Explain why perpendicular lines have negative reciprocal slopes

I am not sure how to explain this. I just know they have negative reciprocals because one one line will have a positive slope while the other negative.

• Two upvotes for 10,000 views. Interesting. Apr 23 '16 at 2:28
• @zz20s Very interesting indeed. We are now at 4 / 13600. If I had to guess, I'd think that many people get here from search engines, but don't have an account. The title is filled with keyword that are frequent in basic analytic geometry, so that may be the reason. Oct 15 '16 at 7:30
• @rubik now at 24 / 34k Nov 15 '21 at 14:00

Translate two lines so that their intersection is the origin and then take two vectors along each line, say $u=(1,k_1), v=(1,k_2)$. The two lines are perpendicular if and only if $u\perp v$, viz $$u\cdot v=1+k_1k_2=0$$ This explains why $k_1$ is the negative reciprocal of $k_2$.

• Could you tell whether we could select x coordinates as 2 instead of 1 and would that be two perpendicular lines. If it is could you telll whether the product of slopes is still -1? Sep 28 '17 at 23:55
• @justin: Yes, the argument holds too. If the x coordinate is $a$, then the vectors along each of the lines are of the form $(a,ak_1)$ and $(a,ak_2)$.
– CAF
Oct 26 '17 at 10:34
• @CAF: Yeah I think for any coordinates this would in fact work unless dot product is non-zero. Oct 26 '17 at 19:01

Draw any line (positive slope works best) other than a horizontal or a vertical. Choose any two points on the line, and let's say the rise between the two points is a and the run is b, so the slope of the line is a/b.

Now rotate your paper 90 degrees.

The same two points on the rotated line have rise b and run (-a), so the slope of the rotated line is -b/a.

Thus the product of the slopes, for the two perpendicular lines, is (a/b)*(-b/a) = -1.

• Hi, I don't quite understand this part: "the rotated line has rise b and run (-a)" - shouldn't the line has rise $-b$ and run $a$ since the rotated line should be "going downwards"? Jul 9 '20 at 1:26
• @Cheryl I believe so Feb 16 '21 at 5:59

Since translation preserves angle we consider two perpendicular straight lines with slopes $m>0$ and $n$ through the origin. Feel free to draw a picture. Consider the triangles $(0,0)$, $(1,0)$, $(1,m)$ and $(0,0)$, $(-m,0)$, $(-m,1)$. Elementary geometry reveals immediately that both triangles are congruent, hence $n=-1/m$.

• This is the best answer, because the simplest. Jan 8 '14 at 18:37
• I believe this proof may assume too much. Specifically it is not clear why the vertex $(-m,1)$ will lie on the line with slope $n$.
– john
May 14 '20 at 6:34
• "Elementary (geometry), Watson!" May 14 '20 at 10:40

Given the equation y=mx + b, we can draw a triangle ABC with the vertical leg length m and the horizontal leg length 1. Next draw triangle ADE with DA perpendicular to AC. ADE is congruent to ABC since angle DAE=angle CAB

We then have slope AC= rise/run = m/1 And slope DA = -1/m

• The triangles are congruent because of SAS, not because of equal angles. You don't know the angles are equal yet. Aug 26 '15 at 16:57
• I'll use three letters to denote angles. $DAE \neq CAB$. Rather, $DAE = ACB$. To see this clearly, $DAE + DAC + CAB = 180^\circ$. Since $DAC = 90^\circ$, then $DAE + CAB = 90^\circ$ and so $DAE$ and $CAB$ are complementary. This means that in right triangles, their opposite angles in the triangle will be complementary. Thus, you can use equal angles to do this proof without requiring the sides of both triangles be $1$ and $m$, which I don't think is a justified assumption.
– JDG
Feb 22 '20 at 15:53

Assuming experience with algebra without calculus background. So, I would suggest keeping to the idea that slope, $m$, is equal to "rise over run." Given a line with slope, lets say $\frac{a}{b}$, that means it rises $a$ in the $y$ direction for every $b$ it goes in the positive $x$ direction in the plane. (I would also encourage to always keep $b>0$ so we always go in the positive $x$ direction.) The way to make the slope the "most opposite" is to flip it and make it negative. Now, that is nowhere close to a proof, but I found one that only uses the fact that the Pythagorean theorem is true when you have a right angle and high school algebra.

Claim: If two lines in the plane, $f(x)=mx+b$ and $g(x)=nx+c$, are perpendicular, then $n=\frac{-1}{m}$.

Two lines $f(x)=mx+b$ and $g(x)=nx+c$ are not parallel, so they intersect. Assume that they do not intersect on the $y$-axis, i.e. $c \neq b$. Then the triangle formed by the graphs of these two lines and the $y$-axis is a right triangle if the Pythagorean theorem holds. WLOG, assume that $c>b$. The lengths of the side on the $y$-axis will be $c-b$. To find the other two sides, we do a little algebra. The intersection of these lines is $mx+b=nx+c$ and solving for $x$, we find that $x=\frac{c-b}{m-n}$. Thus, the point of intersection is

$$\left(\frac{c-b}{m-n},\frac{m(c-b)}{m-n}+b\right)=\left(\frac{c-b}{m-n},\frac{n(c-b)}{m-n}+c\right)$$

which we get by plugging in our $x$ for $f$ and $g$. To find the distance of each of the two legs of our triangle, we just use the distance formula, and find the the distance from

$$\left(\frac{c-b}{m-n},\frac{m(c-b)}{m-n}+b\right) \text{ to } (0,b)$$

is $\frac{(c-b)\sqrt{1+m^2}}{m-n}$. For the other side, we use the distance from

$$\left(\frac{c-b}{m-n},\frac{n(c-b)}{m-n}+c\right) \text{ to } (0,c)$$

which is $\frac{(c-b)\sqrt{1+n^2}}{m-n}$. Now, we set up the Pythagorean theorem, which we can use since the angle between the lines is right, and see that

$$(c-b)^2 = \left[\frac{(c-b)\sqrt{1+m^2}}{m-n}\right]^2 + \left[\frac{(c-b)\sqrt{1+n^2}}{m-n}\right]^2$$

Canceling the $(c-b)^2$ and multiplying by $(m-n)^2$ to both sides, we get

$$(m-n)^2 = 1+m^2 + 1+n^2$$ $$m^2 -2mn + n^2 = 2+m^2 +n^2$$ Canceling and simplifying, we find that $n=\frac{-1}{m}$. You just need to know that

$\tan(x+\pi/2)=-\cot (x)$, and yes

$\cot (x)=1/\tan x$.

Also, you may reckon that in a triangle

$\text{Sum of opposite interior angles = exterior angle}$

• Use $\LaTeX$ please.
– user93957
Nov 28 '13 at 14:09
• Sorry, I am not quite familiar with TEX format. Still Learning! Nov 29 '13 at 17:17
• Is the image being displayed? It is in BMP Nov 29 '13 at 17:21
• Yes, the image is displayed. For some basic information about writing math at this site see e.g. here, here, here and here.
– user93957
Nov 29 '13 at 17:24
• And write $tan$ as $\tan$ instead by adding a '\' to the function. For example cot: $cot$ \cot: $\cot$.
– user93957
Nov 29 '13 at 17:26

Proposition: $\langle (a,b),(x,y) \rangle =0$ iff $(a,b)$ is perpendicular to $(x,y)$, see here.

$$\langle (a,b),(x,y) \rangle =0 \\ ax+by=0$$

Q: For fixed $a,b$, what are the solutions $x,y$ such that the equation is true?

$$\begin{eqnarray*} {x}&=&{-b} & \quad &\quad &\quad &\quad &\quad &\quad &\quad & {x}&=&b \\ {y}&=&{a} & \quad &\quad &\quad &\quad &\quad &\quad &\quad & {y}&=&-a \end{eqnarray*}$$

Now the slope of $(a,b)$ is $\cfrac{b}{a}$. Now for $ax+by=0$, as you know the only possible solutions for $(x,y)$ are the ones given above, the slope is $\cfrac{y}{x}$, just make the substitutions.

This is the core of your proof. You might know that you can parametrize the lines by multiplying the vectors by an scalar and also translate parametrized lines across the space. With some tweaks, you can easily generalize it to $n-$dimensions.

This is one of the sad things happened in high school (mathematics).

Students are often asked to remember the formula m.M = –1 y heart without any explanation.

The proof, however, cannot be fully (and satisfactorily) explained because it requires the knowledge of another higher level topic called compound angle formulaes.

One of these formulas is tan (A – B) = (tan A – tan B) / (1 + tan A tan B). In case A – B = 90 degrees, tan A . tan B has to be –1. tan A happens to be m and tan B happens to be M (or vice versa). Hence, we have mM = –1.

• I don't think it requires trigonometry but you can demonstrate it as such. Oct 9 '13 at 4:13

Two 90 degree rotations will snap back to how it was originally: Suppose you have line A, line A' which is perpendicular to A, and line A'' which is perpendicular to A' (all on the Cartesian plane). A and A'' are parallel (since A'' is A rotated ±90±90 degrees, and even the ± is a silly distinction as rotating 90 degrees clockwise is the same as rotating 90 degrees counterclockwise), which means that if perpendicular slopes have a hard and fast formula, they had better be reciprocals or negative reciprocals (or the same slope, but that's garbage), as otherwise p(p(m)), the perpendicular of the perpendicular of the slope, will not equal the original slope. By inspection, in a pair of perpendicular lines, one has a positive slope and one has a negative slope (I'm not account for the student who asks "What about zero?"). Then the only valid option is negative reciprocals.

In my opinion, this is the best way to show a middle school student that will allows them to teach it to other middle school students (but geometry/trigonometry should eventually show it formally). Rotating a piece of paper 90 degrees and describing how it transforms rise and run is my second favorite.