Question 801277
This question seems to be a mixture of wording from two different problems. I cannot have much faith on my guess about what the problem was supposed to be, but I hope my answer helps anyway.

{{{5 x + 2 y = 5}}} can be written in the form {{{y = mx+b}}} as {{{y=-2.5x+2.5}}}.
That can be found by solving for {{{y}}}:
{{{5x+2y=5}}} --> {{{2y=5-5x}}} --> {{{0.5(2y)=0.5(5-5x)}}} --> {{{y=2.5-2.5x}}}
When the equation of a line is in such a form, the coefficient of {{{x}}} is the slope of the line.
So the slope of {{{5 x + 2 y = 5}}} is {{{-2.5}}}.
 
A line perpendicular to {{{5 x + 2 y = 5}}} must have a slope of
{{{m=(-1)/(-2.5)}}} --> {{{m=0.4}}}
 
All lines perpendicular to {{{5 x + 2 y = 5}}} will have as an equation in the form {{{y = mx+b}}}
an equation represented by {{{y=0.4x+b}}}.
 
The one an only one of those lines perpendicular to {{{5 x + 2 y = 5}}} that passes through point (7,1) will have {{{system (x=7,y=1)}}} as a solution to
{{{y=0.4x+b}}}
So we will have
{{{1=0.4*7+b}}} --> {{{1=2.8+b}}} --> {{{b=1-2.8}}} --> {{{b=-1.8}}}
So the line perpendicular to {{{5 x + 2 y = 5}}} and passing through (7,1) has the equation {{{y=0.4x-1.8}}}.
The same equation can be written in many ways.
For example, multiplying both sides times 5 we get the equivalent equation
{{{5y=2x-9}}}.
And that can be transformed into the equivalent equations
{{{2x-5y-9=0}}} and {{{2x-5y=9}}}.