What is the equation of the line that is perpendicular to the given line and passes through the point (5, 3)

For this case we have that by definition, the equation of the line in the slope-intersection form is given by:[tex]y = mx + b[/tex]Where:m: It's the slopeb: It is the cut-off point with the y axisAccording to the figure, the line goes through the following points:[tex](x_ {1}, y_ {1}) :( 8, -10)\\(x_ {2}, y_ {2}): (- 8,10)[/tex]We found the slope:[tex]m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {10 - (- 10)} {- 8-8} = \frac {10 + 10 } {- 16} = \frac {20} {- 16} = - \frac {5} {4}[/tex]By definition, if two lines are perpendicular then the product of their slopes is -1:[tex]m * - \frac {5} {4} = - 1\\m = \frac {-1} {- \frac {5} {4}}\\m = \frac {4} {5}[/tex]Thus, the equation is of the form:[tex]y = \frac {4} {5} x + b[/tex]If the line goes through [tex](x, y) :( 5,3)[/tex]we have:[tex]3 = \frac {4} {5} (5) + b\\3 = 4 + b\\3-4 = b\\-1 = b[/tex]Finally, the equation is:[tex]y = \frac {4} {5} x-1[/tex]Algebraically manipulating we have:[tex]y + 1 = \frac {4} {5} x\\5y + 5 = 4x\\4x-5y = 5[/tex]Answer:Option A