What is the slope of a line perpendicular to 8x+9y=1

If put into slope intercept form...y=mx+b where m=slope

8x+9y=1

9y=-8x+1

y=(-8x+1)/9

So the slope is -8/9

For a line to be perpendicular to this one, it must have the negative reciprocal slope of the reference line:

Mathematically, for two lines to be perpendicular to each other:

m1*m2=-1  in this case:

-8m/9=-1

m=-9/-8

m=9/8 or if you prefer

m=1.125

For your curiosity...since slope equals dy/dx we can say that:

tanα=(dy/dx) and in this context

tanα=m  so the angle is:

α=arctan(m)

a=arctan(-8/9)≈-41.63

b=arctan(9/8)≈48.37

So the difference between b and a is 48.37--41.63=90°  Since they are at a 90° to each other, they are perpendicular...