start of this section, we see that we can set \(\mathtt{α}\) equal to t to determine the location of point q. So, point q is at \[\begin{bmatrix}\mathtt{0}\\\mathtt{4}\end{bmatrix} \mathtt{+\,\,\, 1.8}\begin{bmatrix}\mathtt{2}\\\mathtt{1}\end{bmatrix} = \begin{bmatrix}\mathtt{3.6}\\\mathtt{5.8}\end{bmatrix}\]

The Midpoint and the Reflection

Now that we have found the location of point q, we can treat it as the midpoint of \(\mathtt{\overline{rr’}}\), or the line segment connecting the point \(\mathtt{r}\) with its reflection across the line.

This is yet another thing we haven’t covered, but the midpoint between \(\mathtt{r}\) and \(\mathtt{r’}\) is \(\mathtt{q = \frac{1}{2}(r + r’)}\). Thus, the equation for the reflection of r (r’) across the given line, when we have figured out the foot of the point q is \[\mathtt{r’ = 2q\,-\,r}\]

I have to say, this makes reflections seem like a lot of work anyway.

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Josh Fisher

Instructional designer, software development in K-12 mathematics education.