I know that the slopes of two lines that are perpendicular have a value of $-1$ when multiplied because they're opposite reciprocals (e.g. $5$ and $-{1\over 5}$), but what if there's a horizontal and a vertical line ($x=3$ and $y=-2$). They're perpendicular, too, but a vertical line has no slope, so about $0\cdot\infty$, that would be undefined, right? Well, infinity is undefined, so that's why a vertical line has no slope. I also think this is why that silly problem is undefined and doesn't always equal zero. Well, what's the slope of a vertical line times the slope of a horizontal line? Tell me what you think!
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2Slope of vertical line is "undefined", so it is true that when you multiply it by zero you do not get $-1$. – GEdgar Jan 06 '15 at 14:40
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That makes me wonder if there is any real-valued function $f$ that can be defined on the real projective line such that there is some constant $c$ so that for all perpendicular $a$ and $b$ one has $f(a)f(b) = c$. I suppose not. – MJD Jan 06 '15 at 14:44
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@MJD sure there is. Any constant function will work. – Matt Samuel Jan 06 '15 at 16:14
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Of course. I originally had $-1$ instead of $c$, which rules out a constant function, and then replaced $-1$ with $c$ without specifying that $c$ should be negative. – MJD Jan 06 '15 at 16:16