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

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.

Answer 3

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 4

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

Answer 5

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 6

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}$

Answer 7

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.

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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.

Answer 8

Since we are only concerned with slopes, let's consider two lines which pass through the origin.

We'll define the first line with the equation $\color{#b33}{y = m_0x}$ where $m_0$ is its slope and pick a point $\color{#b33}a$ on the line where $x = 1$. Given that the slope is $m_0$, the coordinates of $\color{#b33}a$ must be $\color{#b33}{(1, m_0)}$. Now, we want our second line to be perpendicular to the first, so let's rotate this whole setup by $90^\circ$ which gives us a new line $\color{#34d}{y = m_1x}$ and a new point $\color{#34d}b$. We don't yet know how the slope $m_1$ relates to $m_0$ but we can figure out the coordinates of $\color{#34d}b$.

I don't have enough reputation to post images, but here's a diagram

Since the x-coordinate of $\color{#b33}a$ is $1$, clearly the y-coordinate of $\color{#34d}b$ must be $1$ because the x-axis lands precisely on the y-axis if you were to rotate it by $90^\circ$. So we can plug $y = 1$ into our equation for the second line and then solve for $x$:

$$\color{#34d}{1 = m_1 x}$$ $$\color{#34d}{x = \frac{1}{m_1}} $$

We therefore conclude that the coordinates of $\color{#34d}b$ are $\color{#34d}{(\frac{1}{m_1}, 1)}$.

Also note that each quadrant of the cartesian grid is separated from each other by $90^\circ$. Meaning if $\color{#b33}a$ was in quadrant I, then $\color{#34d}b$ would end up in quadrant II when we rotate by $90^\circ$. If $\color{#b33}a$ was in quadrant II, $\color{#34d}b$ would end up in quadrant III, and so on...

Likewise, any line passing through the origin with a positive slope only goes through quadrants I and III because those are the quadrants where $x$ and $y$ have the same sign, and any line with a negative slope goes through quadrants II and IV because those are the quadrants where $x$ and $y$ have opposite signs. According to this line of reasoning, rotating a point by $90^\circ$ must flip the sign of either the $x$ or the $y$ coordinate, because each quadrant differs from adjacent quadrants by the sign of exactly one coordinate.

Since we didn't flip the coordinate that was equal to $1$ when we rotated by $90^\circ$, the other one must have been flipped and we therefore conclude that:

$$ m_0 = -\frac{1}{m_1} $$

Answer 9

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 10

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.