The slope of the given line y=-(2/3)x+2
is m1=-2/3
The slope of any line perpendicular to it is m2=-1/m1=3/2
Since it passes through point (6,3) and the slope m2=3/2 is known, we use the point-slope form to define the new line:
y-y1=m(x-x1) where (x1,y1)=(6,3).
Substituting x1=6, y1=3 into the point slope form,
y-y1=m2(x-x1)
y-3=(3/2)(x-6)
simplify
y=(3/2)x-9+3, =>
y=(3/2)x-6
is the line perpendicular to y=-(2/3)x+2 passing through (6,3)
Check: y=(3/2)6-6=3 ok, and
m1m2=-2/3*(3/2)=-1 ok