Alternative equation for a line

[noxf]

Hi there,

I've been having a problem understanding how to derive a formula. I have here a lemma that states:
A line L is a set of points (x,y) in R^2 where (x,y) satisfy the equation ax + by = c with (a,b) are not trivial and every such equation determines a line. This can be rewritten as:
xsin(r) - ycos(r) + p = 0

Maybe I'm constantly doing something wrong but I've tried to do tricks with the vector form of a line and polar coordinates but to no avail :(

Anyone got an idea?

 

[Mark44]

Hi there,

I've been having a problem understanding how to derive a formula. I have here a lemma that states:
A line L is a set of points (x,y) in R^2 where (x,y) satisfy the equation ax + by = c with (a,b) are not trivial and every such equation determines a line. This can be rewritten as:
xsin(r) - ycos(r) + p = 0

Maybe I'm constantly doing something wrong but I've tried to do tricks with the vector form of a line and polar coordinates but to no avail :(

Anyone got an idea?

What does r represent in the second equation? Pretty obviously it's an angle, but which angle?

 

[noxf]

r is the angle between the line L and x-axis. |p| is the distance from L to (0,0).

 

[Mark44]

From the equation ax + by = c, solve for y to get y = (-a/b) * x + c/b
When x = 0, y = c/b, so the y-intercept is at (0, c/b).

Let P(x, y) be a point on line L. Form a right triangle whose vertices are (0, c/b), (x, y) and (x, y - c/b). (In my drawing I am assuming that line L has a positive slope, so that the point at (x, y) is above and to the right of the y-intercept.)

From this triangle, tan(r) = (y - c/b)/x, so x tan(r) = y - c/b. From here you should be able to get to the equation in the lemma.

 

[noxf]

Thanks a lot for the swift reply :)