# How to explain the perpendicularity of two lines to a High School student?

Today I was teaching my friend from High School about linear functions. One of the exercises we had to do was finding equations of perpendicular and parallel lines. Explaining parallel equations was quite easy, if we have the equation $y = ax + b$ it's not hard to show with a couple of examples that changing the parameter $b$ only "moves" the line up or down but doesn't change the angle, thus lines $k$ and $\ell$ are parallel iff $a_k = a_{\ell}$.

However, I couldn't find a clear way to explain why those lines are perpendicular iff $a_k \times a_{\ell}= -1$. Of course, it's obvious if we use the fact that $a = \tan (\alpha)$ with $\alpha$ being the angle at which line intersects the X axis and that $\tan (\alpha) = - \cot (\frac{\pi}{2} + \alpha)$. But this forces us to introduce trigonometry and rises oh so many questions about the origin of the equation above. Does anyone know a good, simple explanation that's easy to remember?

• Two lines are perpendicular to a High School student, e.g., when they are drawn on the horizontal plane (say, earth surface) and the student stands straight. – Vladimir Jun 7 '14 at 18:16
• @Vladimir, the two lines need to be drawn through the student as well, since any two lines are not necessarily perpendicular to the student. Of course we all learned this in grade school in the related question, "How to explain when two lines are perpendicular to a Grade School student?" – CoolHandLouis Jun 7 '14 at 20:36
• Ask an expert: YouTu.be/BKorP55Aqvg – Jörg W Mittag Jun 8 '14 at 2:08
• @CoolHandLouis In Euclidean geometry the tangent space in one point is naturally identified with the tangent space of another point. So it makes sense to define perpendicularity of lines that do not meet (skew lines). I am not sure if this is done in high schools. For example are the lines $\{(x,y,z)\mid x=0 \wedge y=0\}$ and $\{(x,y,z)\mid x=1 \wedge z=0\}$ perpendicular? – Jeppe Stig Nielsen Jun 8 '14 at 11:49
• On a slight tangent (sorry!), I hope you've also showed your student that there are other equations that can define a straight line besides just $y=ax+b$. Or at least that the equation still defines the same line even if you relabel it as, say, $q=np+k$. (One reason this is relevant is that it's somewhat easier to define perpendicular lines using a representation like, say, $ax+by=c$.) – Ilmari Karonen Jun 9 '14 at 6:46

Draw a line through the origin, $y=mx$, and mark the point $(1,m)$. Rotate the drawing by a quarter turn and observe that the point is now at $(-m,1)$.

Old slope: $m$, new slope: $-\frac1m$. • The obvious ammendum to this is to use concrete numbers like $m=2$. Use $m=1$ to show the operation of reflection across the y axis. Then, try to show by pictures the change from $m$ to $1/m$ by noting that the distance to the x-axis of the first point is the same as the distance to the y-axis. This is essentially trigonometry but pictures might help here. – noobermin Jun 9 '14 at 5:25

Consider a line with slope $a$. This means that as we move along the line, for each 1 unit we move to the right, we move up by $a$ units.

Now imagine rotating the line through 90 degrees, counterclockwise. Now motion to the right becomes motion up, and motion up becomes motion to the left. On this new line, for each 1 unit we move up, we move left by $a$ units. Computing the slope of this new line, we see it's $-1/a$.

A similar argument shows you get the same result if you rotate the line 90 degrees clockwise.

• I think I like this answer better. This way of thinking about slopes is also very helpful in graphing functions – Dunno Jun 8 '14 at 10:06
• "Lines" without further qualification do not carry an orientation, so rotating a line 90 degrees counterclockwise is the same as rotating it 90 degrees clockwise. I can understand that you regard the line as a kind of trajectory with a "direction", and that is instructive, but from high school definitions a line is just a specific set of points. – Jeppe Stig Nielsen Jun 8 '14 at 10:57
• @JeppeStigNielson Forgive me if I am being naive, but even though the $result$ of rotating a line 90 degrees counterclockwise is indistinguishable from clockwise, I think clockwise and counterclockwise still have meaning here. Consider rotating a line 45 degrees in one or the other directions. The results would be very different. – joeA Jun 8 '14 at 22:02

Question: "Explain why two lines are perpendicular iff $a_k \times a_l = -1$. Use a good, simple explanation that's easy to remember."

## A Simple Visual Explanation/Proof

The following animated gif, based on congruent triangles, is the easiest visual-explanation/proof I have ever seen: (Joe Mercer: http://ceemrr.com/Geometry1/EquationsOfLines/EquationsOfLines_print.html)

Note the following:

• The line segments $b^2$ in the above diagram always form an upright square.
• For easy memorization, perpendicular lines with slopes of 1 and -1 form an "X", and obviously $(1 \times -1) = -1$
• You should point out that horizontal and vertical lines have slopes 0 and undefined (aka "infinity") and they don't follow the "-1 rule".

That's it! However, I have two more "helpful points" to consider in the following section.

1. Develop the student's intuition about slope.

Along with the perpendicular lines, it's very important important for a student to have an overall intuition / estimation about what a line's slope is without calculating it. This will help complement and deepen the student's understanding of perpendicular lines as well. Make use of the Internet's fantastic interactive math sites. The following is a good interactive plot for a line using JSXGraph: http://www.intmath.com/plane-analytic-geometry/1b-gradient-slope-line.php

2. Further "slope intuition" can be gained via the concept of "slope at a point on a curve" (aka derivative) using the animated GIF below:

Studies show that people retain more when they can link their understanding to other things. So remember more by learning more. Have them understand the concept of the derivative and then "run through" some other continuous graphs and tell you their estimate of the slope at each point. (If you're brave, show them a rough estimated graph of the derivative. But that's a whole other question!) (GIF shows slope dynamically across a function, aka "the derivative".)

Here's a proof. Start with any two perpendicular lines. Move the intersection to the origin, and pick point $(a,b)$ on the first line, and $(c,d)$ on the second line. These two points, with the origin, make a right triangle. By the distance formula (three times) and the Pythagorean Theorem, $$(a-c)^2+(b-d)^2=(a^2+b^2)+(c^2+d^2)$$ which simplifies to $$-2ac-2bd=0$$ or $$bd=-ac$$ dividing by $ac$ we get $$\frac{b}{a}\frac{d}{c}=-1$$

Note: this fails if $ac=0$, which is exactly when the original statement fails (i.e. slopes do not have product -1).

• @jamesdlin, My answer (submitted after you commented) is a real proof. – CoolHandLouis Jun 8 '14 at 23:59

The vectors $(x, y)$ and $(-y, x)$ are perpendicular. $a_1 = x/y$ and $a_2 = -y/x$.

• Isn't this begging the question? How do you know that those vectors are perpendicular? – jamesdlin Jun 8 '14 at 0:27
• @jamesdlin, it can be shown with scalar product or just by rotating pictures, as Yves Daoust suggests. I was merely pointing out that the concept of directional vectors is more straightforward than that of coefficients. – Karolis Juodelė Jun 8 '14 at 5:28
• Most highschool students who haven't had any trigonometry aren't going to know what a scalar product is. – Tim Seguine Jun 9 '14 at 9:56

This is a variation on @Nate-Eldredge’s answer.

Line #2 is perpendicular to Line #1 if line #1 has slope $s$ and, when turn your head to the left by $90^\circ$, Line #2 looks like it has that same slope, $s$.

Now think about what happens when you turn your head left by $90^\circ$. The $y$-axis becomes the horizontal axis, and the $x$-axis becomes vertical, but with negative numbers in the up direction.

If the apparent slope of Line #2 with your head turned is $s$, then it must be the case that $-x$ (the up direction) equals $sy+b$ (because $y$ is the to-the-right direction).

So $-x=sy+b$ is an equation for Line #2. If you solve this equation for $y$ (which is a good idea, because this will give you the standard form $y=mx+b$, which reveals the real slope of Line #2), you get $y=\left(\frac{-1}{s}\right)x + \text{(a number)}$. This tells you that the slope of Line #2 is $-1$ divided by the slope of Line #1.

If Line #1 is horizontal, this doesn’t quite work, because of division by zero, but if either line is perfectly vertical, you can’t use slopes anyway, because vertical lines don’t have a numerical slope.

The better way is to introduce parametric presentation $(x, y) = (at + x_0, bt + y_0)$ for the line containing the point $(x_0, y_0)$ and being parallel to the vector $(a, b)$. And then it should be checked that two vectors are orthogonal iff its inner product vanishes.

Like most high school teachers,mMy Algebra I teacher in my high school always emphasizes memorizing the classic formula for a perpendicular slope: $$m_{\perp}=-\frac{1}{m}$$where $m$ is the original slope. For example,

• $m=3 \implies m_\perp = -\frac{1}{3}$
• $m=-\frac{2}{5} \implies m_\perp = \frac{5}{2}$

But my teacher also said the key is to also understand where the formula is coming from, before we memorize. What she does is graph the two perpendicular functions, and physically count the units of rise and the units of run, for each graph, with the entire class.

You could explain it as being the opposite reciprocal of the slope in the equation $y=mx+b$, which the slope is represented as $m$. Then once you graph the problem, you will get what you are supposedly wanting to achieve. Two lines forming $90$ degree angles.

You can also give real-life examples like picture frames, door frames, street intersections, a plus sign ($+$), standing up in relation to the surface of the earth, the $"X"$ of the bones on a pirate flag, etc.

• I think you should read the question a little more. A high school student doesn't need real life examples to understand what "perpendicular" is. – Javier Jun 7 '14 at 18:29
• However I still explained the mathematical aspect already – user122519 Jun 7 '14 at 18:30
• The question is about how to explain the fact that the product of the slopes is $-1$. – Javier Jun 7 '14 at 18:34

Let $(d):~y=ax+b$ and $(d'):~y=a'x+b'$ be two secant lines ($a,a'\not=0$ and $a\not=a'$) and let $I\left(\frac{b'-b}{a'-a};\frac{ab'-a'b}{a-a'}\right)$ be their point of intersection. Let $A(0;b)\in (d)$ and $A(0;b')\in (d')$. $$(d)\perp (d')\Longleftrightarrow AI^2+A'I^2=AA'^2\Longleftrightarrow aa'=-1$$ You obtain the last one after some simple algebra.

Here's my favorite approach (not entirely rigorous, but can be made so).

The idea: Do that 90 degree rotation as a composition of two reflections that we can manage easily.

Let's start with a line $L$ through the origin. Assume that it has slope $m$, so its equation is $y=mx$. Let us denote the angle it makes with the $x$-axis by $\theta$. Assume first that $m\neq0$

Let's reflect the line $L$ w.r.t. the line $y=x$. In this process the $x$ and $y$ coordinate switch roles (this is the weak link), so the new line, call it $L'$, has the equation $x=my$ or, equivalently, $y=\dfrac1mx$. Thus it has slope $1/m$. As the line $y=x$ makes the angle $\pi/4$ with the $x$-axis, the line $L'$ is at an angle $\theta'=\pi/4-(\theta-\pi/4)=\pi/2-\theta$ w.r.t. the $x$-axis.

Let's then reflect the line $L'$ w.r.t. the $x$-axis. Call the reflected image $L''$. Its equation is clearly $y=-\dfrac1m x$, so it's slope is $-1/m$. The angle is makes with the $x$-axis is clearly $\theta''=-\theta'=\theta-\pi/2$. Therefore $L''$ is perpendicular to $L$, and we are done.

The case $m=0$ has to be dealt with separately anyway. I'm sure you can manage.

But you need to draw a couple of images to justify the calculations of angles and the effect of that reflection w.r.t. the line $y=x$.

Give a good answer to this question may be also a way to give a rigorous definition of "perpendicular" and of what is the "measure" (in radians) of an angle (a not simple task at elementary level) without using angles measured in degrees.

I sketch the reasoning in steps:

1) Let's $O$ the intersection point of two straight lines $r$ and $s$ and take an orthogonal coordinate system with center $O$ such that the equations of the lines are $r \rightarrow y=ax$; $s \rightarrow y=bx$.

2)Take two points AB in r such that P is the middle point of AB and two points CD in the same way. Using the midpoint formula students can find that: $$A\equiv (x_A,ax_A) \qquad B\equiv (-x_A,-ax_A)$$ $$C\equiv (x_C,bx_C) \qquad D\equiv (-x_C,-bx_C)$$ 3) Define: the stright lines $r$ and $s$ are perpendicular iff $\overline {AC}=\overline{BC}$

4) Proof: $r$ is perpendicular to $s$ iff $ab=-1$. We have: $$\overline {AC}=\overline{BC} \Rightarrow \overline {AC}^2=\overline{BC}^2 \iff$$ $$\iff (x_A-x_C)^2+(ax_A-bx_C)^2=(-x_A-x_C)^2+(-ax_A-bx_C)^2 \iff$$ and, with simple calculations ,we have $$x_Ax_C(1+ab)=0$$ so, if $x_A,x_C \ne 0$, we have $ab=-1$ and we have obtained the result without the use of "angles" and giving a rigorous definition of perpendicularity.

5) At last, using a circle with center $O$ and radius $r=\overline {AB}$, and noting that equal chords subtend equal arcs, we can define the "measure" of the angle $\angle AOC$ to be (the arc) $\wideparen{AB}/r=\pi/2$.

• How to write an arc? – Emilio Novati Jan 28 '15 at 13:33

Let $r$ a line through the origin with slope m. The direction of this line is given by the vector $u = (1, m_r)$. If $s$ is the perpendicular line $r$ Without loss of generality, we can assume that $s$ also passes through the origin. If $v = (a, b)$ is its direction vector, and $r$ is perpendicular a $r$, it follows that $u\cdot v = 0 \ => \ (1,m_r)(a, b) = 0$. Thus, $$a + bm_r = 0 \ => \ m_r = -\dfrac{a}{b} = -\dfrac{1}{\dfrac{b}{a}} = -\dfrac{1}{m_s} \ => \ m_r\cdot m_s = -1$$

• -1 link only answer in a foreign language. – Tim Seguine Jun 9 '14 at 9:58