What is an algebraic explanation of why the product of the slopes of perpendicular lines is $-1$? [duplicate]

Question

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.

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

Answer 1

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.

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?)

Answer 2

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.

Answer 3

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, $$ \boxed{1 = -m_1m_2}. $$

Answer 4

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.)