i think it's important to note that the equation:
only is a line in a space of two dimensions. in more dimensions, say n, what you get is an affine hyperplane (with n-1 dimensions).
let's look at this from a "high-school perspective".
say we have a line given by:
ax + by = c. because of the peculiarities of x as a "preferred independent variable", we have two cases to consider:
case 1) b = 0.
here, our line is: ax + c = 0. again, we have two cases to consider:
case 1a) a = 0.
in this case, c must be 0, as well, and any (x,y) will suffice (0x + 0y + 0 = 0 is true for any point in the plane). so our "line" is actually the entire plane! however, this case is actually ruled out by taking N to be non-zero (if N = 0 were allowed, then every vector in the plane is indeed orthogonal to the origin. strange, but true).
case 1b) a ≠ 0 (which must be true if b = 0).
here, x = c/a, which is a vertical line. so a tangent vector is (x,0) (for any x) and a normal vector is (0,y) for any y. it is perfectly acceptable to choose x = y = a.
case 2a) a = 0, b ≠ 0. this is similar to case 1b) and you can work out the details yourself. (0,b) is a tangent vector, and (-b,0) is a normal vector (you could use (b,0) as well...at every point on a line, there are "two" normals. usually we pick an orientation of the plane, to choose a "positive" one (the one at 90 degrees, instead of -90 degrees)).
case 2b) a ≠ 0, b ≠ 0. this is the "interesting one".
typically, we define two lines:
y = m
_{1}x + b
_{1}
y = m
_{2}x + b
_{2}
to be perpendicular if m
_{1}m
_{2} = -1.
so suppose we have our lines as:
a
_{1}x + b
_{1}y = c
_{1}
a
_{2}x + b
_{2}y = c
_{2}, what does perpendicularity mean in terms of the a's,b's and c's?
putting the two lines in slope-intercept form we have:
y = (-a
_{1}/b
_{1})x + c
_{1}/b
_{1} and
y = (-a
_{2}/b
_{2})x + c
_{2}/b
_{2}
hmm...note that the slopes don't involve "c's" at all, just a's and b's.
thus m
_{1}m
_{2} = -1 becomes:
(-a
_{1}/b
_{1})(-a
_{2}/b
_{2}) = -1 that is:
a
_{1}a
_{2} = -b
_{1}b
_{2} or:
a
_{1}a
_{2} + b
_{1}b
_{2} = 0.
one often identifies the direction of a line with its slope: so a "parallel" vector to the first line going through the origin lies on the line:
y = m
_{1}x = (-a
_{1}/b
_{1})x. we can choose any x we please, so let's choose x = -b
_{1}.
this gives us a parallel vector (x,y) = (-b
_{1},a
_{1}).
now all we need to do is verify that N = (a
_{1},b
_{1}) lies on "the perpendicular line":
y = m
_{2}x. so we plug in the value x = a
_{1} into this formula:
y = m
_{2}a
_{1} = (-a
_{2}/b
_{2})a
_{1} = (-a
_{1}/b
_{1})(-a
_{2}/b
_{2})(-b
_{1})
= (-1)(-b
_{1}) = b
_{1}.
one caveat: these vectors, T and N actually lie in the "tangent space at the point P on the line L" (which is where the "c" comes in). so technically, we should write:
N = (a,b)
_{P}, T = (-b,a)
_{P} to indicate the "tails" of the vectors (arrows) start at the point P, rather than at the origin. geometrically, this means translating the origin to the point P on the line.