How to explain the perpendicularity of two lines to a High School student? Today I was teaching my friend from High School about linear functions. One of the exercises we had to do was finding equations of perpendicular and parallel lines. Explaining parallel equations was quite easy, if we have the equation $y = ax + b$ it's not hard to show with a couple of examples that changing the parameter $b$ only "moves" the line up or down but doesn't change the angle, thus lines $k$ and $\ell$ are parallel iff $a_k = a_{\ell}$. 

However, I couldn't find a clear way to explain why those lines are perpendicular iff $a_k \times a_{\ell}= -1$. Of course, it's obvious if we use the fact that $a = \tan (\alpha)$ with $\alpha$ being the angle at which line intersects the X axis and that $\tan (\alpha) = - \cot (\frac{\pi}{2} + \alpha)$. But this forces us to introduce trigonometry and rises oh so many questions about the origin of the equation above. Does anyone know a good, simple explanation that's easy to remember?
Draw a line through the origin, $y=mx$, and mark the point $(1,m)$. Rotate the drawing by a quarter turn and observe that the point is now at $(-m,1)$.

Old slope: $m$, new slope: $-\frac1m$.
Let $r$ a line through the origin with slope m. The direction of this line is given by the vector $u = (1, m_r)$. If $s$ is the perpendicular line $r$ Without loss of generality, we can assume that $s$ also passes through the origin. If $v = (a, b)$ is its direction vector, and $r$ is perpendicular a $r$, it follows that $u\cdot v = 0 \ =&gt; \ (1,m_r)(a, b) = 0$. Thus, 
$$a + bm_r = 0 \ =&gt; \ m_r = -\dfrac{a}{b} = -\dfrac{1}{\dfrac{b}{a}} = -\dfrac{1}{m_s} \ =&gt; \ m_r\cdot m_s = -1$$