why the product of the perpendicular slopes =-1

Suppose m_1 and m_2 be the slopes of two straight lines L_1, L_2.

Without loss of generality, let A be the angle between x-axis and L_1 and 0 < A < 90. Then, the angle between x-axis and L_2 will be A+90.

Note that m_1 = tan A and m_2 = tan(A+90) = tan(180-(90-A)) = -tan(90-A) = -1/tan A. Thus, m_1*m_2 = -1.

Of course, the above argument is only true for A > 0. For the case A = 0, if you are familiar with the concept of limit, you may consider it is the limiting case of the result, i.e.

m_1*m_2 = lim_{A -> 0} (tan A * tan(A+90)) = lim_{A -> 0} (-1) = -1.

you can use cosine law in a coordinates system to prove it,

a^2 = b^2 + c^2 - 2ab cosC

or using Vector product: cos C = (u x v) / ( | u | x | v | )

where for both methods, cos C = 0 => angle C = 90deg

2007-04-10 03:07:19 補充：
sorry, the first part is c^2 = a^2 + b^2 - 2ab cosC