# Prove that vector n=(a;b) is across with line ax+by=c

• Aug 1st 2005, 01:40 AM
totalnewbie
Prove that vector n=(a;b) is across with line ax+by=c
Can somebody help ? I know the formula (x-x1)/s1=(y-y1)/s2 and using this it could be possible to prove that scalar product is equal to 0, which means that two vectors are across.
• Aug 8th 2005, 12:38 PM
rgep
(x-x1)/s1=(y-y1)/s2 is the equation of a line through the point (x1,y1) with direction vector (s1,s2). That is, the general point (x,y) can be written as (x1,y1) + t(s1,s2) where t is a parameter determining(*) the distance along the line of the general point (x,y) from the starting point (x1,y1). In fact, t is the value of (x-x1)/s1, which is of course also equal to the value of (y-y1)/s2.

Let's take the line ax+by = c. Suppose that (x0,y0) is a point on this line. We have ax + by = ax0 + by0. So a(x-x0) + b(y-y0) = 0, or a(x-x0) = (-b)(y-y0). We write this in the rather odd form (x-x0)/(1/a) = (y-y0)/(-1/b) to see that this is a line with direction vector (1/a, -1/b).

So our lines have direction vectors (a,b) and (1/a,-1/b) respectively. The scalar product of these vectors is a/a-b/b = 0, so they are orthogonal (= perpendicular).

(*) t is the distance times sqrt(s1^2+s2^2)