first we want to write this in standard form which is: y=mx+b (where the slope is m and y-intercept is b)
so 4y=13-3x and so y = (13-3x)/4 = 13/4 - 3/4 x or Y =(-3/4) X + (13/4)
The slope of this line is -3/4
For a line to be perpendicular to that, obviously it has to have the opposite sign. It turns out that a the slope of a perpendicular line is negative reciprocal (inverse) of the slope of the original line. So we can say:
Y' = (4/3) X +b' wher b' is the y intercept of the perpendicular line.
Since the line passes through (2,7), we have:
7 = (4/3) X + 2
So (4/3)X=5 and X = 5*(4/3) =20/3
So Y' = (4/3) X + (20/3)
Note: I have denoted the perpendicular line as Y' since I have already defined the original line as Y.