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?

Questions about the question will be promptly answered in the comments!

  • $\begingroup$ 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. $\endgroup$ Jun 30 '11 at 17:07
  • $\begingroup$ To do this from "first principles", one would need to know the definition of "perpendicular". $\endgroup$
    – GEdgar
    Jun 30 '11 at 19:06
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    $\begingroup$ 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? $\endgroup$ 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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    $\begingroup$ Nice proof; using Pythagoras's theorem as the criterion for perpendicularity is a clever idea. $\endgroup$ Jun 30 '11 at 20:19
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    $\begingroup$ As far as I can see this is the most succinct proof without assuming more than it should. $\endgroup$
    – ttt
    Jun 30 '11 at 22:16
  • $\begingroup$ This is also the simplest way to derive the scalar product in n dimensions using the same trick but for n-dimensional points and without slopes...very nice! $\endgroup$
    – Patricio
    Apr 18 '18 at 14:34
  • $\begingroup$ Can someone please help me understand how $\Vert q-r\Vert^2=(t_1-t_2)^2$? $\endgroup$
    – Rnj
    Sep 18 at 20:54

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.

Illustration for product of slopes = -1 implies lines 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.

  • $\begingroup$ 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. $\endgroup$
    – t.b.
    Jun 30 '11 at 20:08
  • $\begingroup$ @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. $\endgroup$
    – Isaac
    Jun 30 '11 at 20:12
  • $\begingroup$ You beat me to it, I just wanted to add your supplement as a comment. That's indeed even nicer. $\endgroup$
    – t.b.
    Jun 30 '11 at 20:27
  • $\begingroup$ 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! $\endgroup$
    – ttt
    Jun 30 '11 at 22:22
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    $\begingroup$ @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. $\endgroup$
    – 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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    $\begingroup$ 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 $\pi/2$. $\endgroup$ 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.


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 is vertical or horizontal.


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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