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I find the following approach straightforward geometrically — it's more a demonstration than a formal proof, but it explains the intuition well enough. It's similar to Dag Oskar Madsen's answer but isn't based on a congruence argument, which I think makes it easier for students to follow. (Besides, on some syllabuses the perpendicular gradient rule is taught before congruence!)

Simply take a right-angled triangle $ABC$ such that $B$ is to the right of $A$ and $C$ is vertically above $B$. Now rotate by $90^{\circ}$ counter-clockwise about $A$ to produce $A'B'C'$. Since $AB$ was horizontal with $B$ right of $A$, then $A'B'$ will be vertical with $B'$ above $A'$. Similarly, since $BC$ was vertical with $C$ above $B$, then $B'C'$ will be horizontal with $C'$ left of $B'$. And since $A'C'$ was produced by rotating $AC$ by $90^{\circ}$, then $AC$ and $A'C'$ will be perpendicular. Also note that $A'B'$ is the same length as $AB$, and so on. This is a lot of text for something which is obvious from the diagram.

$$\text{Gradient of } AC = \frac{BC}{AB} $$

$$\text{Gradient of } A'C' = -\frac{A'B'}{B'C'} = -\frac{AB}{BC} $$

So the gradients are negative reciprocals of each other, as required.

To make this even clearer, write onto your original diagram that $AB=1$ and $BC=m$ so that the gradient of $AC$ is $m$. Then when you draw the rotated triangle, label on $A'B'$ as $1$ and $B'C'$ as $m$, and it is immediately clear that the gradient of $A'C'$ is $-\frac{1}{m}$.

If you've got a projector in your classroom you can animate the rotation of your gradient triangle. If not, why not use a large cardboard cut-out? I've drawn this on a blank background, but if you're working through a problem in the classroom you might want to superimpose them on the lines you're actually finding the gradients of (similar to Dag Oskar Madsen, but obviously my alignment is different).