Expert: Scott A Wilson - 3/24/2011

Question
Let L be the line with slope 5/12 that contains the point A=(24,-1) and let M be the line perpendicular to line L that contains the point B=(5,6). The original corrdinate axes are erased,and Line L is made the x-axis and Line M the y-axis. In the new coordinate system, point A is on the positive x-axis and point B is on the positive y-axis. The point P with coordinates (-14,27) in the original system coordinates (a,b). Find a+b

Answer
Since the slope of the first line is m = 5/12, the slope of the perpendicular line is -1/m,
and that is -12/5.  The perpendicular line to L has the equation with a slope of -12/5 and
contains the point (5,6).  Given these, the equation is y-6 = (-12/5)(x-5).
If the equation is multiplied by 5, we get 5y - 30 = x - 5.
If 30 is added to both side, the result is 5y = x + 25.

That was for the old line, but for the corrdinates have x as -14 and y as 27.
The x and y coefficients would be the same, but the constant would be different.
That is, 5(27) = -14 + C, and that is 135 + 14 = C,  so C = 149.

The equation is then 5y = x + 149, for that has (-14,27) on it as (x,y).
Point A would be at x = -149 and point B would be at 149/5, so A+B is
-149 + 149/5 = -745/5 + 149/5 = (-745 + 149)/5 = -676/5.