# Show $-a_1a_2=b_1b_2$ for two perpendicular lines

So I was reading through a book I got at the library and it had a question in it that asks readers to prove $a_1a_2=-b_1b_2$ for two lines, $L_1: a_1x+b_1y+c_1 = 0$ and $L_2: a_2x+b_2y+c_2 = 0$.

I think this involves the dot product, but the dot product hasn't been introduced yet in the book, so I tried solving it algebraically by setting the two sides equal to each other. I thought about using the fact that $L_2$'s slope is the negative reciprocal of $L_1$'s, but making that substitution would remove $a_2$ and $b_2$ from the equation, which I didn't want to do.

Does anyone have any ideas? (Or alternatively, if this question has already been answered, could you please direct me to the thread about it?)

Thanks.

• If you know that they are negative reciprocals, $a_1/b_1=-b_2/a_2$. Can you think of any way to manipulate this expression to get the expression you have? (also, I am pretty sure that it should be $-a_1a_2=b_1b_2$) – TokenToucan Nov 29 '15 at 17:45
• Ah, yup. Looks like I was overthinking the question. And it should be, shouldn't it? Maybe it's just a typo in the book. Thanks! – Matt Nov 29 '15 at 17:51