Explain why perpendicular lines have negative reciprocal slopes

Question

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.

Answer 1

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 $k_2$.

Answer 2

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.

Answer 3

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 $n=\frac{-1}{m}$.

Answer 4

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 $n=-1/m$.

Answer 5

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}$