Given a line with slope y/x, find a simple formula for the slope of a second line that forms a 45 degree angle with this line (find slopes for both 45 degrees more and 45 degrees less.)
This can be done without trigonometry.

Find a general formula for any angle.
This requires trigonometry.

OK, I had to draw this for it to make sense - you may need to do the same... :)

For a line A with slope y/x, the line B perpendicular (forming a 90 degree angle) has slope -x/y.  You can join these two perpendicular lines with a third line C that creates an isosceles right triangle.  Line C has a slope 45 degrees "more" than line A.  The slope of line C is found by:

(y + x) / ( x - y)

The line D, perpendicular to line C, will be 45 degrees "less" than line A, and its slope is simply the negative reciprocal of line C:

(y - x) / (y + x)

I'm not 100% sure of the second part, but here's a guess.  I'm pretty sure that the slope of a line is equal to the tangent of its angle.  So, could you do something like this (where m_new is the slope of the new line, m_orig is the slope of the original line)?...

m_new = tan [ (arctan m_orig) + 45 ]    45 deg "more"

m_new = tan [ (arctan m_orig) - 45 ]    45 deg "less"

Posted by tomarken on 2006-03-21 15:59:16