anIf we have an equation of the form
[tex]y=mx+b[/tex]then the equation of the perpendicular will be
[tex]y=-\frac{1}{m}x+c[/tex]where c is any arbitrary constant determined by the point the line must pass through,
Now in our case, we have
[tex]y=-10x+20[/tex]Therefore, the equation of the perpendicular line will be
[tex]y=\frac{1}{10}x+c[/tex]We must choose c such that the above line passes through (-2,2).
Putting in y = 2 and x = 2 from (-2, 2) gives
[tex]2=\frac{1}{10}(-2)+c[/tex][tex]2=-\frac{1}{5}+c[/tex][tex]\therefore c=\frac{11}{5}.[/tex]Hence, the equation of a line perpendicular y = -10x + 20 and passing through (-2, 2) is
[tex]y=\frac{1}{10}+\frac{11}{5}[/tex]Now we find the equation of a line parallel to y = -10x+20 and passing through (-2,2).
Now, the slope of the parallel line is the same as that of the original equation; therefore the equation for the parallel line we have is
[tex]y=-10x+k[/tex]We find k by using the point (-2,2) and substituting x = -2 and y = 2 into the above equation to get
[tex]2=-10(-2)+k[/tex][tex]2=20+k[/tex][tex]\therefore k=-18.[/tex]Hence, the equation of the parallel line is
[tex]y=-10x-18[/tex]