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Q: What is a succinct, clear and purely algebraic explanation of why the product of the slopes of perpendicular lines is $-1$?

Here I am aiming for high-school students (in the U.S.). I have a purely geometric explanation (below), but I would like to supplement it with a purely algebraic explanation.


Rotating a slope triangle by $90^\circ$.

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    $\begingroup$ In a purely algebraic setting, how do you define orthogonality? For my money, two elements of an inner product space are orthogonal iff their inner product is zero. For vectors in $\mathbb{R}^2$, this reduces to $$\langle x_1, y_1 \rangle \cdot \langle x_2, y_2 \rangle = x_1x_2 + y_1y_2 = 0 \implies \frac{y_1}{x_1} = - \frac{x_2}{y_2}, $$ where $y_1/x_1$ is the slope of a line parameterized by the first vector, and $x_2/y_2$ is the reciprocal of the slope of the line parameterized by the second vector. $\endgroup$ – Xander Henderson Nov 14 '20 at 1:41
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    $\begingroup$ But this doesn't seem like a very high school focused explanation, as inner product spaces generally aren't dealt with until much later. Even vectors are typically discussed later on. $\endgroup$ – Xander Henderson Nov 14 '20 at 1:42
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    $\begingroup$ I agree with Xander: without a definition of orthogonality of lines, it is hard to tell what you might be looking for. I think your picture might be about as good as it gets! $\endgroup$ – Steven Gubkin Nov 14 '20 at 2:09
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    $\begingroup$ The fact that the product of the slopes of perpendicular (non-vertical/horizontal) lines is $-1$ is usually what I give as the definition of perpendicular lines. In particular, if $y=mx+b$ then a line of the form $y=-\frac{1}{m}x+c$ is perpendicular to the given line. To justify my definition, I guess we would need a different geometric or algebraic definition of perpendicular. I guess I might go with the lines intersect at a right angle. Maybe that is the definition to anchor this discussion ? $\endgroup$ – James S. Cook Nov 14 '20 at 2:46
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    $\begingroup$ Voting to close as unclear. It's not clear what would even qualify as a purely algebraic definition of orthogonality at the high school level. For high school students, orthogonality is a purely geometric notion. $\endgroup$ – Ben Crowell Nov 14 '20 at 15:44
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Here's the algebra-based proof I've used in a college algebra class. Perpendicular lines are defined as meeting at a right angle. Assume that we know the Pythagorean and distance formulas.

A possible lemma is that slope of a line indicates how much $y$ increases for a 1-unit increase in $x$ on that line. Given that $m = \Delta y / \Delta x$, when $\Delta x = 1$, we have $m = \Delta y / \Delta x = \Delta y / 1 = \Delta y$.

Assume we have two perpendicular lines of defined slope $m _1$ and $m_2$. Call the point of intersection $(x, y)$. Step to the right by $\Delta x = 1$ unit. On one line you'll be at the point $(x+1, y + m_1)$, while on the other you'll be at $(x + 1, y + m_2)$. The three points form a right triangle and we can use the aforementioned formulas.

enter image description here

Call the lengths of the sides of the triangle $a, b, c$. By the distance formula, these lengths are:

$a = \sqrt{ 1^2 + m_1 ^2}$

$b = \sqrt{ 1^2 + m_2 ^2}$

$c = \sqrt{(m_1 - m_2)^2}$

Then by the Pythagorean formula (and the binomial square formula) we get:

$a^2 + b^2 = c^2$

$\implies 1^2 + m_1^2 + 1^2 + m_2^2 = (m_1 - m_2)^2$

$\implies m_1^2 + m_2^2 + 2 = m_1^2 - 2 m_1 m_2 + m_2^2$

$\implies 2 = -2 m_1 m_2$

$\implies -1 = m_1 m_2$

I'm pretty fond of this proof, because it provides an added opportunity to get more practice and experience with the Pythagorean, distance, and binomial-square formulas, which are key topics in this course. (Perhaps the kernel of the idea here is equivalent to James' proof, but more brief?)

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  • $\begingroup$ This doesn't seem any less geometrical than the OP's beautifully simple picture-proof. It's got just as much geometry, plus a large amount of mysterious algebraic manipulation. $\endgroup$ – Ben Crowell Nov 14 '20 at 15:58
  • $\begingroup$ @BenCrowell: The OP's image depends on a rotation and re-interpretation of coordinates, which are not used here. The OP asked for a more algebra-based demonstration, and this seems to satisfy that request with algebra in place of the rotation interpretation. $\endgroup$ – Daniel R. Collins Nov 14 '20 at 17:26
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    $\begingroup$ Yep, this is essentially the same as my answer. Main difference, you fix $(x,y)$ as the point of intersection, whereas I've labeled it $(x_o,y_o)$. Further, my $x_2$ would be your $x+1$. In other terms, $x_2-x_o$ for me is $\triangle x = 1$ for you. Naturally I think your answer is great :) $\endgroup$ – James S. Cook Nov 15 '20 at 1:33
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    $\begingroup$ @JamesS.Cook Another big difference is that, when calculating $c$, this answer recognises that the value of $y$ (or $y_o$ in your notation) cancels out. No $y$-intercepts in sight! $\endgroup$ – Brian Drake Nov 15 '20 at 7:20
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Have these students had trigonometry? If so, they may have seen the formulas for rotating a point around the unit circle. $$ x' = x\cos(\theta) - y\sin(\theta) $$ $$ y' = x\sin(\theta) + y\cos(\theta) $$ If we think of the slope $m=\frac{y}{x}$ relative to an origin of $\langle0,0\rangle$, then $x$ and $y$ can be thought of as just a point that can be rotated around the origin. Now, a perpendicular line can be thought of as a rotation $\theta=90$ or $\theta=-90$ degrees. In either case, $\cos(\theta)$ will be $0$. So now, our above formulas become: $$ x' = -y\sin(\theta) $$ $$ y' = x\sin(\theta) $$ Calculate the slope of our new line: $$ m' = \frac{y'}{x'} $$ $$ m' = \frac{x\sin(\theta)}{-y\sin(\theta)} $$ $$ m' = \frac{x}{-y} $$ Now multiply your slopes: $$ mm' $$ $$ \frac{y}{x}\frac{x}{-y} $$ $$ -1 $$ Note, I'm not a mathematician, so this might not fly as a proof. But this type of trigonometry is fundamental in computer graphics applications like video games, so it has a really cool application that might help to engage students.

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    $\begingroup$ This is a really nice answer. The idea of introducing students to rotations in highschool is great. Rotated coordinates ought to be part of the main curriculum. $\endgroup$ – James S. Cook Nov 15 '20 at 1:26
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    $\begingroup$ @JamesS.Cook: This is a circular answer, because you cannot get the coordinate transformation without using even more involved reasoning than the one shown in the question here..... $\endgroup$ – user21820 Nov 15 '20 at 3:38
  • $\begingroup$ This is a good answer, provided we can assume those formulas for rotations. To make a great answer, we should deal with lines that don’t pass through the origin or are parallel to the axes (both minor problems in this context). But we are left with the question of where those rotation formulas come from to begin with… $\endgroup$ – Brian Drake Nov 15 '20 at 7:23
  • $\begingroup$ … I agree with user21820 that this looks kind of circular. Also, if you are looking for a succint and clear explanation for high-school students, I am not sure this fits. But perhaps the high-school curriculum is a bit different in the US. $\endgroup$ – Brian Drake Nov 15 '20 at 7:25
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Since you used " $90^\circ$ ", let me suggest something motivated by trigonometry.

Define the relative-slope between two lines by $$m_{rel}=\frac{m_2-m_1}{1+m_2m_1},$$ from the trigonometric identity $$\tan(\theta_2-\theta_1)=\frac{\tan\theta_2-\tan\theta_1}{1+\tan\theta_2\tan\theta_1}.$$

If the lines are parallel [i.e. $\theta_2-\theta_1=0^\circ$], then $m_{rel}=0$ implies $m_2=m_1$.

If the lines are perpendicular [i.e. $\theta_2-\theta_1=90^\circ$], then $\displaystyle\frac{1}{m_{rel}}=0$ implies $m_2m_1=-1$.

(The analogous argument for special relativity implies that the product of the slopes equals 1 for the spacelike-axis Minkowski-orthogonal to a timelike-axis.)

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    $\begingroup$ This argument seeks to justify a very simple and intuitive geometrical fact using a very complicated and nonintuitive geometrical fact. Any explanation of the identity for high school students will rely on geometry, so this is no less geometrical than the method given in the OP's question. Realistically, this would be a disaster for high school students. They would have no clue what the mysterious identity was about. $\endgroup$ – Ben Crowell Nov 14 '20 at 15:56
  • $\begingroup$ @BenCrowell Yes, I agree. The result for the relative-slope should be possible by purely-algebraic methods (without appealing to trig-functions explicitly)... but I haven't found one that is simple enough. Xander's comment to the OP seemed like the most direct for the OP. In my opinion, one needs to DEFINE perpendicular, then find a condition in terms of the desired slopes. If one uses an angle-measure [like "90 degrees"], then trigonometry is fair game. $\endgroup$ – robphy Nov 14 '20 at 16:45
  • $\begingroup$ @BenCrowell +100, if I could. I am not even sure how to explain what a “relative slope” is without using geometry, never mind that crazy $\tan$ identity (which seems to implicate unit circles and other identities involving $\sin$ and $\cos$, which themselves require geometric proofs – or at least that’s what I learned in school). $\endgroup$ – Brian Drake Nov 15 '20 at 7:33
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Let us define two lines $L_1: y=m_1x+b_1$ and $L_2: y = m_2x+b_2$ to be perpendicular if their intersection exists and forms a right angle. Clearly $m_1 \neq m_2$. Let $P=(x_o,y_o)$ be the point of intersection. Then, $$ m_1x_o+b_1 = y_o = m_2x_o+b_2 $$ Observe $b_2-b_1 = (m_1-m_2)x_o$ this will be important later. Furthermore, select $x_2 > x_o$ and notice $Q = (x_2, m_1x_2+b_1) \in L_1$ and $R =(x_2, m_2x_2+b_2) \in L_2$. The line-segments $PQ$ and $PR$ form adjacent legs of a right triangle with hypotenuse $QR$ hence by the Pythogorean Theorem we have: $$ \| PQ \|^2+\|PR \|^2 = \| QR \|^2 $$ Notice, \begin{align} \| PQ \| &= \sqrt{(x_2-x_o)^2+(m_1x_2+b_1-y_o)^2} \\ &= \sqrt{(x_2-x_o)^2+(m_1x_2+b_1-m_1x_o-b_1)^2} \\ &= \sqrt{(x_2-x_o)^2+m_1^2(x_2-x_o)^2} \\ &= |x_2-x_o|\sqrt{1+m_1^2} \\ \end{align} Also, \begin{align} \| PR \| &= \sqrt{(x_2-x_o)^2+(m_2x_2+b_2-y_o)^2} \\ &= \sqrt{(x_2-x_o)^2+(m_2x_2+b_2-m_2x_o-b_2)^2} \\ &= \sqrt{(x_2-x_o)^2+m_2^2(x_2-x_o)^2} \\ &= |x_2-x_o|\sqrt{1+m_2^2} \\ \end{align} Finally, recall $b_2-b_1 = (m_1-m_2)x_o$ and calculate \begin{align} \| QR \| &= \sqrt{(x_2-x_2)^2+(m_2x_2+b_2-(m_1x_2+b_1))^2} \\ &= |m_2x_2+b_2-m_1x_2-b_1| \\ &= |(m_2-m_1)x_2+(m_1-m_2)x_o| \\ &= |x_2-x_o||m_2-m_1| \end{align} Hence, by Pythagorean Theorem, $$ |x_2-x_o|^2(1+m_1^2)+|x_2-x_o|^2(1+m_2^2) =|x_2-x_o|^2|m_2-m_1|^2$$ Thus, as $|x_2-x_o| \neq 0$ and $|a|^2 = a^2$ we find $$ 2+m_1^2+m_2^2 = (m_2-m_1)^2 = m_2^2-2m_1m_2+m_1^2 $$ Therefore, $$ \boxed{1 = -m_1m_2}. $$

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    $\begingroup$ I'm not sure this is clear, and I'm certain it's not succinct. But, I believe you can cull from this what you want for the audience you know... $\endgroup$ – James S. Cook Nov 14 '20 at 3:19
  • $\begingroup$ Even without comparing this to a simpler answer, there are some obvious questions. Why do you use the letter $o$ instead of the digit $0$ in the subscripts for the coordinates of $P$? Why is there an $x_2$ but no $x_1$? $\endgroup$ – Brian Drake Nov 14 '20 at 11:23
  • $\begingroup$ @BrianDrake well, I happen to like the sub o. I mean, $why_o$ ? So far as $x_1$ goes, I'm saving it for another problem. $\endgroup$ – James S. Cook Nov 15 '20 at 1:37

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