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

Question

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?

Answer 1

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: $-\frac1m$.

Answer 2

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.

Answer 3

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.

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Additional Helpful Points

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".)}

Answer 4

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).

Answer 5

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

Answer 6

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.

Answer 7

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.

Answer 8

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.

Answer 9

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.

Answer 10

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.

Answer 11

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 $y=x$.

Answer 12

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) $\wideparen{AB}/r=\pi/2$.

Answer 13

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, $$a + bm_r = 0 \ => \ m_r = -\dfrac{a}{b} = -\dfrac{1}{\dfrac{b}{a}} = -\dfrac{1}{m_s} \ => \ m_r\cdot m_s = -1$$