The line [tex] y = 10x - 45 [/tex] is already written in the [tex] y = mx+q [/tex] form. This means that [tex] m [/tex] is the slope. If two lines are perpendicular, their slopes are the anti-inverse of each other, i.e. their product is -1.
So, our perpendicular line has a slope of [tex] -\frac{1}{10} [/tex]
Finally, we want a line passing through (1,1) with slope [tex] -\frac{1}{10} [/tex]:
[tex] y-1 = -\dfrac{1}{10}(x-1) \iff y = -\dfrac{x}{10} + \dfrac{1}{10}+1 = -\dfrac{x}{10} + \dfrac{11}{10} [/tex]
