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?

Questions about the question will be promptly answered in the comments!
 A: 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.
A: 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 $t_1t_2 = -1$
A: 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.
A: 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.
A: 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 $\|p-q\|^2+\|p-r\|^2=\|q-r\|^2\iff 1+t_1^2+1+t_2^2=(t_1-t_2)^2\iff t_1t_2=-1$.
A: 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, $b_1,b_2\neq0$).
A: Let the two slopes be $m_1$ and $m_2$
We know, $m = \tan(x)$, where $x$ is the angle made by the line with positive x-axis
Now, $$m_1 \cdot m_2= \tan(x) \cdot \tan(x+90^\circ) = \tan(x) \cdot (-\cot(x)) = -\frac{\tan(x)}{\tan(x)} =-1.$$
