# How do you explain why perpendicular lines have negative reciprocated slopes?

For my purposes, I am interested mostly in a medium-sized liberal art college setting. My students have mostly seen this before, but it is not something they understand. When discussing parallel lines, I have them try to find a point of intersection, fail to find such a point, and conclude that the lines never meet, which I think is convincing.

I am trying to find ways that they can convince themselves that two lines with negative reciprocated slopes are perpendicular. Do you have a method that you like?

(also: not sure how to tag this question)

• This proof uses the converse of the Pythagorean Theorem to derive $m_1=-1/m_2$. Sep 30 '15 at 21:51
• @DavidEbert Isn't the OP asking about the converse of that statement? Anyway, I guess all steps of the proof are logical equivalences. Sep 30 '15 at 22:20
• The question in the title is different from the question in the body. Sep 30 '15 at 23:26
• Whoops, sorry for the confusion. I am interested in either/both directions Oct 1 '15 at 1:37
• @DavidEbert if you posted the entire pythagorean answer, I would accept it. Oct 1 '15 at 17:38

A proof following from the Pythagorean Theorem:

Claim: If two lines are perpendicular, then their slopes are negative reciprocals of one another.

Proof: Assume WLOG that two nonvertical lines with slopes $$m_1$$ and $$m_2$$ intersect at the origin and form a right angle. Construct the line $$x=1$$. Then we have a triangle with coordinates $$(0,0)$$, $$(1,m_1)$$, and $$(1,m_2)$$. Moreover, the triangle is a right triangle with legs $$\sqrt{1^2+m_1^2}$$ and $$\sqrt{1^2+m_2^2}$$, and hypotenuse $$(m_1-m_2)$$.

By the Pythagorean theorem:

$$\sqrt{1^2+m_1^2}^2+\sqrt{1^2+m_2^2}^2=(m_1-m_2)^2$$ $$2+m_1^2+m_2^2=m_1^2-2m_1m_2+m_2^2$$ $$2=-2m_1m_2$$ $$\frac{-1}{m_1}=m_2$$ • @NickC Just edit it! :) Nov 16 '20 at 15:10

If we have two lines $l$ and $l'$ with slopes $m>0$ and $m'= - \tfrac 1 m$ respectively, then we can always make the following diagram where the blue line is parallel to the $x$-axis and the green lines are parallel to the $y$-axis. That the lines $l$ and $l'$ are perpendicular now follows from that the two triangles are congruent and that the sum of angles in a triangle is $180^{\circ}$. The Pythagorean Theorem is not needed.

For the converse, we can use almost the same diagram. Assume the lines $l$ and $l'$ are perpendicular and $l$ has slope $m>0$. We can then make the diagram where the blue line is still parallel to the $x$-axis and the green lines are parallel to the $y$-axis. This time an Angle-Side-Angle comparison shows that the two triangles are congruent. Therefore $BD=1$, and the slopes are negative reciprocals of each other.

• I don't have a reference, just some experience with teaching Euclidean geometry (to school teachers) where such arguments come naturally. Oct 1 '15 at 22:45
• The previous comment was an answer to @BenjaminDickman Oct 1 '15 at 23:17
• I should point out that if $0 < m \leq 1$, then the diagrams will look slightly different, but the arguments are the same. Oct 1 '15 at 23:44

One approach is to start with the fact that a 90-degree rotation results from a reflection across the $x$-axis followed by a reflection across the line $y=x$. (In general, reflecting across two intersecting lines results in a rotation through an angle $2\theta$ where $\theta$ is the angle between the two lines.)

If you can get students to this place, then the rest is easy: reflecting across the $x$-axis takes $y$ to $-y$, and reflecting across the diagonal line exchanges $x$ and $y$. So $\frac{y}{x}$ is transformed into $\frac{x}{-y}$.

One slightly different perspective, but with most of the same mechanics, is to imagine any set of perpendicular lines as having been translated and/or rotated from the x- and y-axes. If you pick any point on one line and form a triangle with the nearest axis, then because they've both rotated the same angle, the other line must form the same triangle with the other axis. This triangle, though has its rise and run swapped and a left-to-right run has become a top-to-bottom drop. So the slope must be the negative reciprocal.

A similar rotational reasoning can be applied to two parallel lines. Any two parallel lines can be thought of as being rotated from two horizontal lines. Since they're rotated the same way, they'll have the same slopes. (Less compelling than the perpendicular lines, but I thought I'd put it in for completeness.)

I find the following approach straightforward geometrically — it's more a demonstration than a formal proof, but it explains the intuition well enough. It's similar to Dag Oskar Madsen's answer but isn't based on a congruence argument, which I think makes it easier for students to follow. (Besides, on some syllabuses the perpendicular gradient rule is taught before congruence!)

Simply take a right-angled triangle $ABC$ such that $B$ is to the right of $A$ and $C$ is vertically above $B$. Now rotate by $90^{\circ}$ counter-clockwise about $A$ to produce $A'B'C'$. Since $AB$ was horizontal with $B$ right of $A$, then $A'B'$ will be vertical with $B'$ above $A'$. Similarly, since $BC$ was vertical with $C$ above $B$, then $B'C'$ will be horizontal with $C'$ left of $B'$. And since $A'C'$ was produced by rotating $AC$ by $90^{\circ}$, then $AC$ and $A'C'$ will be perpendicular. Also note that $A'B'$ is the same length as $AB$, and so on. This is a lot of text for something which is obvious from the diagram. $$\text{Gradient of } AC = \frac{BC}{AB}$$

$$\text{Gradient of } A'C' = -\frac{A'B'}{B'C'} = -\frac{AB}{BC}$$

So the gradients are negative reciprocals of each other, as required.

To make this even clearer, write onto your original diagram that $AB=1$ and $BC=m$ so that the gradient of $AC$ is $m$. Then when you draw the rotated triangle, label on $A'B'$ as $1$ and $B'C'$ as $m$, and it is immediately clear that the gradient of $A'C'$ is $-\frac{1}{m}$.

If you've got a projector in your classroom you can animate the rotation of your gradient triangle. If not, why not use a large cardboard cut-out? I've drawn this on a blank background, but if you're working through a problem in the classroom you might want to superimpose them on the lines you're actually finding the gradients of (similar to Dag Oskar Madsen, but obviously my alignment is different).

I am interpreting the question as: negative reciprocal slopes $\implies$ perpendicular (rather than the converse).

(1) If they understand the dot product already, as the projection of one vector on another, then the following might convince.

Suppose you have two lines with negative reciprocal slopes. They are not parallel, so they cross. Translate the lines so the crossing point is the origin. Now take two points on the lines, $a=(x,y)$ and $b=(-y,x)$, so their slopes are $y/x$ and $-x/y$ respectively. Viewing $a$ and $b$ as vectors from the origin, we have $a \cdot b = -xy + xy = 0$.

(2) If they don't understand the dot product, they might be persuaded by the calculation $$| a-b |^2 = (x+y)^2 + (y-x)^2 = (x^2+2xy+y^2) + (y^2 -2xy + x^2) = 2 x^2 + 2 y^2 \;.$$ We could approach defining perpendicularity by the notion of minimizing the distance from a point $$P$$ to a line $$l$$.

Then we will have an optimization problem with a constraint.

If $$P=(x_0,y_0)$$ and if $$l$$ is given by $$y=mx+b$$, then we must minimize $$f(x,y)=(x-x_0)^2+(y-y_0)^2$$ subject to the constraint $$mx-y+b=0$$.

Setting this up using the formalism of Lagrange multipliers, we work with $$F(x,y,\lambda)=(x-x_0)^2+(y-y_0)^2+\lambda(mx-y+b).$$

Next, $$\begin{array}{tcl} \dfrac{\partial F}{\partial x}&=&2(x-x_0)+\lambda m=0\\ \dfrac{\partial F}{\partial y}&=&2(y-y_0)-\lambda =0\\ \end{array}$$

leads to $$\lambda=2(y-y_0)=-\dfrac{2(x-x_0)}{m}$$ and hence $$-\dfrac{1}{m}=\dfrac{y-y_0}{x-x_0}.$$ Summarizing, if point $$Q=(x,y)$$ on the line $$l$$ minimizes $$d(P,Q)$$, then the slope of $$\overline{PQ}$$, which is equal to $$\dfrac{y-y_0}{x-x_0}$$, must in turn be equal to $$\dfrac{-1}{m}.$$

Evidently, this approach would not be appropriate in a precalculus class. But as an instructor thinking about how to present this idea, it is heartening to know that there is in fact an approach that relies on neither trigonometry nor diagrams.