# Product of slopes is -1 iff perpendicular proof from first principles

Once again I'm working through Stillwell's Four Pillars of Geometry. I'm on Chapter 3 where he first introduces coordinates. The question reads,

3.5.1 Show that lines of slopes $t_1$ and $t_2$ are perpendicular just in case $t_1t_2=-1$.

I read that as, "Line 1 and Line 2 Perpendicular $\Leftrightarrow$ $t_1t_2=-1$". From what I've tried I can say that using contrapositives isn't very useful since in algebra having something not equal something else doesn't tell you much.

I also tried assuming you could move out from the intersection by 1 on both lines. Then draw two right triangles and go from there. (So, both hypotenuses being 1 and sides $a$ and $b$.) I couldn't finish this idea - there are a couple cases, and it involves "moving" the intersection to the origin which, although allowable, isn't quite allowed yet in Four Pillars

Is there an elegant way to show 3.5.1?

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The reformulation has a small potential problem when one of the slopes does not exist. I do not dare post a solution since I do not know the Stillwell ground rules. –  André Nicolas Jun 30 '11 at 17:07
To do this from "first principles", one would need to know the definition of "perpendicular". –  GEdgar Jun 30 '11 at 19:06
It seems there ought to be a geometric way of doing this - the 'dot product' is invariant with change of axes. But defining the "slopes" identifies axes, and seems to land every proof with a special case (lines parallel to axes). Is there a way of avoiding this, or an axis free way of posing the question? –  Mark Bennet Jun 30 '11 at 20:28

Here's an elementary, trig-free proof.

Suppose the slopes $t_1,t_2$ of the lines $L_1,L_2$, respectively, are both defined (real numbers) and the lines intersect.

Let $p$ be their intersection. Then $q=p+(1,t_1)\in L_1$ and $r=p+(1,t_2)\in L_2$.

Now $L_1$ is perpendicular to $L_2$ if and only if the triangle $pqr$ has a right angle at $p$. By Pythagoras' theorem, this is equivalent to $\|p-q\|^2+\|p-r\|^2=\|q-r\|^2\iff 1+t_1^2+1+t_2^2=(t_1-t_2)^2\iff t_1t_2=-1$.

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Nice proof; using Pythagoras's theorem as the criterion for perpendicularity is a clever idea. –  ShreevatsaR Jun 30 '11 at 20:19
As far as I can see this is the most succinct proof without assuming more than it should. –  ttt Jun 30 '11 at 22:16

I'm not sure what geometric properties you're allowed to use as yet, but here's an attempt at a purely-geometric proof (trig-free).

Let's suppose, for convenience, that the point of intersection of the two lines is not on the $x$-axis and that neither line is horizontal or vertical. Call the point of intersection of each line with the $x$-axis $A$ and $B$ and the intersection point of the two lines $C$. Call the intersection of the vertical line through $C$ with the $x$-axis $D$. Looking at $\triangle ADC$, $\frac{DC}{AD}$ is the absolute value of the slope of the line that contains $A$ and $C$; similarly, $\frac{DC}{BD}$ is the absolute value of the slope of the line that contains $B$ and $C$.

If the lines are perpendicular, than $\angle ACB$ is a right angle, so $\triangle ABC$ is a right triangle, and $CD$ is the geometric mean of $AD$ and $BD$, so $AD\cdot BD=CD^2$, from which $\frac{DC}{AD}\cdot\frac{DC}{BD}=1$, so the product of the absolute values of the slopes is $1$. Since the slopes clearly have opposite signs, their product is $-1$.

If the product of the slopes is $-1$, then $\frac{DC}{AD}\cdot\frac{DC}{BD}=1$ or $AD\cdot BD=CD^2$. If you reflect point $C$ over the $x$-axis to $C'$, $CD=C'D$ and $AD\cdot BD=CD\cdot C'D$, so by the power of a point theorem, $A$, $B$, $C$, and $C'$ lie on a circle and since $AB$ is the perpendicular bisector of $CC'$, $AB$ is a diameter of the circle, so $\angle ACB$ is a right angle. Hence, the lines are perpendicular.

Alternately, if the product of the slopes is $-1$, then $\frac{DC}{AD}\cdot\frac{DC}{BD}=1$ or $\frac{DC}{AD}=\frac{BD}{DC}$ and since $\angle ADC$ and $\angle BDC$ are both right angles, $\triangle ADC\sim\triangle CDB$, so $\angle DAC\cong\angle DCB$ and $\angle DCA\cong\angle DBC$. Now, looking at the measures of the interior angles of $\triangle ABC$, their sum must be $180°$, but $\angle ACB$ is the sum of two angles that are congruent to $\angle ABC$ and $\angle BAC$, so the measure of $\angle ACB$ must be half of $180°$, which is $90°$, so $\angle ACB$ is a right angle. Hence, the lines are perpendicular.

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This is beautiful! I never saw this argument before. I added a primitive picture illustrating your second argument. I hope you don't mind. If you don't like it, just remove it. –  t.b. Jun 30 '11 at 20:08
@Theo: Thanks for adding the picture. I was actually thinking it seemed rather clumsy to have to use the power of a point theorem backwards to get that the angle is inscribed in a semi-circle to get the right angle... I've been pondering a similar-triangles argument to replace it with—I guess now, if that argument comes together, I'll just add it as a supplement. –  Isaac Jun 30 '11 at 20:12
You beat me to it, I just wanted to add your supplement as a comment. That's indeed even nicer. –  t.b. Jun 30 '11 at 20:27
I think @mac answer is what Stillwell is looking for but these are the proofs I wish I could give! The power of a point and the altitude being a geometric mean is actually news to me! –  ttt Jun 30 '11 at 22:22
@Tony: Yeah, mac's answer does seem the most likely fit for that early in a geometry book, provided that you have the Pythagorean Theorem in both directions ($a^2+b^2=c^2\Leftrightarrow\angle ACB\text{ is right}$). The altitude to the hypotenuse of a right triangle being the geometric mean of the two pieces is a consequence of the similar triangles that I used in the alternate proof of the other direction. The power of a point theorem is a really nice a surprisingly-general tool for intersecting chords, secants, etc., with a circle. –  Isaac Jun 30 '11 at 22:28

The slopes $t_1$ and $t_2$ are the tangents of the angles $\alpha_1$ and $\alpha_2$ the two lines make with the $x$-axis.

We have $\alpha_1 - \alpha_2 = \pi/2$

Therefore loosely $\tan{(\alpha_1 - \alpha_2)} = \tan{(\pi/2)} = \infty$

And $\displaystyle\tan{(\alpha_1 - \alpha_2)} = \frac{t_1-t_2}{1+t_1t_2}$ by the formula for the sum/difference of tangents.

So we see that $t_1t_2 = -1$

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+1 Nice, intuitive argument. –  Beni Bogosel Jun 30 '11 at 17:18
To make this less loose, use the contrapositive: if $t_1t_2 \neq -1$, then $\tan(\alpha_1 - \alpha_2) = (t_1 - t_2)/(1 + t_1t_2)$ is some well-defined finite number (the denominator is not zero), so $|\alpha_1 - \alpha_2|$ is not $\pi/2$. –  ShreevatsaR Jun 30 '11 at 17:25

A quick way of seeing this is the following. A group theoretically minded reader will realize that I get a 90 degree rotation as a composition of two reflections, with respect to two lines with a 45 degree angle between them.

Let's assume that one of the lines is `pointing in the direction' $\alpha$ (= the angle between the line and the $x$-axis). If we reflect this line with respect to the line $y=x$, then the new line is pointing in the direction $\beta=\pi/2-\alpha$, because the original line and the new line both form an angle $\pi/4-\alpha$ with the line $y=x$, but they are on the opposite sides. If the slope of the original line was $k$, then the slope of the reflected line is $k_2=1/k$, because this reflection simply swaps the roles of the coordinates $x$ and $y$, and $y=kx+b \Leftrightarrow x=\frac1k (y-b)$.

In the second step we reflect the new line with respect to the $x$-axes. The twice reflected line is pointing in the direction $-\beta=\alpha-\pi/2$, so it is perpendicular to the original line. In this reflection the sign of the slope is toggled, so the slope of this perpendicular line is $k_3=-k_2=-1/k$.

There are some special cases ($\alpha=0, \alpha=\pi/2$) not covered by this argument, but in that case one line is horizontal and the other vertical, and their respective slopes are $0$ and $\infty$, so their product doesn't really make sense.

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Suppose that none of the slopes is $\infty$. For every line there exists a unit vector direction $v(a,b)$. The slope of the line is the tangent of the angle made by $v$ to the $Ox$ axis. The slope of the line is $\frac{b}{a}$.

Now, two lines with slopes $t_1,t_2$ are perpendicular if and only if their direction vectors $v(a,b),w(c,d)$ are orthogonal, i.e. $\langle v,w\rangle=ac+bd=0$ ($\langle \cdot,\cdot \rangle$ is the usual dot product). This means that $\frac{a}{b}=-\frac{d}{c}$ which means that $\frac{a}{b} \cdot \frac{c}{d}=-1$, and this is exactly $t_1t_2=-1$.

If one of the slopes is $\infty$, then that line is vertical, and the orthogonal line to it has slope $0$. If the relation would hold always, then we would have $0 \cdot \infty=-1$, which is not true. The relation between the slopes of perpendicular lines in the form $t_1t_2=-1$ is used when no line vertical or horizontal.

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lets say you have $a_1x+b_1y=c_1, a_2x+b_2y=c_2$. then the lines are perpendicular iff $a_1a_2+b_1b_2=0$ i.e. $(a_1/b_1)(a_2/b_2)=-1$ (when defined, $b_1,b_2\neq0$).

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